MATH 55203–55303

Theory of Functions of a Complex Variable I–II

Fall 2025–Spring 2026

(Fall) SCEN 404, MWF 2:00–2:50 p.m.

Next class day: Monday, October 13, 2025

Contents

1.1. Elementary Properties of the Complex Numbers

[Definition: The complex numbers] The set of complex numbers is \(\R ^2\), denoted \(\C \). (In this class, you may use everything you know about \(\R \) and \(\R ^2\)—in particular, that \(\R ^2\) is an abelian group and a normed vector space.)

[Definition: Real and imaginary parts] If \((x,y)\) is a complex number, then \(\re (x,y)=x\) and \(\im (x,y)=y\).

[Definition: Addition and multiplication] If \((x,y)\) and \((\xi ,\eta )\) are two complex numbers, we define \begin {align*} (x,y)+(\xi ,\eta )&=(x+\xi ,y+\eta ),\\ (x,y)\cdot (\xi ,\eta )&=(x\xi -y\eta ,x\eta +y\xi ). \end {align*}

(Problem 10) Show that multiplication in the complex numbers is commutative.

Let \((x,y)\) and \((\xi ,\eta )\) be two complex numbers. Then \begin {align*} (x,y)\cdot (\xi ,\eta )&=(x\xi -y\eta ,x\eta +y\xi ),\\ (\xi ,\eta )\cdot (x,y)&=(\xi x-\eta y,\xi y+\eta x). \end {align*}

Because multiplication in the real numbers is commutative, we have that \begin {align*} (\xi ,\eta )\cdot (x,y)&=(x\xi -y\eta ,y\xi + x\eta ). \end {align*}

Because addition in the real numbers is commutative, we have that \begin {align*} (\xi ,\eta )\cdot (x,y)&=(x\xi -y\eta , x\eta +y\xi )=(x,y)\cdot (\xi ,\eta ) \end {align*}

as desired.

(Fact 20) This notion of addition and multiplication makes the complex numbers a ring—thus, multiplication is also associative and distributes over addition.

(Problem 30) What is the multiplicative identity?

(Problem 40) Let \(r\) be a real number. Recall that \(\C =\R ^2\) is a vector space over \(\R \), so we can multiply vectors (complex numbers) by scalars (real numbers). Is there a complex number \((\xi ,\eta )\) such that \(r(x,y)=(\xi ,\eta )\cdot (x,y)\) for all \((x,y)\in \C \)?

[Definition: Notation for the complex numbers]

• If \(r\in \R \), we identify \(r\) with the number \((r,0)\in \C \).

• We let \(i\) denote \((0,1)\).

(Problem 50) If \(x\), \(y\) are real numbers, what complex number is \(x+iy\)?

(Problem 60) If \(z=x+iy\) for \(x\), \(y\) real, what are \(\re z\) and \(\im z\)?

(Problem 70) If \(z\in \C \) and \(r\) is real, what are \(\re (zr)\) and \(\im (zr)\)?

(Problem 80) If \(z\), \(w\in \C \), what are \(\re (z+w)\), \(\im (z+w)\) in terms of \(\re z\), \(\re w\), \(\im z\), and \(\im w\)?

(Problem 90) If \(z\), \(w\in \C \), what are \(\re (zw)\), \(\im (zw)\) in terms of \(\re z\), \(\re w\), \(\im z\), and \(\im w\)?

[Definition: Conjugate] The conjugate to the complex number \(x+iy\), where \(x\), \(y\) are real, is \(\overline {x+iy}=x-iy\).¹

(Problem 100) If \(z\) and \(w\) are complex numbers, show that \(\bar z+\overline w=\overline {z+w}\).

(Problem 110) Show that \(\bar z\cdot \overline {w}=\overline {zw}\).

(Problem 120) Write \(\re z\) and \(\im z\) in terms of \(z\) and \(\bar z\).

(Problem 130) Show that \(z\bar z\) is always real and nonnegative. If \(z\bar z=0\), what can you say about \(z\)?

(Problem 140) If \(z\) is a complex number with \(z\neq 0\), show that there exists another complex number \(w\) such that \(zw=1\). Give a formula for \(w\) in terms of \(z\). We will write \(w=\frac {1}{z}\).

\(z\bar z\) is a positive real number, and we know from real analysis that positive real numbers have reciprocals. Thus \(\frac {1}{z\bar z}\in \R \). We can multiply complex numbers by real numbers, so \(\frac {1}{z\bar z}\bar z\) is a complex number and it is the \(w\) of the problem statement.

[Definition: Modulus] If \(z\) is a complex number, we define its modulus \(|z|\) as \(|z|=\sqrt {z\bar z}\).

(Fact 150) \(|\re z|\leq |z|\) and \(|\im z|\leq |z|\) (where the first \(|\,\cdot \,|\) denotes the absolute value in the real numbers and the second \(|\,\cdot \,|\) denotes the modulus in the complex numbers.)

(Problem 160) If \(z\) and \(w\) are complex numbers, show that \(|zw|=|z|\,|w|\).

(Problem 170) Give an example of a non-constant polynomial that has no roots (solutions) that are real numbers. Find a root (solution) to your polynomial that is a complex number.

1.2. Real Analysis

(Fact 180) If \(z=x+iy=(x,y)\), then the complex modulus \(|z|\) is equal to the vector space norm \(\|(x,y)\|\) in \(\R ^2\).

(Fact 190) \(\C \) is complete as a metric space if we use the expected metric \(d(z,w)=|z-w|\).

(Bashar, Problem 200) Recall that \((\R ^2,d)\) is a metric space, where \(d(u,v)=\|u-v\|\). In particular, this metric satisfies the triangle inequality. Write the triangle inequality as a statement about moduli of complex numbers. Simplify your statement as much as possible.

The conclusion is that \(|z+w|\leq |z|+|w|\) for all \(z\), \(w\in \C \). This is Proposition 1.2.3 in your textbook.

(Memory 210) If \(\{a_n\}_{n=1}^\infty \) is a sequence of points in \(\R ^p\), \(a\in \R ^p\), and we write \(a_n=(a_n^1,a_n^2,\dots ,a_n^p)\), \(a=(a^1,\dots a^p)\), then \(a_n\to a\) (in the metric space sense) if and only if \(a_n^k\to a^k\) for each \(1\leq k\leq p\).

(Dibyendu, Problem 220) What does this tell you about the complex numbers?

If \(\{z_n\}_{n=1}^\infty \) is a sequence of points in \(\C \) and \(z\in \C \), then \(z_n\to z\) if and only if both \(\re z_n\to \re z\) and \(\im z_n\to \im z\).

[Definition: Maclaurin series] If \(f:\R \to \R \) is an infinitely differentiable function, then the Maclaurin series for \(f\) is the power series \begin {equation*} \sum _{n=0}^\infty \frac {f^{(n)}(0)}{n!}x^n \end {equation*} with the convention that \(0^0=1\).

(Memory 221) If \(x\) is real, then the Maclaurin series for \(\exp x\), \(\sin x\), or \(\cos x\) converges to \(\exp x\), \(\sin x\), or \(\cos x\), respectively.

(Memory 230) The Maclaurin series for the \(\exp \) function is \(\sum _{k=0}^\infty \frac {x^k}{k!}\).

(Memory 240) The Maclaurin series for the \(\sin \) function is \(\sum _{k=0}^\infty (-1)^{k}\frac {x^{2k+1}}{(2k+1)!}\).

(Memory 250) The Maclaurin series for the \(\cos \) function is \(\sum _{k=0}^\infty (-1)^{k}\frac {x^{2k}}{(2k)!}\).

(Memory 270) If \(x\) and \(t\) are real numbers then \begin {align*} \sin (x+t)&=\sin x\cos t+\sin t\cos x,\\ \cos (x+t)&=\cos x\cos t-\sin x\sin t. \end {align*}

(Memory 280) The Cauchy-Schwarz inequality for real numbers states that if \(n\in \N \) is a positive integer, and if for each \(k\) with \(1\leq k\leq n\) the numbers \(x_k\), \(\xi _k\) are real, then \begin {equation*} \Bigl (\sum _{k=1}^n x_k\,\xi _k\Bigr )^2\leq \Bigl (\sum _{k=1}^n x_k^2 \Bigr )\Bigl (\sum _{k=1}^n \xi _k^2\Bigr ). \end {equation*}

1.2. Further Properties of the Complex Numbers

(Hope, Problem 290) State the Cauchy-Schwarz inequality for complex numbers and prove that it is valid.

This is Proposition 1.2.4 in your book. If \(n\in \N \), and if \(z_1\), \(z_2,\dots ,z_n\) and \(w_1\), \(w_2,\dots ,w_n\) are complex numbers, then \begin {equation*} \Bigl |\sum _{k=1}^n z_k\,w_k\Bigr |^2\leq \sum _{k=1}^n |z_k|^2 \sum _k |w_k|^2. \end {equation*}

We can prove this as follows. By the triangle inequality, \(|z_1w_1+z_2w_2|\leq |z_1w_1|+|z_2w_2|=|z_1||w_1|+|z_2||w_2|\). A straightforward induction argument yields that \begin {equation*} \Bigl |\sum _{k=1}^n z_k\,w_k\Bigr | \leq \sum _{k=1}^n|z_k||w_k|. \end {equation*} Applying the real Cauchy-Schwarz inequality with \(x_k=|z_k|\) and \(\xi _k=|w_k|\) completes the proof.

(James, Problem 300) Let \(z\in \C \). Consider the series \(\sum _{k=0}^\infty \frac {z^k}{k!}\), that is, the sequence of complex numbers \(\bigl \{\sum _{k=0}^n \frac {z^k}{k!}\bigr \}_{n=0}^\infty \). Show that this sequence is a Cauchy sequence.

(Problem 310) Since \(\C \) is complete, the series converges. If \(z=x\) is a real number, to what number does the series converge?

It converges to \(e^x\).

(Micah, Problem 320) If \(z=iy\) is purely imaginary (that is, if \(y\in \R \)), show that \(\sum _{k=0}^\infty \frac {(iy)^k}{k!}\) converges to \(\cos y+i\sin y\).

An induction argument establishes that \begin {align*} \re i^k &= \begin {cases} 0, &k\text { is odd},\\1,&k\text { is even and a multiple of~$4$},\\-1,&k\text { is even and not a multiple of~$4$},\end {cases} \end {align*}

and \begin {align*} \im i^k &= \begin {cases} 0, &k\text { is even},\\1,&k\text { is odd and one more than a multiple of~$4$},\\-1,&k\text { is even and one less than a multiple of~$4$}.\end {cases} \end {align*}

We then see that we may write the Maclaurin series for \(\cos \) and \(\sin \) as \begin {equation*} \cos (y)=\sum _{k=0}^\infty \re i^k \frac {y^k}{k!}, \qquad \sin (y)=\sum _{k=0}^\infty \im i^k \frac {y^k}{k!} . \end {equation*} We then have that \begin {equation*} \re \Bigl (\sum _{k=0}^n \frac {(iy)^k}{k!}\Bigr ) =\sum _{k=0}^n \re \biggl (\frac {(iy)^k}{k!}\biggr ) =\sum _{k=0}^n \re i^k\frac {y^k}{k!} \end {equation*} which converges to \(\cos y\) as \(n\to \infty \). Similarly \begin {equation*} \im \Bigl (\sum _{k=0}^n \frac {(iy)^k}{k!}\Bigr ) =\sum _{k=0}^n \im \biggl (\frac {(iy)^k}{k!}\biggr ) =\sum _{k=0}^n \im i^k\frac {y^k}{k!} \end {equation*} converges to \(\sin y\) as \(n\to \infty \). Thus the series \(\sum _{k=0}^\infty \frac {(iy)^k}{k!}\) converges to \(\cos y+i\sin y\), as desired.

(Bonus Problem 330) If \(z=x+iy\), show that \(\sum _{j=0}^\infty \frac {z^j}{j!}\) converges to the product \(\bigl (\sum _{j=0}^\infty \frac {x^j}{j!}\bigr )\bigl (\sum _{j=0}^\infty \frac {(iy)^j}{j!}\bigr )\).

[Definition: The complex exponential] If \(x\) is real, we define \begin {equation*} \exp (x)=\sum _{j=0}^\infty \frac {x^j}{j!}\quad \text { and }\quad \exp (ix)=\sum _{j=0}^\infty \frac {(ix)^j}{j!}. \end {equation*} If \(z=x+iy\) is a complex number, we define \begin {equation*} \exp (z)=\exp (x)\cdot \exp (iy). \end {equation*}

(Muhammad, Problem 340) If \(y\), \(\eta \) are real, show that \(\exp (iy+i\eta )=\exp (iy)\cdot \exp (i\eta )\).

Using the sum angle identities for sine and cosine, we compute \begin {align*} \exp (iy+i\eta ) &=\exp (i(y+\eta )) = \cos (y+\eta )+i\sin (y+\eta ) \\&= \cos y\cos \eta - \sin y\sin \eta + i\sin y\cos \eta +i \cos y\sin \eta \end {align*}

and \begin {align*} \exp (iy)\exp (i\eta )&= (\cos y+i\sin y)(\cos \eta +i\sin \eta ) \\&= \cos y\cos \eta - \sin y\sin \eta + i\sin y\cos \eta +i \cos y\sin \eta \end {align*}

and observe that they are equal.

(Robert, Problem 350) If \(z\), \(w\) are any complex numbers, show that \(\exp (z+w)=\exp (z)\cdot \exp (w)\).

There are real numbers \(x\), \(y\), \(\xi \), \(\eta \) such that \(z=x+iy\) and \(w=\xi +i\eta \).

By definition \begin {equation*} \exp (z)=\exp (x)\exp (iy),\qquad \exp (w)=\exp (\xi )\exp (i\eta ). \end {equation*} Because multiplication in the complex numbers is associative and commutative, \begin {align*} \exp (z)\exp (w) &=[\exp (x)\exp (iy)][\exp (\xi )\exp (i\eta )] =[\exp (x)\exp (\xi )][\exp (iy)\exp (i\eta )] . \end {align*}

By properties of exponentials in the real numbers and by the previous problem, we see that \begin {align*} \exp (z)\exp (w) &=\exp (x+\xi )\exp (iy+i\eta ) . \end {align*}

By definition of the complex exponential, \begin {align*} \exp (z)\exp (w)&=\exp ((x+\xi )+i(y+\eta ))=\exp (z+w) \end {align*}

as desired.

(Sam, Problem 360) Suppose that \(z\) is a complex number and that \(|z|=1\). Show that there is a number \(\theta \in \R \) with \(\exp (i\theta )=z\). How many such numbers \(\theta \) exist? (Use only undergraduate real analysis and methods established so far in this course.)

We know from real analysis that, if \((x,y)\) lies on the unit circle, then \((x,y)=(\cos \theta ,\sin \theta )\) for some real number \(\theta \). By definition of complex modulus, if \(|z|=1\) and \(z=x+iy\) then \((x,y)\) lies on the unit circle. Thus \(z=\cos \theta +i\sin \theta =\exp (i\theta )\) for some \(\theta \in \R \).

Infinitely many such numbers \(\theta \) exist.

[Chapter 1, Problem 25] If \(\theta \), \(\varpi \in \R \), then \(e^{i\theta }=e^{i\varpi }\) if and only if \((\theta -\varpi )/(2\pi )\) is an integer.

(William, Problem 370) Suppose that \(z\) is a complex number. Show that there exist numbers \(r\in [0,\infty )\) and \(\theta \in \R \) such that \(z=r\exp (i\theta )\). How many possible values of \(r\) exist? How many possible values of \(\theta \) exist? (Use only undergraduate real analysis and methods established so far in this course.)

Observe that \(|re^{i\theta }|=r|e^{i\theta }|\) because \(r\geq 0\) and because the modulus distributes over products. But \(|e^{i\theta }|=|\cos \theta +i\sin \theta |=\sqrt {\cos ^2\theta +\sin ^2\theta }=1\), and so the only choice for \(r\) is \(r=|z|\).

If \(z=0\) then we must have that \(r=0\) and can take any real number for \(\theta \).

If \(z\neq 0\), let \(r=|z|\). Then \(w=\frac {1}{r}z\) is a complex number with \(|z|=1\), and so there exist infinitely many values \(\theta \) with \(e^{i\theta }=w\) and thus \(z=re^{i\theta }\).

(Wilson, Problem 380) Find all solutions to the equation \(z^6=i\). Use only undergraduate real analysis and methods established so far in this course.

Suppose that \(z=re^{i\theta }\) for some \(r\geq 0\), \(\theta \in \R \).

Then \(z^6=r^6 e^{6i\theta }\). If \(z^6=i\), then \(1=|i|=|z^6|=r^6\) and so \(r=1\) because \(r\geq 0\). We must then have that \(i=e^{6i\theta }\). Observe that \(i=e^{i\pi /2}\). By Homework 1.25, we must have that \(6\theta =\pi /2+2\pi n\) for some \(n\in \Z \), and so \((e^{i\theta })^6=i\) if and only if \(\theta =\pi /12+n \pi /3\). Thus the solutions are \begin {equation*} e^{\pi /12},\quad e^{5\pi /12},\quad e^{9\pi /12},\quad e^{13\pi /12},\quad e^{17\pi /12},\quad e^{21\pi /12}. \end {equation*} Any other solution is of the form \(e^{i\theta }\), where \(\theta \) differs from one of the listed numbers by \(2\pi \).

1.3. Real Analysis

(Problem 390) Give an example of a function that can be written in two different ways.

[Definition: Real polynomial] Let \(p:\R \to \R \) be a function. We say that \(p\) is a (real) polynomial in one (real) variable if there is a \(n\in \N _0\) and constants \(a_0\), \(a_1,\dots ,a_n\in \R \) such that \(p(x)=\sum _{k=0}^n a_k x^k\) for all \(x\in \R \).

[Definition: Real polynomial in two variables] Let \(p:\R ^2\to \R \) be a function. We say that \(p\) is a (real) polynomial in two (real) variables if there is a \(n\in \N _0\) and constants \(a_{k,\ell }\in \R \) such that \(p(x,y)=\sum _{k=0}^n \sum _{\ell =0}^n a_{k,\ell } x^k y^\ell \) for all \(x\), \(y\in \R \).

(Adam, Problem 400) Let \(p(x)=\sum _{k=0}^n a_k\,x^k\) and let \(q(x)=\sum _{k=0}^n b_k\,x^k\) be two polynomials in one variable, with \(a_k\), \(b_k\in \R \). Show that if \(p(x)=q(x)\) for all \(x\in \R \) then \(a_k=b_k\) for all \(k\in \N _0\).

\(p\) and \(q\) are infinitely differentiable functions from \(\R \) to \(\R \), and because \(p(x)=q(x)\) for all \(x\in \R \), we must have that \(p'=q'\), \(p''=q'',\dots ,p^{(k)}=q^{(k)}\) for all \(k\in \N \).

We compute \(p^{(k)}(0)=k!a_k\) and \(q^{(k)}(0)=k!b_k\). Setting them equal we see that \(a_k=b_k\).

[Definition: Degree] If \(p(z)=\sum _{k=0}^n a_k\,z^k\), then the degree of \(p\) is the largest nonnegative integer \(m\) such that \(a_m\neq 0\). (The degree of the zero polynomial \(p(z)=0\) is either undefined, \(-1\), or \(-\infty \).)

(Amani, Problem 410) Let \(p\) be a polynomial. Suppose that \(x_0\in \R \) and that \(p(x_0)=0\). Show that there exists a polynomial \(q\) such that \(p(x)=(x-x_0)q(x)\) for all \(x\in \R \). Further show that, if \(p\) is a polynomial of degree \(m\geq 0\), then \(q\) is a polynomial of degree \(m-1\). Hint: Use induction.

If \(p\) is the zero polynomial we may take \(q\) to also be the zero polynomial. If \(p\) is a nonzero constant polynomial then no such \(x_0\) can exist. We therefore need only consider the case where \(p\) is a polynomial of degree \(m\geq 1\).

If \(m=1\), then \(p(x)=a_1x+a_0\) for some \(a_1\), \(a_0\); if \(p(x_0)=0\) then \(a_0=-a_1x_0\) and so \(p(x)=a_1(x-x_0)\). Then \(q(x)=a_1\) is a polynomial of degree \(0=m-1\).

Suppose that the statement is true for all polynomials of degree at most \(m-1\), \(m\geq 2\). Let \(p\) be a polynomial of degree \(m\). Then \(p(x)=a_m x^m +r(x)\) where \(r\) is a polynomial of degree at most \(m-1\). We add and subtract \(a_m x_0 x^{m-1}\) to see that \begin {equation*} p(x)=a_m x^{m-1} (x-x_0) + a_mx_0 x^{m-1}+r(x). \end {equation*} Then \(s(x)=a_mx_0 x^{m-1}+r(x)\) is a polynomial of degree at most \(m-1\). If \(s\) is a constant then \(0=p(x_0)=a_m x_0^{m-1}(x-x_0)+s\) and so \(s=0\); taking \(q(x)=a_mx^{m-1}\) we are done.

Otherwise, \(s(x)\) is a polynomial of degree at least one and at most \(m-1\). Also, \(s(x_0)=p(x_0)-a_mx_0^m(x_0-x_0)=0\), so by the induction hypothesis \(s(x)=(x-x_0)t(x)\) for a polynomial \(t\) of degree at most \(m-2\). Taking \(q(x)=a_mx^{m-1}+t(x)\) we are done.

(Bashar, Problem 420) Let \(p(x)=\sum _{k=0}^n a_k\,x^k\) and let \(q(x)=\sum _{k=0}^n b_k\,x^k\) be two polynomials of degree at most \(n\), with \(a_k\), \(b_k\in \R \) and \(n\in \N _0\). Suppose that there are \(n+1\) distinct numbers \(x_0,x_1,\dots ,x_n\in \R \) such that \(p(x_j)=q(x_j)\) for all \(0\leq j\leq n\). Show that \(a_k=b_k\) for all \(k\in \N _0\). Hint: Consider the polynomial \(r(x)=p(x)-q(x)\).

Let \(r(x)=p(x)-q(x)\). Then \(r(x_j)=p(x_j)-q(x_j)=0\) for all \(0\leq j\leq n\) and \(r\) is a polynomial of degree at most \(n\). Furthermore, \(r(x_j)=0\) for all \(0\leq j\leq n\).

Suppose for the sake of contradiction that \(r\) is not identically equal to zero. Then \(r\) is a polynomial of degree \(m\), \(0\leq m\leq n\). By Problem 410, \begin {equation*} r(x)=(x-x_1)(x-x_2)\dots (x-x_m)r_m(x) \end {equation*} where \(r_m\) is a polynomial of degree \(m-m\), that is, a constant. But \begin {equation*} 0=p(x_0)-q(x_0)=r(x_0)=(x_0-x_1)(x_0-x_2)\dots (x_0-x_m) r_m(x_0). \end {equation*} Since \(x_j\neq x_0\) for all \(j\geq 1\) we must have that \(r_m(x_0)=0\); thus \(r_m\) is the constant function zero and so \(r\) is the constant function zero, as was to be proven. (This is technically a contradiction to the assumption \(m\geq 0\) because if \(m\geq 0\) then \(r\) is not the zero polynomial.)

(Dibyendu, Problem 430) Let \(p(x,y)=\sum _{j=0}^n\sum _{k=0}^n a_{j,k}\,x^j\,y^k\) and let \(q(x,y)=\sum _{j=0}^n\sum _{k=0}^n b_{j,k}\,x^j\,y^k\) be two polynomials of two variables, with \(a_{j,k}\), \(b_{j,k}\in \R \). Show that if \(p(x,y)=q(x,y)\) for all \((x,y)\in \R ^2\) then \(a_{j,k}=b_{j,k}\) for all \(j\), \(k\in \N _0\).

Fix a \(y\in \R \). Then \(p_y(x)=\sum _{j=0}^n \Bigl (\sum _{k=0}^n a_{j,k}y^k\Bigr ) x^j\) and \(q_y(x)=\sum _{j=0}^n \Bigl (\sum _{k=0}^n b_{j,k}y^k\Bigr ) x^j\) are both polynomials in one variable that are equal for all \(x\). So by Problem 400 their coefficients must be equal, so \(\Bigl (\sum _{k=0}^n a_{j,k}y^k\Bigr )=\Bigl (\sum _{k=0}^n b_{j,k}y^k\Bigr )\). This is true for all \(y\in \R \); another application of Problem 400 shows that \(a_{j,k}=b_{j,k}\) for all \(j\) and \(k\).

(Problem 431) Give an example of a polynomial of two variables \(p\) such that \(p(x,y)=0\) for infinitely many values of \((x,y)\), but such that \(p\) is not the zero polynomial.

The polynomial \(p(x,y)=xy\) is such a polynomial, because \(p(0,y)=0\) and \(p(x,0)=0\) for all \(x\in \R \) or \(y\in \R \), and there are infinitely many \(x\in \R \) and infinitely many \(y\in \R \).

(Memory 440) If \(\Omega \subseteq \R ^2\) is both open and connected, then \(\Omega \) is path connected: for every \(z\), \(w\in \Omega \) there is a continuous function \(\gamma :[0,1]\to \Omega \) such that \(\gamma (0)=z\) and \(\gamma (1)=w\).

(Memory 450) If \(\Omega \subseteq \R ^2\) is open and connected, we may require the paths in the definition of path connectedness to be \(C^1\).

(Memory 460) If \(\Omega \subseteq \R ^2\) is open and connected, we may require the paths in the definition of path connectedness to consist of finitely many horizontal or vertical line segments.

Definition 1.3.1 (part 1). Let \(\Omega \subseteq \R ^2\) be open. Suppose that \(f:\Omega \to \R \). We say that \(f\) is continuously differentiable, or \(f\in C^1(\Omega )\), if the two partial derivatives \(\frac {\partial f}{\partial x}\) and \(\frac {\partial f}{\partial y}\) exist everywhere in \(\Omega \) and \(f\), \(\frac {\partial f}{\partial x}\), and \(\frac {\partial f}{\partial y}\) are all continuous on \(\Omega \).

(Hope, Problem 470) Let \(B=B(z,r)\) be a ball in \(\R ^2\). Let \(f\in C^1(B)\) and suppose that \(\frac {\partial f}{\partial y}=\frac {\partial f}{\partial x}=0\) everywhere in \(B\). Show that \(f\) is a constant.

Let \(z=(x,y)\). Let \((\xi ,\eta )\in B((x,y),r)\).

We consider the case \(\xi \geq x\) and \(\eta \geq y\); the cases \(\xi <x\) or \(\eta <y\) are similar. Then \(\{(t,y):x\leq t\leq \xi \}\subset B((x,y),r)\), and if we let \(F_y(x)=F(x,y)\), then \(F_y\) is a continuously differentiable function on \([x,\xi ]\) with \(F_y'(t)=0\) for all \(x\leq t\leq \xi \); by the Mean Value Theorem, \(F_y(x)=F_y(\xi )\) and so \(f(x,y)=f(\xi ,y)\). Similarly, \(\{(\xi ,t):y\leq t\leq \eta \}\subset B((x,y),r)\), and so \(f(x,y)=f(\xi ,y)=f(\xi ,\eta )\).

Thus \(f\) is a constant in \(B((x,y),r)\).

(James, Problem 480) Suppose that \(\Omega \subseteq \R ^2\) is open and connected. Let \(f\in C^1(\Omega )\) and suppose that \(\frac {\partial f}{\partial y}=\frac {\partial f}{\partial x}=0\) everywhere in \(\Omega \). Show that \(f\) is a constant.

1.3. Complex Polynomials

[Definition: Complex polynomials in one variable] Let \(p:\C \to \C \) be a function. We say that \(p\) is a polynomial in one (complex) variable if there is a \(n\in \N _0\) and constants \(a_0\), \(a_1,\dots ,a_n\in \C \) such that \(p(z)=\sum _{k=0}^n a_k z^k\) for all \(z\in \C \).

[Definition: Complex polynomial in two variables] Let \(p:\C \to \C \) be a function. We say that \(p\) is a polynomial in two real variables if there is a \(n\in \N _0\) and constants \(a_{k,\ell }\in \C \) such that \(p(x+iy)=\sum _{k=0}^n \sum _{\ell =0}^n a_{k,\ell } x^k y^\ell \) for all \(x\), \(y\in \R \).

(Fact 490) Problem 430 is true for complex polynomials of two real variables; that is, if \(p(x+iy)=\sum _{k,\ell =0}^n a_{k,\ell } x^ky^\ell \), \(q(x+iy)=\sum _{k,\ell =0}^n c_{k,\ell } x^ky^\ell \), and \(p(z)=q(z)\) for all \(z\in \C \), then \(a_{k,\ell }=c_{k,\ell }\) for all \(k\) and \(\ell \).

(Fact 500) Problem 400, Problem 410, and Problem 420 are valid for complex polynomials of one complex variable. There are complex polynomials of two real variables with infinitely many zeroes, as in Problem 431, that are not the zero polynomial.

[Chapter 1, Problem 35] The functions \(p(x+iy)=x\) and \(q(x+iy)=x^2\) are both clearly polynomials of two real variables. Prove that neither is a polynomial of one complex variable.

(Micah, Problem 510) Show that \(p\) is a polynomial in two real variables if and only if there are constants \(b_{k,\ell }\in \C \) such that \(p(z)=\sum _{k=0}^n \sum _{\ell =0}^n b_{k,\ell } z^k \bar z^\ell \) for all \(z\in \C \).

(Problem 511) Problem 430 is true for complex polynomials of two real variables written in this form; that is, if \(p(z)=\sum _{k,\ell =0}^n b_{k,\ell } z^k\bar z^\ell \), \(q(z)=\sum _{k,\ell =0}^n d_{k,\ell } z^k\bar z^\ell \), and \(p(z)=q(z)\) for all \(z\in \C \), then \(b_{k,\ell }=d_{k,\ell }\) for all \(k\) and \(\ell \).

1.3. The complex derivatives \(\p {z}\) and \(\p {\bar z}\)

Definition 1.3.1 (part 2). Let \(\Omega \subseteq \C \) be an open set. Recall \(\C =\R ^2\). Let \(f:\Omega \to \C \) be a function. Then \(f\in C^1(\Omega )\) if \(\re f\), \(\im f\in C^1(\Omega )\).

[Definition: Derivative of a complex function] Let \(f\in C^1(\Omega )\). Let \(u(z)=\re f(z)\) and let \(v(z)=\im f(z)\). Then \begin {equation*} \frac {\partial f}{\partial x} = \frac {\partial u}{\partial x}+i\frac {\partial v}{\partial x}, \quad \frac {\partial f}{\partial y} = \frac {\partial u}{\partial y}+i\frac {\partial v}{\partial y}. \end {equation*}

(Muhammad, Problem 520) Let \(\Psi \), \(\Omega \subseteq \C \) be two open sets, and let \(f:\Psi \to \Omega \) and \(g:\Omega \to \C \) be two \(C^1\) functions. What is \(\p {x}(g\circ f)\)?

(Nisa, Problem 530) Establish the Leibniz rules \begin {equation*} \frac {\partial }{\partial x} (fg)=\frac {\partial f}{\partial x} g+f\frac {\partial g}{\partial x},\qquad \frac {\partial }{\partial y} (fg)=\frac {\partial f}{\partial y} g+f\frac {\partial g}{\partial y} \end {equation*} for \(f\), \(g\in C^1(\Omega )\).

Let \(f=u+iv\), \(g=w+i\varpi \), where \(u\), \(v\), \(w\), and \(\varpi \) are real-valued functions in \(C^1(\Omega )\).

Then \(fg=(uw-v\varpi )+i(vw+u\varpi )\), where \((uw-v\varpi )\) and \((vw+u\varpi )\) are both real-valued \(C^1\) functions.

Then \begin {align*} \frac {\partial }{\partial x} (fg) &=\frac {\partial }{\partial x} [(uw-v\varpi )+i(vw+u\varpi )] \\&= \frac {\partial }{\partial x}(uw-v\varpi )+i\frac {\partial }{\partial x}(vw+u\varpi ). \end {align*}

Applying the Leibniz (product) rule for real-valued functions, we see that \begin {align*} \frac {\partial }{\partial x} (fg) &=\frac {\partial u}{\partial x}w + u\frac {\partial w}{\partial x}-\frac {\partial v}{\partial x}\varpi -v\frac {\partial \varpi }{\partial x} \\&\qquad + i\frac {\partial v}{\partial x}w + iv\frac {\partial w}{\partial x}+i\frac {\partial u}{\partial x}\varpi +iu\frac {\partial \varpi }{\partial x}. \end {align*}

Furthermore, \begin {align*} f\frac {\partial g}{\partial x}+\frac {\partial f}{\partial x} g &= (u+iv) \biggl (\frac {\partial w}{\partial x}+i\frac {\partial \varpi }{\partial x}\biggr ) \\&\qquad +\biggl (\frac {\partial u}{\partial x}+i\frac {\partial v}{\partial x}\biggr )(w+i\varpi ) \\&= u\frac {\partial w}{\partial x}+iu\frac {\partial \varpi }{\partial x}+iv\frac {\partial w}{\partial x}-v\frac {\partial \varpi }{\partial x}\\&\qquad +\frac {\partial u}{\partial x}w+i\frac {\partial u}{\partial x}\varpi +i\frac {\partial v}{\partial x}w -\frac {\partial v}{\partial x}\varpi . \end {align*}

Rearranging, we see that the two terms are the same.

[Definition: Complex derivative] Let \(f\in C^1(\Omega )\). Then \begin {equation*} \frac {\partial f}{\partial z} = \frac {1}{2}\frac {\partial f}{\partial x}+\frac {1}{2i}\frac {\partial f}{\partial y}, \quad \frac {\partial f}{\partial \bar z} = \frac {1}{2}\frac {\partial f}{\partial x}-\frac {1}{2i}\frac {\partial f}{\partial y}. \end {equation*}

(Robert, Problem 540) Let \(f(z)=z\) and let \(g(z)=\bar z\). Show that \(\frac {\partial f}{\partial z}=1\), \(\frac {\partial f}{\partial \bar z}=0\), \(\frac {\partial g}{\partial z}=0\), \(\frac {\partial g}{\partial \bar z}=1\).

Recall that \(z=x+iy\). Thus, \begin {equation*} \frac {\partial }{\partial z} (z) = \frac {1}{2}\frac {\partial }{\partial x}(x+iy)+\frac {1}{2i}\frac {\partial }{\partial y}(x+iy) =\frac {1}{2}+\frac {i}{2i}=1 \end {equation*} and \begin {equation*} \frac {\partial }{\partial \bar z} (z) = \frac {1}{2}\frac {\partial }{\partial x}(x+iy)-\frac {1}{2i}\frac {\partial }{\partial y}(x+iy) =\frac {1}{2}-\frac {i}{2i}=0 . \end {equation*}

Recall that \(\bar z=x-iy\). Thus, \begin {equation*} \frac {\partial }{\partial z} (\bar z) = \frac {1}{2}\frac {\partial }{\partial x}(x-iy)+\frac {1}{2i}\frac {\partial }{\partial y}(x-iy) =\frac {1}{2}-\frac {i}{2i}=0 \end {equation*} and \begin {equation*} \frac {\partial }{\partial \bar z} (\bar z) = \frac {1}{2}\frac {\partial }{\partial x}(x-iy)-\frac {1}{2i}\frac {\partial }{\partial y}(x-iy) =\frac {1}{2}+\frac {i}{2i}=1 . \end {equation*}

(Fact 550) \(\frac {\partial }{\partial z}\) and \(\frac {\partial }{\partial \bar z}\) are linear operators.

This follows immediately from linearity of the differential operators \(\frac {\partial }{\partial x}\) and \(\frac {\partial }{\partial y}\).

(Fact 560) \(\frac {\partial }{\partial z}\) and \(\frac {\partial }{\partial \bar z}\) commute in the sense that, if \(\Omega \subseteq \C \) is open and \(f\in C^2(\Omega )\), then \(\frac {\partial }{\partial z}\left (\frac {\partial }{\partial \bar z} f\right )=\frac {\partial }{\partial \bar z}\left (\frac {\partial }{\partial z} f\right )\).

(Fact 570) The following Leibniz rules are valid: \begin {equation*} \frac {\partial }{\partial z} (fg)=\frac {\partial f}{\partial z} g+f\frac {\partial g}{\partial z}, \qquad \frac {\partial }{\partial \bar z} (fg)=\frac {\partial f}{\partial \bar z} g+f\frac {\partial g}{\partial \bar z} . \end {equation*}

We have that \begin {align*} \frac {\partial }{\partial z} (fg) =\frac {1}{2}\frac {\partial }{\partial x} (fg) +\frac {1}{2i}\frac {\partial }{\partial y} (fg) . \end {align*}

Using the Leibniz rules for \(\frac {\partial }{\partial x}\) and \(\frac {\partial }{\partial y}\), we see that \begin {align*} \frac {\partial }{\partial z} (fg) &=\frac {1}{2}f\frac {\partial g}{\partial x} +\frac {1}{2}\frac {\partial f}{\partial x} g +\frac {1}{2i}f\frac {\partial g}{\partial y}+\frac {1}{2i}\frac {\partial f}{\partial y}g \\&= f\biggl (\frac 12\frac {\partial g}{\partial x}+\frac {1}{2i}\frac {\partial g}{\partial y}\biggr ) +\biggl (\frac {1}{2}\frac {\partial f}{\partial x}+\frac {1}{2i}\frac {\partial f}{\partial y}\biggr )g \\&=f\frac {\partial g}{\partial z}+\frac {\partial f}{\partial z}g . \end {align*}

The argument for \(\frac {\partial }{\partial \bar z}\) is similar.

(Sam, Problem 580) Show that \(\frac {\partial }{\partial z} (z^\ell \bar z^m)=\ell z^{\ell -1}\bar z^m\) and \(\frac {\partial }{\partial \bar z} (z^\ell \bar z^m)=mz^\ell \bar z^{m-1}\) for all nonnegative integers \(m\) and \(\ell \) (with the minor abuse of notation that we take \(z^0\equiv \bar z^0\equiv 1\) and \(0z^{-1}\equiv 0\bar z^{-1}\equiv 0\), that is, we ignore the singularities at \(z=0\)).

If \(\ell =m=0\), then \(z^\ell \bar z^m=1\) and \(\ell z^{\ell -1}\bar z^m=0=mz^\ell \bar z^{m-1}\). The result is obvious in this case.

Suppose now that \(m=0\), \(\ell \geq 1\) and the result is true for \(\ell -1\). (The result is true for \(\ell -1\) if \(\ell =1\) by the above argument.) By Fact 570, \begin {align*} \p { z} z^\ell &=\p { z}(z\cdot z^{\ell -1}) = \biggl (\p { z}z\biggr )z^{\ell -1}+z\biggl (\p { z}z^{\ell -1}\biggr ), \\ \p {\bar z} z^\ell &=\p {\bar z}(z\cdot z^{\ell -1}) = \biggl (\p {\bar z}z\biggr )z^{\ell -1}+z\biggl (\p {\bar z}z^{\ell -1}\biggr ), \end {align*}

which by the induction hypothesis and Problem 540 equal \begin {align*} \p { z} z^\ell = z^{\ell -1}+z\cdot (\ell -1)z^{\ell -2}, \\ \p {\bar z} z^\ell = 0\cdot z^{\ell -1}+z\cdot 0, \end {align*}

which simplify to the desired result. Thus by induction the result is true whenever \(m=0\). A similar induction argument yields the result whenever \(\ell =0\). Finally, the general case follows from Fact 570: \begin {align*} \frac {\partial }{\partial z} (z^\ell \bar z^m)&= \biggl (\p {z}z^\ell \biggr )\bar z^m+z^\ell \biggl (\p { z}\bar z^m\biggr ) = \ell z^{\ell -1}\bar z^m+0,\\ \frac {\partial }{\partial \bar z} (z^\ell \bar z^m)&= \biggl (\p {\bar z}z^\ell \biggr )\bar z^m+z^\ell \biggl (\p {\bar z}\bar z^m\biggr ) = 0+ z^{\ell }\cdot m\bar z^{m-1} \end {align*}

as desired.

(William, Problem 590) Let \(p\) be a complex polynomial in two real variables. Show that \(p\) is a complex polynomial in one complex variable if and only if \(\frac {\partial p}{\partial \bar z}=0\) everywhere in \(\C \).

If \(p\) is a polynomial in one complex variable, then by definition there are constants \(a_k\in \C \) and \(n\in \N \) such that \(p(z)=\sum _{k=0}^n a_k z^k\). By linearity of the complex derivative operator and by Problem 580, we have that \(\p {\bar z} p(z)=0\), as desired.

Now suppose that \(p\) is a complex polynomial in two real variables and \(\p {\bar z} p(z)=0\) for all \(z\in \C \). By Problem 510, there are constants \(b_{k,\ell }\in \C \) and \(m\in \N \) such that \begin {equation*} p(z)=\sum _{k=0}^m \sum _{\ell =0}^m b_{k,\ell } z^k\bar z^\ell \end {equation*} for all \(z\in \C \). By linearity of the complex derivative operator and by Problem 580, we have that \begin {equation*} \p {\bar z} p(z)=\sum _{k=0}^m \sum _{\ell =0}^m \ell b_{k,\ell } z^k\bar z^{\ell -1}. \end {equation*} This polynomial is identically equal to zero. By Problem 511, this implies that \(\ell b_{k,\ell }=0\) for all \(k\) and \(\ell \). In particular, if \(\ell \geq 1\) then \(b_{k,\ell }=0\). Thus \begin {equation*} p(z)=\sum _{k=0}^m b_{k,0} z^k \end {equation*} as desired.

Definition 1.4.1. Let \(\Omega \subseteq \C \) be open and let \(f\in C^1(\Omega )\). We say that \(f\) is holomorphic in \(\Omega \) if \begin {equation*} \frac {\partial f}{\partial \bar z}=0 \end {equation*} everywhere in \(\Omega \).

(Fact 600) A polynomial in two real variables is a polynomial in one complex variable if and only if it is holomorphic.

(Wilson, Problem 610) Suppose that \(\Omega \subseteq \C \) is open and connected, that \(f\in C^1(\Omega )\), and that \(\frac {\partial f}{\partial z}=0=\frac {\partial f}{\partial \bar z}\) in \(\Omega \). Show that \(f\) is constant in \(\Omega \).

We observe that \begin {equation*} \p [f]{x}=\p [f]{z}+\p [f]{{\bar z}},\qquad \p [f]{y}=i\p [f]{z}-i\p [f]{{\bar z}}. \end {equation*} Thus \(\p [f]{x}=\p [f]{y}=0\) in \(\Omega \) and the result follows from Problem 480.

[Chapter 1, Problem 34] Suppose that \(\Omega \subseteq \C \) is open and that \(f\in C^1(\Omega )\). Show that \begin {equation*} \frac {\partial f}{\partial z}(w) = \overline {\left (\frac {\partial \overline f}{\partial \bar z}(w)\right )} \end {equation*} for all \(w\in \Omega \).

(Adam, Problem 620) Show that \(\frac {\partial }{\partial z}\frac {1}{z}=-\frac {1}{z^2}\) and \(\frac {\partial }{\partial \bar z}\frac {1}{z}=0\) if \(z\neq 0\). Then compute \(\frac {\partial }{\partial z}\frac {1}{z^n}\) and \(\frac {\partial }{\partial \bar z}\frac {1}{z^n}\) for any positive integer \(n\).

Observe that \(\frac {1}{z}=\frac {\bar z}{z\bar z}\) and so if \(z=x+iy\), \(x\), \(y\in \R \), then \(\frac {1}{z}=\frac {x-iy}{x^2+y^2}\). We compute \begin {align*} \p {x}\frac {1}{z}&= \p {x}\re \frac {1}{z} +i\p {x}\im \frac {1}{z}= \p {x}\frac {x}{x^2+y^2} +i\p {x}\frac {-y}{x^2+y^2} = \frac {y^2-x^2+2ixy}{(x^2+y^2)^2}, \\ \p {y}\frac {1}{z}&= \p {y}\frac {x}{x^2+y^2} +i\p {y}\frac {-y}{x^2+y^2} = \frac {-ix^2+iy^2-2xy}{(x^2+y^2)^2}. \end {align*}

Thus \begin {align*} \p {z}\frac {1}{z} &=\frac {1}{2} \p {x}\frac {1}{z}-\frac {i}{2}\p {y}\frac {1}{z} = \frac {y^2-x^2+2ixy}{(x^2+y^2)^2} \\&= \frac {-(x-iy)^2}{(x+iy)^2(x-iy)^2} =-\frac {1}{z^2} \end {align*}

and \begin {equation*} \p {z}\frac {1}{z} =\frac {1}{2} \p {x}\frac {1}{z}+\frac {i}{2}\p {y}\frac {1}{z} = 0. \end {equation*} Using the Leibniz rule for the inductive step, a straightforward induction argument shows that \begin {equation*} \p {z}\frac {1}{z^n}=-\frac {n}{z^{n+1}}, \qquad \p {\bar z}\frac {1}{z^n} =0. \end {equation*}

[Chapter 1, Problem 49] Let \(\Omega \), \(W\subseteq \C \) be open and let \(g:\Omega \to W\), \(f:W\to \C \) be two \(C^1\) functions. The following chain rules are valid: \begin {gather*} \frac {\partial }{\partial z} (f\circ g)= \frac {\partial f}{\partial g}\frac {\partial g}{\partial z} +\frac {\partial f}{\partial \overline g}\frac {\partial \overline g}{\partial z}, \\ \frac {\partial }{\partial \bar z} (f\circ g)= \frac {\partial f}{\partial g}\frac {\partial g}{\partial \overline z} +\frac {\partial f}{\partial \overline g}\frac {\partial \overline g}{\partial \overline z}, \end {gather*} where \(\frac {\partial f}{\partial g} = \left .\frac {\partial f}{\partial z}\right |_{z\to g(z)}\), \(\frac {\partial f}{\partial \overline g} = \left .\frac {\partial f}{\partial \bar z}\right |_{z\to g(z)}\).

In particular, if \(f\) and \(g\) are both holomorphic then so is \(f\circ g\).

1.4. Real Analysis

(Memory 630) Let \(f\) be a \(C^2\) function in an open set in \(\R ^2\). Show that \(\frac {\partial }{\partial x} \frac {\partial f}{\partial y}=\frac {\partial }{\partial y} \frac {\partial f}{\partial x}\) everywhere in the domain.

1.4. Holomorphic Functions, the Cauchy-Riemann Equations, and Harmonic Functions

Lemma 1.4.2. Let \(f\in C^1(\Omega )\), let \(u=\re f\), and let \(v=\im f\). Then \(f\) is holomorphic in \(\Omega \) if and only if \begin {equation*} \frac {\partial u}{\partial x}=\frac {\partial v}{\partial y}\quad \text {and}\quad \frac {\partial u}{\partial y}=-\frac {\partial v}{\partial x} \end {equation*} everywhere in \(\Omega \). (These equations are called the Cauchy-Riemann equations.)

(Amani, Problem 640) Prove the “only if” direction of Lemma 1.4.2: show that if \(f\) is holomorphic in \(\Omega \), \(\Omega \subseteq \C \) open, then the Cauchy-Riemann equations hold for \(u=\re f\) and \(v=\im f\).

(Bashar, Problem 650) Prove the “if” direction of Lemma 1.4.2: show that if \(u=\re f\) and \(v=\im f\) are \(C^1\) in \(\Omega \) and satisfy the Cauchy-Riemann equations, then \(f\) is holomorphic in \(\Omega \).

Recall that \begin {align*} 2\p [f]{\bar z} = \p [f]{x} +i\p [f]{y} \end {align*}

by definition of \(\p {\bar z}\). Applying the fact that \(f=u+iv\), we see that \begin {align*} 2\p [f]{\bar z}&=\p [u]{x}+i\p [v]{x} +i \biggl (\p [u]{y}+i\p [v]{y}\biggr ) \\&= \biggl (\p [u]{x}-\p [v]y\biggr )+ i\biggl (\p [v]x+\p [u]y\biggr ). \end {align*}

Because \(u\) and \(v\) are real-valued, so are their derivatives. Thus, the real and imaginary parts of the right hand side, respectively, are \(\p [u]x-\p [v]y\) and \(\p [u]y+\p [v]x\).

Thus, \(\p [f]{\bar z} = 0\) if and only if the Cauchy-Riemann equations hold.

Proposition 1.4.3. [Slight generalization.] Let \(f\in C^1(\Omega )\). Then \(f\) is holomorphic at \(p\in \Omega \) if and only if \(\frac {\partial f}{\partial x}(p)=\frac {1}{i}\frac {\partial f}{\partial y}(p)\) and that in this case \begin {equation*} \frac {\partial f}{\partial z}(p)=\frac {\partial f}{\partial x}(p) =\frac {1}{i}\frac {\partial f}{\partial y}(p). \end {equation*}

(Dibyendu, Problem 660) Begin the proof of Proposition 1.4.3 by showing that if \(f\) is holomorphic then \(\frac {\partial f}{\partial z}=\frac {\partial f}{\partial x}=\frac {1}{i}\frac {\partial f}{\partial y}\).

By definition of \(\p {z}\) and \(\p {\bar z}\), if \(f\in C^1(\Omega )\) then \(\p [f]{x}=\p [f]{z}+\p [f]{\bar z}\) and \(\p [f]{y}=i\p [f]{z}-i\p [f]{\bar z}\). Thus, if \(\p [f]{\bar z}(p)=0\) then \(\p [f]{x}(p)=\p [f]{z}(p)\) and \(\p [f]{y}(p)=i\p [f]{z}(p)=i\p [f]{x}(p)\), as desired.

(Hope, Problem 670) Complete the proof of Proposition 1.4.3 by showing that if \(f\in C^1(\Omega )\) and \(\frac {\partial f}{\partial x}=\frac {1}{i}\frac {\partial f}{\partial y}\), then \(f\) is holomorphic.

Recall \begin {equation*} \p [f]{\bar z} = \frac 12\biggl (\p [f]x- \frac {1}{i}\p [f]{y}\biggr ). \end {equation*} Thus, if \(\p [f]x=\frac {1}{i}\p [f]y\) then \(\p [f]{\bar z}=0\), as desired.

Definition 1.4.4. We let \(\triangle =\frac {\partial ^2 }{\partial x^2}+\frac {\partial ^2 }{\partial y^2}\). If \(\Omega \subseteq \C \) is open and \(u\in C^2(\Omega )\), then \(u\) is harmonic if \begin {equation*} \triangle u=\frac {\partial ^2 u}{\partial x^2}+\frac {\partial ^2 u}{\partial y^2}=0 \end {equation*} everywhere in \(\Omega \).

(Fact 680) Show that if \(f\in C^1(\Omega )\) then \(\triangle f=4\frac {\partial }{\partial z}\frac {\partial f} {\partial \bar z} =4\frac {\partial }{\partial \bar z} \frac {\partial f}{\partial z}\).

We compute that \begin {align*} \p {z}\p [f]{\bar z} &= \frac {1}{4} \biggl (\p x+\frac {1}{i}\p y\biggr ) \biggl (\p [f] x-\frac {1}{i} \p [f] y\biggr ) \\&=\frac {1}{4}\biggl (\frac {\partial ^2f}{\partial x^2} +\frac {\partial ^2f}{\partial y^2} +\frac {1}{i}\frac {\partial ^2 f}{\partial y\,\partial x} -\frac {1}{i}\frac {\partial ^2 f}{\partial x\,\partial y}\biggr ) . \end {align*}

If \(f\in C^1\) then \(\frac {\partial ^2 f}{\partial y\,\partial x} =\frac {\partial ^2 f}{\partial x\,\partial y}\) and the proof is complete. The argument for \(\p {\bar z}\p [f]{z} \) is similar.

(Micah, Problem 690) Suppose that \(f\) is holomorphic and \(C^2\) in an open set \(\Omega \) and that \(u=\re f\) and \(v=\im f\). Compute \(\triangle u\) and \(\triangle v\).

Because \(f\) is holomorphic, \begin {equation*} \triangle f = 4\frac {\partial }{\partial z}\frac {\partial f} {\partial \bar z}=4\frac {\partial }{\partial z}0=0. \end {equation*} But \begin {equation*} \triangle f = (\triangle u) + i(\triangle v) \end {equation*} and \(\triangle u\) and \(\triangle v\) are both real-valued, so because \(\triangle f=0\) we must have \(\triangle u=0=\triangle v\) as well.

(Muhammad, Problem 700) Let \(f\) be a holomorphic polynomial. Show that there is a holomorphic polynomial \(F\) such that \(\frac {\partial F}{\partial z} = f\). How many such polynomials are there?

By Fact 600, \(f\) is a polynomial in one complex variable; that is, there is a \(n\in \N \) and constants \(a_k\in \C \) such that \begin {equation*} f(z)=\sum _{k=0}^n a_k\,z^k \end {equation*} for all \(z\in \C \). Let \(b\in \C \) and let \begin {equation*} F(z)=b+\sum _{k=0}^n \frac {a_k}{k+1}\,z^{k+1}. \end {equation*} By Problem 580, we have that \(\p {z}F=f\).

There are infinitely many such polynomials, one for each choice of \(b\). By Problem 610, any two such antiderivatives \(F_1\) and \(F_2\) must differ by a constant.

(Nisa, Problem 710) Show that if \(u\) is a harmonic polynomial (of two real variables) then \(u(z)=p(z)+q(\bar z)\) for some polynomials \(p\), \(q\) of one complex variable.

Because \(u\) is harmonic, by Fact 680 we have that \begin {equation*} 0=\triangle u=4\frac {\partial }{\partial \bar z} \frac {\partial u}{\partial z} \end {equation*} and so \(f=\p [u]{z}\) is holomorphic. By Problem 580 we have that \(f\) is a polynomial. Thus by Problem 700 there is a holomorphic polynomial \(F\) with \(\p [F]{z}=f=\p [u]{z}\).

Let \(p=F\). Let \(g(z)=u(z)-p(z)\). Then \(g\) is a polynomial and \(\p [g]{z}=0\), so by Problem 1.34 \(\p [\overline g]{\bar z}=0\). Thus \begin {equation*} \overline {g(z)}=\sum _{k=0}^m b_k \,z^k \end {equation*} for some \(m\in \N \) and some constants \(b_k\in \C \). Taking the complex conjugate yields that \begin {equation*} g(z)=\sum _{k=0}^m \overline {b_k}\,\bar z^k. \end {equation*} This completes the proof with \(q(z)=\sum _{k=0}^m \overline {b_k}\, z^k\).

Lemma 1.4.5. Let \(u\) be harmonic and real valued in \(\C \). Suppose in addition that \(u\) is a polynomial of two real variables. Then there is a holomorphic polynomial \(f\) such that \(u(z)=\re f(z)\).

(Robert, Problem 720) Prove Lemma 1.4.5.

By Problem 710, we have that \(u(z)=p(z)+q(\bar z)\) for some polynomials \(p\) and \(q\). We may write \begin {equation*} u(z)=\sum _{k=0}^n a_k z^k+\sum _{k=0}^n b_k \bar z^k =(a_0+b_0)+\sum _{k=1}^n a_k z^k+\sum _{k=1}^n b_k \bar z^k \end {equation*} for some \(n\in \N \) and some \(a_k\), \(b_k\in \C \). Because \(u\) is real-valued, we have that \(u(z)=\overline {u(z)}\) and so \begin {equation*} (a_0+b_0)+\sum _{k=1}^n a_k z^k+\sum _{k=1}^n b_k \bar z^k = \overline {(a_0+b_0)}+\sum _{k=1}^n \overline {b_k} z^k+\sum _{k=1}^n \overline {a_k} \bar z^k. \end {equation*} By Problem 511, this implies that \(a_0+b_0\) is real and that \(a_k=\overline {b_k}\) for all \(k\geq 1\).

Thus \begin {align*} u(z) &=(a_0+b_0)+ \sum _{k=1}^n (a_k z^k+\overline {a_k z^k}) =\re (a_0+b_0)+ \sum _{k=1}^n 2\Re (a_k z^k) \\&=\re \Bigl ((a_0+b_0)+ \sum _{k=1}^n 2a_k z^k\Bigr ) \end {align*}

as desired.

1.5. Real Analysis

(Memory 730) State Green’s theorem.

Let \(\gamma :[0,1]\to \R ^2\) be a piecewise \(C^1\) simple closed curve.² Let \(\Omega \subset \R ^2\) be the bounded open set that satisfies \(\partial \Omega =\gamma ([0,1])\); by the Jordan curve theorem, exactly one such \(\Omega \) exists. Let \(W\) be open and satisfy \(\overline \Omega \subset W\). Let \(\vec F:W\to \R ^2\) be a \(C^1\) function.

Then \begin {equation*} \int _0^1 \vec F(\gamma (t))\cdot \gamma '(t)\,dt = \pm \int _\Omega \frac {\partial F_2}{\partial x}-\frac {\partial F_1}{\partial y}\,dx\,dy \end {equation*} where the sign is determined by the orientation of \(\gamma \).

(Memory 740) State the Mean Value Theorem.

(Memory 750) If \(a<b\), if each \(f_n\) is bounded and Riemann integrable on \([a,b]\), and if \(f_n\to f\) uniformly on \([a,b]\), then \(f\) is also Riemann integrable on \([a,b]\), \(\lim _{n\to \infty } \int _a^b f_n\) exists, and \(\int _a^b f = \lim _{n\to \infty } \int _a^b f_n\).

(Problem 760) Let \(f:[a,b]\times [c,d]\to \R \). Suppose that \(f\) is continuous on \([a,b]\times [c,d]\). Define \(F:[a,b]\to \R \) by \(F(x)=\int _{c}^d f(x,y)\,dy\). Show that \(F\) is continuous on \([a,b]\).

(Problem 770) Let \(f:(a,b)\times [c,d]\to \R \). Suppose that \(f\) is continuous on \((a,b)\times [c,d]\) and the partial derivative \(\partial _xf=\frac {\partial f}{\partial x}\) exists and is continuous everywhere on \((a,b)\times [c,d]\). Define \(F(x)=\int _c^d f(x,y)\,dy\). Show that \(F\) is differentiable on \((a,b)\) and that \begin {equation*} F'(x)=\frac {d}{dx} \int _c^d f(x,y)\,dy=\int _c^d \p {x} f(x,y)\,dy \end {equation*} for all \(a<x<b\).

This will be proven as homework.

1.5. Real and Holomorphic Antiderivatives

(Sam, Problem 780) Prove the converse to Clairaut’s theorem. That is, suppose that there are two \(C^1\) functions \(g\) and \(h\) defined in an open rectangle or disc \(\Omega \) such that \(\frac {\partial }{\partial x}g=\frac {\partial }{\partial y}h\) everywhere in \(\Omega \). Show that there is a function \(f\in C^2(\Omega )\) such that \(\frac {\partial f}{\partial y}=g\) and \( \frac {\partial f}{\partial x}=h\).

(Bonus Problem 790) State the definition of a simply connected set and then generalize Problem 780 to any simply connected open set.

(Problem 800) Let \(\Omega =\R ^2\setminus \{(0,0)\}\). Let \(g(x,y)=\frac {x}{x^2+y^2}\) and \(h(x,y)=\frac {-y}{x^2+y^2}\). Show that \(\frac {\partial }{\partial x}g=\frac {\partial }{\partial y}h\).

This is routine calculation. By the quotient rule of undergraduate calculus, \begin {equation*} \p {x}g = \frac {1(x^2+y^2)-x(2x)}{(x^2+y^2)^2} = \frac {y^2-x^2}{(x^2+y^2)^2} \end {equation*} and \begin {equation*} \p {y}h = \frac {-1(x^2+y^2)-(-y)(2y)}{(x^2+y^2)^2} = \frac {y^2-x^2}{(x^2+y^2)^2} \end {equation*} which are equal.

(William, Problem 810) Show that there is no function \(f\in C^1(\Omega )\) such that \(\frac {\partial f}{\partial y}=g\) and \( \frac {\partial f}{\partial x}=h\).

(Wilson, Problem 820) Why doesn’t this contradict Bonus Problem 790?

The domain \(\Omega \) is not simply connected.

(Adam, Problem 830) Suppose that \(u\) is real-valued and harmonic (and not necessarily a polynomial) in an open rectangle or disc \(\Omega \). Show that there is a function \(f\) that is holomorphic in \(\Omega \) such that \(u=\re f\).

Let \(g=\frac {\partial u}{\partial x}\) and let \(h=-\frac {\partial u}{\partial y}\). Then \begin {equation*} \p [g]{x}=\frac {\partial ^2 u}{\partial x^2} = -\frac {\partial ^2 u}{\partial y^2}=\p [h]{y} \end {equation*} by definition of \(g\) and \(h\) and because \(u\) is harmonic. Thus by Bonus Problem 790 there is a \(v:\Omega \to \R \) such that \(\p [v]{y}=g=\frac {\partial u}{\partial x}\) and \(\p [v]{x}=h=-\frac {\partial u}{\partial y}\).

Then \(u\) and \(v\) satisfy the Cauchy-Riemann equations, and so by Problem 650, \(f=u+iv\) is holomorphic in \(\Omega \).

(Amani, Problem 840) Suppose that \(f\) is holomorphic in an open rectangle or disc \(\Omega \). Show that there is a function \(F\) that is holomorphic in \(\Omega \) such that \(f=\frac {\partial F}{\partial z}\). (We call \(F\) the holomorphic antiderivative of \(f\).)

[Chapter 1, Problem 52] The function \(f(z)=1/z\) is holomorphic on \(\Omega =\{z\in \C :1<|z|<2\}\) but has no holomorphic antiderivative on \(\Omega \).

2.1. Real Analysis

(Memory 841) Let \((a,b)\subset \R \) be an open interval, let \(\Omega \subseteq \R ^2\) be open, let \(\gamma :(a,b)\to \Omega \) be \(C^1\), and let \(f:\Omega \to \R \) be \(C^1\). Then \begin {align*} \frac {d}{dt} f(\gamma (t))&= \nabla f(\gamma (t))\cdot \gamma '(t) = \frac {\partial f}{\partial x}\bigg \vert _{\gamma (t)}\times \gamma _1'(t)+ \frac {\partial f}{\partial y}\bigg \vert _{\gamma (t)}\times \gamma _2'(t) \end {align*}

where \(\cdot \) is the dot product in the vector space \(\R ^2\), and \(\gamma _1\), \(\gamma _2:(a,b)\to \R \) are the \(C^1\) functions such that \(\gamma (t)=(\gamma _1(t),\gamma _2(t))\).

(Memory 850) State the Intermediate Value Theorem.

(Memory 860) State the change of variables theorem for integrals over real intervals.

[Definition: Continuous] Let \((X,d)\) and \((Z,\rho )\) be two metric spaces and let \(f:X\to Z\). We say that \(f\) is continuous at \(x\in X\) if, for all \(\varepsilon >0\), there is a \(\delta >0\) such that if \(d(x,y)<\delta \) and \(y\in X\) then \(\rho (f(x),f(y))<\varepsilon \).

(Memory 870) Let \(a<b\) and let \(\varphi :[a,b]\to \R \) be continuous. Then \(\left |\int _a^b \varphi \right |\leq \int _a^b |\varphi |\leq (b-a)\sup _{[a,b]}|\varphi |\).

(Memory 871) Let \(X\) be a compact metric space and let \(f:X\to Z\) be a continuous function. Then \(f\) is uniformly continuous.

(Memory 872) Let \(X\) be a compact metric space and let \(f:X\to \R \) be a continuous function. Then \(f\) attains its maximum and minimum, that is, there are points \(m\) and \(M\) in \(X\) such that \(f(m)\leq f(x)\leq f(M)\) for all \(x\in X\). In particular, \(f\) is bounded on \(X\).

(Problem 873) If \(\varphi :[c,d]\to [a,b]\) is a continuous bijection for \(a\), \(b\), \(c\), \(d\in \R \) with \(a<b\) and \(c<d\), show that either \(\varphi \) is strictly increasing, \(\varphi (c)=a\), and \(\varphi (d)=b\), or \(\varphi \) is strictly decreasing, \(\varphi (c)=b\), and \(\varphi (d)=a\).

Let \(m\), \(M\in [c,d]\) with \(\varphi (m)\leq \varphi (t)\leq \varphi (M)\) for all \(t\in [c,d]\). Such \(m\) and \(M\) exist by Memory 872.

Suppose for the sake of contradiction that \(\varphi (m)<\varphi (c)\) and \(\varphi (m)<\varphi (d)\). Then \(c\neq m\neq d\) and so \(c<m<d\). Let \(y\) satisfy \(\varphi (m)<y<\min (\varphi (c),\varphi (d))\). By the intermediate value theorem, there are points \(t_1\) and \(t_2\) with \(c<t_1<m<t_2<d\) (in particular, \(t_1\neq t_2\)) and \(\varphi (t_1)=y=\varphi (t_2)\). But then \(\varphi \) is not injective. This contradicts our assumption on \(\varphi \), and so either \(\varphi (m)=\varphi (c)\) and so \(m=c\), or \(\varphi (m)=\varphi (d)\) and so \(m=d\).

Similarly, \(M\in \{c,d\}\). Since \(\varphi \) is injective, \(c<d\), and so \([c,d]\) contains more than one point, we have that \(M\neq m\) and so either \(M=c\), \(m=d\) or \(m=c\), \(M=d\).

If \(m=c\), \(M=d\), then \(\varphi (c)<\varphi (t)<\varphi (d)\) for all \(t\in (c,d)\). If \(c\leq t_1<t_2\leq d\) and \(\varphi (t_1)>\varphi (t_2)\), then another application of the intermediate value theorem contradicts injectivity of \(\varphi \), and so \(\varphi \) must be increasing; because \(\varphi \) is injective it must be strictly increasing. Similarly, if \(m=d\), \(M=c\), then \(\varphi \) is strictly decreasing, as desired.

(Memory 880) Let \(X\) be a compact metric space and let \(f:X\to Z\) be a continuous function. Then \(f(X)\) is compact.

(Memory 890) Let \(X\) be a compact metric space and let \(f:X\to Z\) be a continuous bijection. Then \(f^{-1}\) is also continuous.

(Bashar, Problem 900) Is the previous problem true if \(X\) is not compact?

No. Let \(X=(-\pi /2,0)\cup (0,\pi /2]\subset \R \) with the usual metric on \(\R \). Then the function \(\cot :X\to \R \) is continuous on \(X\) and is a bijection, but \(\cot ^{-1}(0)=\frac {\pi }{2}\) and \(\lim _{x\to 0^-}\cot ^{-1}(x)=-\frac {\pi }{2}\), and so \(\cot ^{-1}\) (with the given range) is discontinuous at \(0\).

(Memory 910) If \(\gamma :X\to \R ^2\) and \(\gamma (t)=(\gamma _1(t),\gamma _2(t))\) for all \(t\in X\), then \(\gamma \) is continuous if and only if \(\gamma _1\) and \(\gamma _2\) are continuous.

Definition 2.1.1. (\(C^1\) on a closed set.) Let \([a,b]\subseteq \R \) be a closed bounded interval and let \(f:[a,b]\to \R \). We say that \(f\in C^1([a,b])\), or \(f\) is continuously differentiable on \([a,b]\), if

(a)

\(f\) is continuous on \([a,b]\),

(b)

\(f\) is differentiable on \((a,b)\),

(c)

The derivative \(f'\) is continuous on \((a,b)\),

(d)

\(\lim _{t\to a^+} f'(t)\) and \(\lim _{t\to b^-} f'(t)\) both exist and are finite.

(Memory 920) If the conditions (a), (b) and (c) hold, then the condition (d) holds if and only if the two limits \(\lim _{t\to a^+} \frac {f(t)-f(a)}{t-a}\) and \(\lim _{t\to b^-} \frac {f(b)-f(t)}{b-t}\) exist, and in this case \(\lim _{t\to a^+} \frac {f(t)-f(a)}{t-a}=\lim _{t\to a^+} f'(t)\) and \(\lim _{t\to b^-} \frac {f(b)-f(t)}{b-t}=\lim _{t\to b^-} f'(t)\).

[Definition: One-sided derivative] If \(f:[a,b]\to \R \), we define \(f'(a)=\lim _{t\to a^+} \frac {f(t)-f(a)}{t-a}\) and \(f'(b)=\lim _{t\to b^-} \frac {f(b)-f(t)}{b-t}\), if these limits exist.

[Definition: Curve] A curve in \(\R ^2\) is a continuous function \(\gamma :[a,b]\to \C \), where \([a,b]\subseteq \R \) is a closed and bounded interval. The trace (or image) of \(\gamma \) is \(\widetilde \gamma =\gamma ([a,b])=\{\gamma (t):t\in [a,b]\}\).

[Definition: Closed; simple] A curve \(\gamma :[a,b]\to \R ^2\) is closed if \(\gamma (a)=\gamma (b)\). A closed curve is simple if \(\gamma (b)=\gamma (a)\) and \(\gamma \) is injective on \([a,b)\) (equivalently on \((a,b]\)).

[Definition: \(C^1\) curve in \(\R ^2\)] A curve \(\gamma :[a,b]\to \R ^2\) is \(C^1\) (or continuously differentiable) if \(\gamma (t)=(\gamma _1(t),\gamma _2(t))\) for all \(t\in [a,b]\) and both \(\gamma _1\), \(\gamma _2\) are \(C^1\) on \([a,b]\). We write \begin {equation*} \gamma '(t)=\frac {d\gamma }{dt}=\biggl (\frac {d\gamma _1}{dt},\frac {d\gamma _2}{dt}\biggr ). \end {equation*}

[Definition: Arc length] If \(\gamma :[a,b]\to \R ^2\) is a \(C^1\) curve, then its length (or arc length) is \(\int _a^b \|\gamma '(t)\|\,dt\).

Proposition 2.1.4. Let \(\gamma \in C^1([a,b])\), \(\gamma :[a,b]\to \Omega \) for some open set \(\Omega \subseteq \R ^2\) and let \(f:\Omega \to \R \) with \(f\in C^1(\Omega )\). Then \begin {equation*} f(\gamma (b))-f(\gamma (a))=\int _a^b \frac {\partial f}{\partial x}\Big \vert _{(x,y)=\gamma (t)} \frac {\partial \gamma _1}{\partial t} +\frac {\partial f}{\partial y}\Big \vert _{(x,y)=\gamma (t)} \frac {\partial \gamma _2}{\partial t}\,dt. \end {equation*}

(Dibyendu, Problem 930) Prove Proposition 2.1.4. Hint: Start by computing \(\frac {d(f\circ \gamma )}{dt}\).

By the multivariable chain rule, \begin {equation*} \frac {d(f\circ \gamma )}{dt}= \frac {\partial f}{\partial x}\Big \vert _{(x,y)=\gamma (t)}\frac {\partial \gamma _1}{\partial t} +\frac {\partial f}{\partial y}\Big \vert _{(x,y)=\gamma (t)}\frac {\partial \gamma _2}{\partial t}. \end {equation*} The result then follows from the fundamental theorem of calculus.

[Definition: Real line integral] Let \(\gamma \in C^1([a,b])\), \(\gamma :[a,b]\to \Omega \) for some open set \(\Omega \subseteq \R ^2\) and \(F:\Omega \to \R \) be continuous on \(\Omega \). We define \begin {equation*} \int _\gamma F\,ds=\int _a^b F(\gamma (t))\,\|\gamma '(t)\|\,dt. \end {equation*} Let \(\vec F:\Omega \to \R ^2\) be continuous on \(\Omega \). We define \begin {equation*} \int _\gamma \vec F\cdot \tau \,ds=\int _a^b \vec F(\gamma (t))\cdot \gamma '(t)\,dt \end {equation*} where we use a dot product in the second integral.

2.1. Real and Complex Line Integrals

Definition 2.1.3. (Integral of a complex function.) If \(f:[a,b]\to \C \), and both \(\re f\) and \(\im f\) are integrable on \([a,b]\), we define \(\int _a^b f=\int _a^b\re f+i\int _a^b \im f\).

(Problem 931) If \(\alpha \), \(\beta \in \C \) are constants and \(f\), \(g:[a,b]\to \C \) are both continuous, show that \(\int _a^b(\alpha f+\beta g)=\alpha \int _a^b f+\beta \int _a^b g\).

Proposition 2.1.7. Suppose that \(a<b\) and that \(f:[a,b]\to \C \) is continuous. Then \(|\int _a^b f|\leq \int _a^b |f|\leq (b-a)\sup _{[a,b]}|f|\).

(Hope, Problem 940) Prove Proposition 2.1.7. Hint: Start by showing that the integral is finite.

First, \begin {align*} \biggl |\int _a^b f\biggr | &\leq \biggl |\re \int _a^b f\biggr |+\biggl |\im \int _a^b f\biggr | = \biggl |\int _a^b \re f\biggr |+\biggl |\int _a^b \im f\biggr |. \end {align*}

By continuity of \(\re f\) and \(\im f\) and compactness of \([a,b]\), we have that \(\sup _{[a,b]}|\re f|<\infty \) and \(\sup _{[a,b]}|\im f|<\infty \) and so the integral is finite by Memory 870.

If \(\int _a^b f=0\) we are done. Otherwise, let \(\theta \in \R \) be such that \(e^{i\theta }\int _a^b f\) is a nonnegative real number. Then \(|\int _a^b f|=\int e^{i\theta } f\). But \(|\int _a^b f|\) is real and so \(\im \int e^{i\theta } f=0\). Therefore \(|\int _a^b f|=\re \int e^{i\theta } f =\int \re (e^{i\theta } f)\). The result then follows from the corresponding result for real integrals.

Definition 2.1.4. (\(C^1\) curve in \(\C \).) A curve \(\gamma :[a,b]\to \C \) is a \(C^1\) curve (in \(\C \)) if \((\re \gamma ,\im \gamma )\) is a \(C^1\) curve (in \(\R ^2\)). We write \begin {equation*} \gamma '(t)=\frac {d\gamma }{dt}=\frac {d(\re \gamma )}{dt}+i\frac {d(\im \gamma )}{dt}. \end {equation*}

(Micah, Problem 950) If \(t\in (a,b)\) and \(\gamma :[a,b]\to \C \) is \(C^1\), show that \(\gamma '(t)=\lim _{s\to t}\frac {\gamma (s)-\gamma (t)}{s-t}\).

We compute that \begin {align*} \lim _{s\to t}\frac {\gamma (s)-\gamma (t)}{s-t} &=\lim _{s\to t}\biggl ( \frac {\re \gamma (s)-\re \gamma (t)}{s-t} +i\frac {\im \gamma (s)-\im \gamma (t)}{s-t}\biggr ). \end {align*}

By linearity of limits \begin {align*} \lim _{s\to t}\frac {\gamma (s)-\gamma (t)}{s-t} &=\left ( \lim _{s\to t}\frac {\re \gamma (s)-\re \gamma (t)}{s-t}\right )+i\left (\lim _{s\to t}\frac {\im \gamma (s)-\im \gamma (t)}{s-t}\right ) = (\re \gamma )'(t)+i(\im \gamma )'(t) \end {align*}

as desired.

Definition 2.1.5. (Complex line integral.) Let \(\gamma \in C^1([a,b])\), \(\gamma :[a,b]\to \Omega \) for some \(\Omega \subseteq \C \), and let \(F:\Omega \to \C \) be continuous on \(\Omega \). We define \begin {equation*} \oint _\gamma F(z)\,dz=\int _a^b F(\gamma (t))\,\gamma '(t)\,dt \end {equation*} where we use complex multiplication in the second integral.

(Problem 951) Let \(a<b\) be real numbers and let \(f:[a,b]\to \C \) be a continuous function.

(a)

Let \(\gamma _1:[a,b]\to \C \) be given by \(\gamma _1(t)=t\). Let \(\Omega =\widetilde \gamma _1\) and let \(F:\Omega \to \C \) be given by \(F(z)=f(z)\). Show that \(\oint _{\gamma _1} F(z)\,dz=\int _a^b f\).

(b)

Let \(\gamma _2:[a,b]\to \C \) be given by \(\gamma _1(t)=t+i\). Let \(\Omega =\widetilde \gamma _2\) and let \(F:\Omega \to \C \) be given by \(F(t+i)=f(t)\). Show that \(\oint _{\gamma _1} F(z)\,dz=\int _a^b f\).

(c)

Let \(\gamma _3:[a,b]\to \C \) be given by \(\gamma _3(t)=it\). Let \(\Omega =\widetilde \gamma _2\) and let \(F:\Omega \to \C \) be given by \(F(it)=f(t)\). Show that \(\oint _{\gamma _3} F(z)\,dz=i\int _a^b f\).

(Muhammad, Problem 960) Let \( \gamma :[0,1]\to \Omega \subset \R ^2\) be a \(C^1\) curve and let \(\vec F:\Omega \to \R ^2\) be a vector-valued function. Recall that we identify \(\R ^2\) with \(\C \), so that we identify \( \gamma =(\gamma _1,\gamma _2)\) with \(\gamma _1+i\gamma _2\) and \(\vec F=(F_1,F_2)\) with \(F=F_1+iF_2\).

Show that \begin {equation*} \int _\gamma \vec F\cdot \tau \,ds = \re \oint _\gamma \overline {F}(z)\,dz, \qquad \int _\gamma \vec F\cdot \nu \,ds = \im \oint _\gamma \overline {F}(z)\,dz, \end {equation*} where \(\nu =\begin {pmatrix}0&1\\-1&0\end {pmatrix}\tau \) is the unit rightward normal vector to \(\gamma \).

Proposition 2.1.6. Let \(\gamma :[a,b]\to \Omega \subseteq \C \) be \(C^1\), where \(\Omega \) is open, and let \(f\) be holomorphic in \(\Omega \). Then \begin {equation*} \oint _\gamma \frac {\partial f}{\partial z}\,dz=f(\gamma (b))-f(\gamma (a)). \end {equation*}

(Problem 961) Give an example of a \(C^1\) curve \(\gamma \) and a \(C^1\) function \(f\) defined in an open neighborhood of \(\widetilde \gamma \) such that \begin {equation*} \oint _\gamma \frac {\partial f}{\partial z}\,dz\neq f(\gamma (b))-f(\gamma (a)). \end {equation*}

Let \(f(z)=\overline {z}\). Then \(\frac {\partial f}{\partial z}=0\) and so the left hand side is zero, but \(\overline z\) is not a constant and so the right hand side is not zero unless \(\gamma \) is closed; for example, if \(\gamma (t)=t\), \(a=0\), \(b=1\), then \(f(\gamma (b))-f(\gamma (a))=1\neq 0\).

(Nisa, Problem 970) Prove Proposition 2.1.6.

By the fundamental theorem of calculus, we have that \begin {align*} f(\gamma (b))-f(\gamma (a)) &=\bigl [\re f(\gamma (b))-\re f(\gamma (a))\bigr ] + i\bigl [\im f(\gamma (b))-\im f(\gamma (a))\bigr ] \\&= \int _a^b \frac {d}{dt} \re f(\gamma (t)) \,dt + i \int _a^b \frac {d}{dt} \im f(\gamma (t)) \,dt \\&= \int _a^b \frac {d}{dt} \re f(\gamma (t)) + i \frac {d}{dt} \im f(\gamma (t)) \,dt \\&= \int _a^b \frac {d}{dt} f(\gamma (t)) \,dt . \end {align*}

By Memory 841 (with \(\gamma '(t)\) viewed as a vector in \(\R ^2\)) \begin {align*} \frac {d}{dt} \re f(\gamma (t)) &= \nabla (\re f)(\gamma (t)) \cdot \gamma '(t) \,dt . \end {align*}

Let \(\gamma (t)=\gamma _1(t)+i\gamma _2(t)=(\gamma _1(t),\gamma _2(t))\) where \(\gamma _1\), \(\gamma _2\) are real valued functions. Then \begin {align*} \frac {d}{dt} \re f(\gamma (t)) &= \p [(\re f)]{x}(\gamma (t))\,\gamma _1'(t) +\p [(\re f)]{y}(\gamma (t))\,\gamma _2'(t) . \end {align*}

Similarly \begin {align*} \frac {d}{dt} \im f(\gamma (t)) &= \p [(\im f)]{x}(\gamma (t))\,\gamma _1'(t) +\p [(\im f)]{y}(\gamma (t))\,\gamma _2'(t) \end {align*}

and so \begin {align*} \frac {d}{dt} f(\gamma (t)) &= \frac {d}{dt} \re f(\gamma (t)) + i \frac {d}{dt} \im f(\gamma (t)) \\&= \p [ f]{x}(\gamma (t))\,\gamma _1'(t) +\p [f]{y}(\gamma (t))\,\gamma _2'(t) . \end {align*}

By Problem 660, and because \(f\) is holomorphic, we have that \begin {align*} \frac {d}{dt} f(\gamma (t)) &= \p [ f]{z}(\gamma (t))\,\gamma _1'(t) +i\p [f]{z}(\gamma (t))\,\gamma _2'(t) \\&=\p [ f]{z}(\gamma (t))\,\gamma '(t) . \end {align*}

Thus \begin {align*} f(\gamma (b))-f(\gamma (a)) &= \int _a^b \frac {d}{dt} f(\gamma (t)) \,dt \\&= \int _a^b \p [ f]{z}(\gamma (t))\,\gamma '(t) \,dt . \end {align*}

By definition \begin {equation*} \oint _\gamma \frac {\partial f}{\partial z}\,dz = \int _a^b \p [f]{z}(\gamma (t))\,\gamma '(t)\,dt. \end {equation*} This completes the proof.

Proposition 2.1.8. If \(\gamma :[a,b]\to \Omega \subseteq \C \) is a \(C^1\) curve and \(f:\Omega \to \C \) is continuous, then \(\displaystyle \left |\oint _\gamma f(z)\,dz\right |\leq \sup _{[a,b]} |f\circ \gamma | \cdot \ell (\gamma )=\sup _{\widetilde \gamma } |f| \cdot \ell (\gamma )\), where \(\ell (\gamma )=\int _a^b |\gamma '|\).

(Robert, Problem 980) Prove Proposition 2.1.8.

By definition \begin {equation*} \oint _\gamma f(z)\,dz = \int _a^b f(\gamma (t))\,\gamma '(t)\,dt. \end {equation*} By Problem 940 \begin {align*} \biggl |\oint _\gamma f(z)\,dz\biggr | &\leq \int _a^b |f(\gamma (t))|\,|\gamma '(t)|\,dt \leq \sup _{[a,b]} |f\circ \gamma | \int _a^b |\gamma '(t)|\,dt. \end {align*}

Recalling the definition of arc length completes the proof.

Proposition 2.1.9. Let \(\Omega \subseteq \C \), let \(F:\Omega \to \C \) be continuous, let \(\gamma _1:[a,b]\to \Omega \) be a \(C^1\) curve, and let \(\varphi :[c,d]\to [a,b]\) be \(C^1\) and bijective. Define \(\gamma _2=\gamma _1\circ \varphi \).

If \(\varphi \) is strictly increasing, then \(\oint _{\gamma _1} F(z)\,dz=\oint _{\gamma _2} F(z)\,dz\).

(Sam, Problem 990) In this problem we begin the proof of Proposition 2.1.9. Let \(\gamma _1:[a,b]\to \C \) be a \(C^1\) curve. Let \(\varphi :[c,d]\to [a,b]\) be a continuous bijection. Define \(\gamma _2=\gamma _1\circ \varphi \). Compute \(\gamma _2'(t)\) in terms of \(\gamma _1\), \(\gamma _1'\), \(\varphi \), and \(\varphi '\). Then show that \(\widetilde \gamma _1=\widetilde \gamma _2\). (Recall \(\widetilde \gamma \) denotes the image of \(\gamma \).)

(William, Problem 1000) Prove Proposition 2.1.9. Do not assume that \(F\) is holomorphic.

(Problem 1010) Let \(\gamma _1:[-a,a]\to \C \). Let \(\gamma _2:[-a,a]\to \C \) be given by \(\gamma _2(t)=\gamma _1(-t)\). Show that if \(F\) is continuous in a neighborhood of \(\widetilde \gamma _1\), then \(\oint _{\gamma _1} F(z)\,dz=-\oint _{\gamma _2} F(z)\,dz\).

(Problem 1011) Suppose that \(\gamma _1:[a,b]\to \Omega \subseteq \C \) is a simple closed \(C^1\) curve, and that \(\gamma _1'(a)=\gamma _1'(b)\). If \(z_0\in \widetilde \gamma _1\), show that there is a simple closed curve \(\gamma _2:[0,1]\to \C \) such that \(\gamma _2(c)=\gamma _2(d)=z_0\), \(\widetilde \gamma _2=\widetilde \gamma _1\), and \(\oint _{\gamma _1} F(z)\,dz=\oint _{\gamma _2} F(z)\,dz\) for every \(F\) continuous on \(\Omega \).

(Problem 1012) Prove that the previous problem still holds even if \(\gamma _1'(a)\neq \gamma _1'(b)\).

(Bonus Problem 1013) If \(\gamma _1:[a,b]\to \C \) is injective, and \(\gamma _2:[a,b]\to \C \) is a simple closed curve, then \(\widetilde \gamma _1\neq \widetilde \gamma _2\).

(Proposition 1014) Given a set in \(C^1\) that can be written as the image of a \(C^1\) curve, and an orientation of that set, any two parameterizations of the set will yield the same line integrals.

More precisely, let \(\widetilde \gamma \subseteq \Omega \subseteq \C \). Suppose that there is at least one injective or simple closed \(C^1\) curve \(\gamma _1:[a,b]\to \Omega \) such that \(\widetilde \gamma _1=\widetilde \gamma \).

If \(\gamma _2:[c,d]\to \Omega \) is any other injective or simple closed \(C^1\) curve with \(\widetilde \gamma _2=\widetilde \gamma \), then either \(\oint _{\gamma _1} F(z)\,dz=\oint _{\gamma _2} F(z)\,dz\) for all functions \(F:\Omega \to \C \) continuous, or \(\oint _{\gamma _1} F(z)\,dz=-\oint _{\gamma _2} F(z)\,dz\) for all functions \(F:\Omega \to \C \) continuous.

Furthermore, if \(\gamma _1\) and \(\gamma _2\) are injective, if \(z_1\), \(z_2\in \widetilde \gamma \), and if \(\gamma _1^{-1}(z_1)<\gamma _1^{-1}(z_2)\) and \(\gamma _2^{-1}(z_1)<\gamma _2^{-1}(z_2)\), then \(\oint _{\gamma _1} F(z)\,dz=\oint _{\gamma _2} F(z)\,dz\) for all functions \(F:\Omega \to \C \) continuous.

Finally, if \(\gamma _1\) and \(\gamma _2\) are simple closed curves, if \(z_1\), \(z_2\), \(z_3\in \widetilde \gamma \), if \(\gamma _1(a)=\gamma _1(b)=\gamma _2(c)=\gamma _2(d)=z_3\), and if \(\gamma _1^{-1}(z_1)<\gamma _1^{-1}(z_2)\) and \(\gamma _2^{-1}(z_1)<\gamma _2^{-1}(z_2)\), then \(\oint _{\gamma _1} F(z)\,dz=\oint _{\gamma _2} F(z)\,dz\) for all functions \(F:\Omega \to \C \) continuous.

(Wilson, Problem 1020) Let \(\gamma _1:[a,b]\to \C \) and \(\gamma _2:[c,d]\to \C \) be two curves. Suppose further that \(\widetilde \gamma _1=\widetilde \gamma _2\), \(\gamma _1(a)=\gamma _2(c)\), \(\gamma _1(b)=\gamma _2(d)\), and that \(\gamma _1\) and \(\gamma _2\) are injective. Show that there is a continuous strictly increasing function \(\varphi :[c,d]\to [a,b]\) such that \(\gamma _2=\gamma _1\circ \varphi \).

By definition \(\gamma _1:[a,b]\to \widetilde \gamma _1\) is a bijection. Memory 880 and Memory 890, we have that the inverse \(\gamma _1^{-1}\) is also continuous. Thus \(\varphi =\gamma _1^{-1}\circ \gamma _2\) is a continuous bijection from \([c,d]\) to \([a,b]\) that satisfies \(\gamma _2=\gamma _1\circ \varphi \). By Problem 873, and because \(\varphi (c)=\gamma _1^{-1}(\gamma _2(c))=a\), we have that \(\varphi \) is strictly increasing.

(Adam, Problem 1030) If \(\gamma _1:[a,b]\to \C \) and \(\gamma _2:[c,d]\to \C \) are simple closed curves rather than injective functions, with \(\widetilde \gamma _1=\widetilde \gamma _2\) and \(\gamma _1(a)=\gamma _1(b)=\gamma _2(c)=\gamma _2(d)\), is it necessarily the case that \(\gamma _2=\gamma _1\circ \varphi \) for a continuous strictly increasing function \(\varphi :[c,d]\to [a,b]\)?

No. Let \(\gamma _1(t)=e^{it}\) and let \(\gamma _2=e^{-it}\), both with domain \([-\pi ,\pi ]\). If \(\gamma _2(t)=\gamma _1(\varphi (t))\), then \(\varphi (t)=-t\) for all \(t\in (-\pi ,\pi )\), and so in particular \(\varphi \) cannot be strictly increasing.

(Bonus Problem 1040) Prove Proposition 1014.

(Amani, Problem 1050) Let \(\gamma _1:[a,b]\to \C \) and \(\gamma _2:[c,d]\to \C \) be two \(C^1\) curves. Suppose that \(\gamma _1(b)=\gamma _2(c)\). Show that there is a \(C^1\) curve \(\gamma _3:[-1,1]\to \C \) such that \(\gamma _3\big \vert _{[-1,0]}\) is a reparameterization of \(\gamma _1\) and \(\gamma _3\big \vert _{[0,1]}\) is a reparameterization of \(\gamma _2\). (We will write \(\gamma _3=\gamma _1*\gamma _2\). This means that \(\widetilde \gamma _3=\widetilde \gamma _1\cup \widetilde \gamma _2\) and \(\oint _{\gamma _3} F(z)\,dz=\oint _{\gamma _1} F(z)\,dz+\oint _{\gamma _2} F(z)\,dz\) for all \(F\) continuous in a neighborhood of \(\widetilde \gamma _3\).)

The function \begin {equation*} \gamma _3(t)=\begin {cases} \gamma _1(b+(b-a)t^3), & -1\leq t\leq 0,\\ \gamma _2(c+(d-c)t^3),& 0\leq t\leq 1\end {cases} \end {equation*} satisfies the given conditions. Note in particular that \(\gamma _3'(0)=0\).

Addendum: Change of variables

(Bashar, Problem 1060) Let \(\Omega \), \(W\subseteq \C \) be open and let \(u:\Omega \to W\) be holomorphic. Let \(\gamma :[0,1]\to \Omega \) be a \(C^1\) closed curve. Let \(f:W\to \C \) be continuous. Show that \begin {equation*} \oint _{u\circ \gamma } f(w)\,dw = \oint _\gamma f(u(z))\,\frac {\partial u}{\partial z}\,dz. \end {equation*}

By definition \begin {equation*} \oint _{u\circ \gamma } f(w)\,dw=\int _0^1 f(u\circ \gamma (t)) \, (u\circ \gamma )'(t)\,dt, \qquad \oint _\gamma f(u(z))\,\frac {\partial u}{\partial z}\,dz =\int _0^1 f(u(\gamma (t))) \,\frac {\partial u}{\partial z}\big \vert _{z=\gamma (t)} \,\gamma '(t)\,dt. \end {equation*} As in the proof of Problem 970, because \(u\) is holomorphic we have that \((u\circ \gamma )'(t)=\frac {\partial u}{\partial z}\big \vert _{z=\gamma (t)} \,\gamma '(t)\). This completes the proof.

(Dibyendu, Problem 1070) I want to compute \(\int _{-1}^1 \frac {(t+i)^3}{(t+i)^4+1}dt\). A naïve student uses the \(u\)-substitution \(u=(t+i)^4\) and converts the integral to \(\int _{-4}^{-4} \frac {1}{4} \frac {1}{u+1} du=0\). But when I compute \(\int _{-1}^1 \frac {(t+i)^3}{(t+i)^4+1}dt\) using a numerical solver, I get \(-i\pi /2\). What went wrong?

2.2. Real Analysis

[Definition: Limit in metric spaces] If \((X,d)\) and \((Z,\rho )\) are metric spaces, \(p\in Z\), and \(f:Z\setminus \{p\}\to X\), we say that \(\lim _{z\to p}f(z)=\ell \) if, for all \(\varepsilon >0\), there is a \(\delta >0\) such that if \(z\in Z\) and \(0<\rho (z,p)<\delta \), then \(d(f(z),f(p))<\varepsilon \).

[Definition: Continuous function on metric spaces] If \((X,d)\) and \((Z,\rho )\) are metric spaces and \(f:Z\to X\), we say that \(f\) is continuous at \(p\in Z\) if \(f(p)=\lim _{z\to p} f(z)\).

(Memory 1071) If \(\Omega \subset \R ^d\) is open, \(p\in \Omega \), and \(f:\Omega \to \R \) is a \(C^1\) function, then \begin {equation*} \lim _{x\to p} \frac {1}{\|x-p\|}\left | f(x)-f(p) - \nabla f(p)\cdot (x-p)\right |=0 \end {equation*} where \(\cdot \) denotes the dot product and \(\nabla f(p)\) is the vector of partial derivatives \((\frac {\partial f}{\partial x_1},\p [f]{x_2},\dots ,\p [f]{x_d})\) evaluated at the point \(p\).

2.2. Complex Differentiability and Conformality

[Definition: Disc] The open disc (or ball) in \(\C \) of radius \(r\) and center \(p\) is \(D(p,r)=B(p,r)=\{z\in \C :|z-p|<r\}\). The closed disc (or ball) in \(\C \) of radius \(r\) and center \(p\) is \(\overline D(p,r)=\overline B(p,r)=\{z\in \C :|z-p|\leq r\}\).

(Hope, Problem 1080) Let \(\gamma :[0,1]\to \C \) be a (parameterization of a) nondegenerate scalene triangle of your choice. You don’t need to find a formula for \(\gamma (t)\). Sketch the trace of \(\gamma \) and of \(f\circ \gamma \) for the following choices of \(f\):

(a)

\(f(z)=z-3+i\)

(b)

\(f(z)=(\frac {1}{2}+\frac {\sqrt {3}}{2}i)z\)

(c)

\(f(z)=2z\)

(d)

\(f(z)=(1+i)z\)

(e)

\(f(z)=(1+i)z-3+i\)

(f)

\(f(z)=\bar z\)

(g)

\(f(z)=z+2\bar z\)

We chose \(\gamma \) to be a parameterization of:

Then:

(a)

\(f(z)=z-3+i\)

(b)

\(f(z)=(\frac {1}{2}+\frac {\sqrt {3}}{2}i)z\)

(c)

\(f(z)=2z\)

(d)

\(f(z)=(1+i)z\)

(e)

\(f(z)=(1+i)z-3+i\)

(f)

\(f(z)=\bar z\)

(g)

\(f(z)=z+2\bar z\)

[Definition: Complex derivative] Let \(p\in \Omega \subseteq \C \), where \(\Omega \) is open. Let \(f:\Omega \to \C \). Suppose that \(\lim _{z\to p} \frac {f(z)-f(p)}{z-p}\) exists. Then we say that \(f\) has a complex derivative at \(p\) and write \(f'(p)=\lim _{z\to p} \frac {f(z)-f(p)}{z-p}\).

(Problem 1081) Show that \(f(z)=\overline z\) does not have a complex derivative.

(Problem 1082) Show that if \(f\) is constant then \(f'(p)=0\) for all \(p\).

(Problem 1083) Show that if \(f(z)=z\) then \(f'(p)=1\) for all \(p\).

(Fact 1090) If \(\Omega \subseteq \C \) is open, \(p\in \Omega \), and \(f\), \(g:\Omega \setminus \{p\}\to \C \) are such that \(\lim _{z\to p} f(z)\) and \(\lim _{z\to p} g(z)\) exist (as complex numbers), then we have the usual formulas \begin {gather*} \lim _{z\to p} \bigl (f(z)+g(z)\bigr )=\bigl (\lim _{z\to p} f(z)\bigr ) +\bigl (\lim _{z\to p} g(z)\bigr ), \\ \lim _{z\to p} \bigl (f(z)-g(z)\bigr )=\bigl (\lim _{z\to p} f(z)\bigr )-\bigl (\lim _{z\to p} g(z)\bigr ), \\ \lim _{z\to p} \bigl (f(z)g(z)\bigr )=\bigl (\lim _{z\to p} f(z)\bigr )\bigl (\lim _{z\to p} g(z)\bigr ) \end {gather*} and (if \(\lim _{z\to p} g(z)\neq 0\)) \begin {equation*} \lim _{z\to p} \frac {f(z)}{g(z)}=\frac {\lim _{z\to p} f(z)}{\lim _{z\to p} g(z)}. \end {equation*}

[Chapter 2, Problem 10] If \(f\) has a complex derivative at \(p\), then \(f\) is continuous at \(p\).

[Chapter 2, Problem 13] If \(\Omega \subseteq \C \) is open, \(p\in \Omega \), and \(f\), \(g:\Omega \to \C \) are functions that have complex derivatives at \(p\), and if \(\alpha \), \(\beta \in \C \), then \(\alpha f+\beta g\) and \(fg\) have complex derivatives at \(p\) and \begin {align*} (\alpha f+\beta g)'(p)&=\alpha f'(p)+\beta g'(p),\\ (fg)'(p)&=f'(p)g(p)+f(p)g'(p). \end {align*}

If in addition \(g(p)\neq 0\) then \(f/g\) has a complex derivative at \(p\) and \begin {equation*} (f/g)'(p)=\frac {f'(p)g(p)-f(p)g'(p)}{(g(p))^2}. \end {equation*}

(Fact 1100) If \(\Omega \subseteq \C \) and \(W\subseteq \C \) are open, \(p\in \Omega \), \(f:\Omega \setminus \{p\}\to W\) is such that \(L=\lim _{z\to p} f(z)\) exists, \(L\in W\), and \(g:W\to \C \) is continuous at \(L\), then \begin {equation*} \lim _{z\to p} g(f(z))=g(L). \end {equation*} Observe that we do require \(g(L)\) to exist, not only \(\lim _{w\to L} g(w)\).

Theorem 2.2.2. Suppose that \(f\) has a complex derivative at \(p\). Then \(\frac {\partial f}{\partial z}\big \vert _{z=p}=f'(p)\).

(Problem 1110) Suppose that \(f\) has a complex derivative at \(p\). Prove Theorem 2.2.2 and also show that \(\frac {\partial f}{\partial \bar z}\big \vert _{z=p}=0\).

First, observe that \begin {equation*} \frac {\partial f}{\partial x}\Big \vert _{x+iy=p} = \lim _{\substack {s\to 0\\s\in \R }} \frac {f(p+s)-f(p)}{s}. \end {equation*} Because \(f'(p)=\lim _{z\to p}\frac {f(z)-f(p)}{z-p}\), by definition of limit, for every \(\varepsilon >0\) there is a \(\delta >0\) such that if \(z\in \C \) and \(0<|z|<\delta \) then \(\left |\frac {f(p+z)-f(p)}{z}-f'(p)\right |<\varepsilon \). This in particular is true if \(s\) is real and \(0<|s|<\delta \), as real numbers are complex numbers. Thus we must have that \(f'(p)=\frac {\partial f}{\partial x}\big \vert _{x+iy=p}\).

Similarly, \begin {equation*} \frac {\partial f}{\partial y}\Big \vert _{x+iy=p} = \lim _{\substack {s\to 0\\s\in \R }} \frac {f(p+is)-f(p)}{s} = i\lim _{\substack {s\to 0\\s\in \R }} \frac {f(p+is)-f(p)}{is}=if'(p) . \end {equation*} We thus compute \begin {equation*} \frac {\partial f}{\partial z}\bigg \vert _{z=p} =\frac {1}{2}\biggl (\p [f]{x}\bigg \vert _{x+iy=p}+\frac {1}{i} \p [f]{y}\bigg \vert _{x+iy=p}\biggr ) =f'(p) \end {equation*} and \begin {equation*} \frac {\partial f}{\partial \bar z}\bigg \vert _{z=p} =\frac {1}{2}\biggl (\p [f]{x}\bigg \vert _{x+iy=p}-\frac {1}{i} \p [f]{y}\bigg \vert _{x+iy=p}\biggr ) =0. \end {equation*}

(Micah, Problem 1120) Let \(\Omega \subseteq \C \) be open and let \(f:\Omega \to \C \) be continuous. Let \(W\subseteq \C \) be open. Let \(g:W\to \Omega \) be continuous. Then \(f\circ g:W\to \C \) is continuous. Suppose that \(z_0\in W\) and that \(g'(z_0)\) and \(f'(g(z_0))\) exist (in the sense of limits as above). Show that \((f\circ g)'(z_0)\) exists and that \((f\circ g)'(z_0)=f'(g(z_0))\,g'(z_0)\).

Let \begin {equation*} F(z)=\begin {cases} \frac {f(z)-f(g(t_0))}{z-g(t_0)},& z\in \Omega \setminus \{g(t_0)\},\\ f'(g(t_0)),&z=g(t_0).\end {cases} \end {equation*} Because \(f'(g(t_0))\) exists, \(\lim _{z\to g(t_0)} F(z)=F(g(t_0))\).

Now, \begin {align*} \lim _{s \to t_0} \frac {f\circ g(s)-f\circ g(t_0)}{s-t_0} &=\lim _{s \to t_0} F(g(s)) \frac {g(s)-g(t_0)}{s-t_0} \\&=f'(g(t_0))\,g'(t_0). \end {align*}

Theorem 2.2.1. (Generalization.) Suppose that \(\Omega \subseteq \C \) is open and that \(f\) is \(C^1\) on \(\Omega \). Let \(p\in \Omega \) and suppose \(\p [f]{\bar z}\big \vert _{z=p}=0\). Then \(f\) has a complex derivative at \(p\) and \(f'(p)=\frac {\partial f}{\partial z}\big \vert _{z=p}\).

(Muhammad, Problem 1130) Prove this generalization of Theorem 2.2.1.

(Nisa, Problem 1140) Let \(F:\R ^2\to \R ^2\). Suppose that \(\nabla F_1\) and \(\nabla F_2\) are constants. Show that \(F(x,y)=F(0,0)+(\partial _1 F_1,\partial _1 F_2)x+(\partial _2 F_1,\partial _2 F_2)y\) for all \((x,y)\in \R ^2\).

We compute that (for \(j=1\) or \(j=2\)) \begin {align*} F_j(x,y)&= F_j(x,y)-F_j(x,0)+F_j(x,0)-F_j(0,0)+F_j(0,0) \\&= F_j(0,0)+\int _0^x \frac {\partial }{\partial s} F_j(s,0)\,ds + \int _0^y \p {t} F_j(x,t)\,dt. \end {align*}

Because the integrands are constants, they may easily be evaluated to yield the desired result.

(Robert, Problem 1150) Suppose that \(f:\C \to \C \). Suppose that \(f'\) exists everywhere and is a constant. Show that \(f(z)=f(0)+f'(0)z\) for all \(z\in \C \). Conclude that if \(z\), \(\omega \), \(w\in \C \) with \(\omega \neq z\neq w\), then \(\frac {|f(\omega )-f(z)|}{|\omega -z|}=\frac {|f(w)-f(z)|}{|w-z|}\).

Define \(\gamma :[0,1]\to \Omega \) by \(\gamma (t)=p+t(z-p)\). Then by Proposition 2.1.6 (Problem 970), we have that \begin {equation*} f(\zeta )-f(p)=\oint _{\gamma } \frac {\partial f}{\partial z}\,dz. \end {equation*} By Theorem 2.2.2 (Problem 1110), \(\frac {\partial f}{\partial z}=f'(z)=f'(0)\) because \(f'\) is constant. The result follows by definition of line integral.

(Sam, Problem 1160) Let \(F:\R ^2\to \R ^2\). Let \(F=(F_1,F_2)\). Suppose that \(\nabla F_1\) and \(\nabla F_2\) exist everywhere and are constant. Show that if \(C\) is a circle, then \(F(C)\) is an ellipse, and if \(S\) is a square, show that \(F(S)\) is a parallelogram.

(Problem 1161) Let \(f:\C \to \C \). Suppose that \(f'\) exists everywhere and is constant. Show that if \(C\) is a circle, then \(F(C)\) is also a circle, and if \(S\) is a square, show that \(F(S)\) is also a square.

Theorem 2.2.3.1. Let \(z_0\in \Omega \subseteq \C \) for some open set \(\Omega \). Let \(f:\Omega \to \C \). Let \(w_1\), \(w_2\in \C \) with \(w_1\), \(w_2\neq 0\). Suppose that \(f'(z_0)\) exists. Then \(\lim _{t\to 0} \frac {|f(z_0+tw_1)-f(z_0)|}{|tw_1|}=\lim _{t\to 0} \frac {|f(z_0+tw_2)-f(z_0)|}{|tw_2|}\).

(Wilson, Problem 1170) Prove Theorem 2.2.3.1. How does this relate to the result of Problem 1150?

[Chapter 2, Problem 12] Let \(z_0\in \Omega \subseteq \C \) for some open set \(\Omega \). Let \(f:\Omega \to \C \). Suppose that \(\lim _{t\to 0} \frac {|f(z_0+tw_1)-f(z_0)|}{|tw_1|}=\lim _{t\to 0} \frac {|f(z_0+tw_2)-f(z_0)|}{|tw_2|}\) for all \(w_1\), \(w_2\in \C \setminus \{0\}\). Then either \(f'(z_0)\) exists or \((\overline f)'(z_0)\) exists.

[Definition: Unit vector] If \(z\in \C \setminus \{0\}\), define \(\hat z=\frac {1}{|z|}z\).

(Problem 1171) If \(z\in \C \setminus \{0\}\), then \(\hat z=e^{i\theta }\) for some real number \(\theta \), and if \(\hat z=e^{i\psi }\) then \(\theta -\psi \) is an integer multiple of \(2\pi \).

[Definition: Directed angle] Let \(z\), \(w\in \C \setminus \{0\}\). Let \(\angle (z,w)\) be the directed angle from \(z\) to \(w\) (that is, the angle between the line from \(0\) to \(z\) and the line from \(0\) to \(w\), chosen such that you move from \(z\) to \(w\) counterclockwise).

(Problem 1172) Show that \(\exp (i\angle (z,w)) = \frac {\hat w}{\hat z}\). (This provides an algebraically simpler but geometrically les intuitive definition of the directed angle, as the unique real number in \([0,2\pi )\) such that the above equation is true.)

[Definition: Angle preserving] Let \(z_0\in \Omega \subseteq \C \) for some open set \(\Omega \). Let \(f:\Omega \to \C \). We say that \(f\) preserves angles at \(z_0\) if, for all \(w_1\), \(w_2\in \C \setminus \{0\}\), we have that \begin {equation*} \lim _{t\to 0^+} \exp \bigl (i\angle (f(z_0+tw_1)-f(z_0),f(z_0+tw_2)-f(z_0)\bigr ) =\exp \bigl (i\angle (w_1,w_2\bigr ) \end {equation*} and in particular that \(f(z_0+tw_1)-f(z_0)\) and \(f(z_0+tw_2)-f(z_0)\) are not zero when \(t\) is sufficiently close to \(0\). [This is not the definition in the book.]

Theorem 2.2.3.2. If \(f'(z_0)\) exists and is not zero, then \(f\) preserves angles at \(z_0\).

(Adam, Problem 1180) Prove Theorem 2.2.3.2.

[Chapter 2, Problem 9a] If \(f\) is \(C^1\) and preserves angles at \(z_0\), and if the Jacobian matrix of \(f\) at \(z_0\) is nonsingular, then \(f'(z_0)\) exists.

(Bonus Problem 1181) If \(f\) is \(C^1\) and preserves angles at \(z_0\), and if the Jacobian matrix of \(f\) at \(z_0\) is singular, must \(f'(z_0)\) exist?

(Amani, Problem 1190) Consider the following figures. On the left is shown the traces of \(\gamma _j\) for several values of \(j\). On the right is shown the traces of \(f\circ \gamma _j\), \(g\circ \gamma _j\), or \(h\circ \gamma _j\) for the same \(\gamma _j\). You are given that exactly two of the quantities \(f'(0)\), \(g'(0)\), and \(h'(0)\) exist and that exactly one of those quantities is zero. Based on the images, which function do you think has nonzero derivative, which has zero derivative, and which does not have a derivative?

2.3. Real Analysis

(Memory 1200) Let \(f(x)=x^2\sin (1/x)\) if \(x\neq 0\) and let \(f(0)=0\). Then \(f\) is continuous on \((-\infty ,\infty )\), continuously differentiable on \((-\infty ,0)\) and \((0,\infty )\), and \(f'(0)\) exists, but the limit \(\lim _{x\to 0} f'(x)\) does not exist.

Lemma 2.3.1. Suppose that \(a<p<b\) and let \(H:(a,b)\to \R \) be continuous. Suppose that \(H\) is differentiable on both \((a,p)\) and \((p,b)\), and that \(\lim _{x\to p} H'(x)=h\) for some \(h\in \R \). Then \(H'(p)\) exists and \(H'(p)=\lim _{x\to p} H'(x)\).

2.3. Antiderivatives Revisited

Recall [Problem 780]: Suppose that there are two \(C^1\) functions \(g\) and \(h\) defined in an open rectangle or disc \(\Omega \) such that \(\frac {\partial }{\partial x}g=\frac {\partial }{\partial y}h\). Then there is a function \(f\in C^2(\Omega )\) such that \(\frac {\partial f}{\partial y}=g\) and \( \frac {\partial f}{\partial x}=h\).

Theorem 2.3.2. Let \(\Omega \subset \R ^2\) be an open rectangle or disc and let \(P\in \Omega \). Suppose that there are two functions \(g\) and \(h\) that are continuous on \(\Omega \), continuously differentiable on \(\Omega \setminus \{P\}\), and such that \(\frac {\partial }{\partial x}g=\frac {\partial }{\partial y}h\) on \(\Omega \setminus \{P\}\) for some \(P\in \Omega \). Then there is a function \(f\in C^1(\Omega )\) such that \(\frac {\partial f}{\partial y}=g\) and \( \frac {\partial f}{\partial x}=h\) everywhere in \(\Omega \) (including at \(P\)).

(Bashar, Problem 1220) Prove Theorem 2.3.2.

Let \(P=(x_0,y_0)\) and let \(f(x,y)=\int _{x_0}^x h(s,y_0)\,ds+\int _{y_0}^y g(x,t)\,dt\). Observe that if \((x,y)\in \Omega \) then so is \((s,y_0)\) and \((x,t)\) for all \(s\) between \(x_0\) and \(x\) and all \(t\) between \(y_0\) and \(y\). Thus \(g\) and \(h\) are defined at all required values. Because \(g\) and \(h\) are continuous, the integrals exist.

Furthermore, I claim \(f\) is continuous. Let \((x,y)\in \Omega \) and let \(\delta _1>0\) be such that \(B((x,y),\delta _1)\subset \Omega \). By continuity of \(g\) and \(h\) and compactness of \(\overline B((x,y),\delta _1/2)\), \(g\) and \(h\) are bounded on \(\overline B((x,y),\delta _1/2)\). If \((\xi ,\eta )\in B((x,y),\delta _1/2)\), then \begin {align*} f(\xi ,\eta )-f(x,y) &= \int _x^{\xi } h(s,y_0)\,ds + \int _y^{\eta } g(\xi ,t)\,dt + \int _{y_0}^y g(\xi ,t)-g(x,t)\,dt \end {align*}

and so \begin {align*} |f(\xi &,\eta )-f(x,y)| \\&\leq |\xi -x|\sup _{\overline B((x,y),\delta _1/2)} |h| +|\eta -y|\sup _{\overline B((x,y),\delta _1/2)} |g| + \biggl |\int _{y_0}^y g(x,t)-g(\xi ,t)\,dt\biggr | . \end {align*}

Furthermore, \(g\) must be uniformly continuous on \(\overline B((x,y),\delta _1/2)\). Choose \(\varepsilon >0\) and let \(\delta _2\) be such that if \(|(x,t)-(\xi ,t)|<\delta _2\) then \(|g(x,t)-g(\xi ,t)|<\varepsilon \). We then have that \begin {align*} |f(x&,y)-f(\xi ,\eta )| \leq |\xi -x|\sup _{\overline B((x,y),\delta _1/2)} |h| +|\eta -y|\sup _{\overline B((x,y),\delta _1/2)} |g| +|y_0-y|\varepsilon . \end {align*}

There is then a \(\delta _3>0\) such that if \(|(\xi ,\eta )-(x,y)|<\delta _3\) then \(|f(x,y)-f(\xi ,\eta )|<(1+|y_0-y|)\varepsilon \), and so \(f\) is continuous at \((x,y)\), as desired.

By the fundamental theorem of calculus, we have that \(\frac {\partial f}{\partial y}=g\) everywhere in \(\Omega \), including at \((x,y)=(x_0,y_0)=P\).

Furthermore, by Problem 770 and the fundamental theorem of calculus, we have that if \(x\neq x_0\) then \begin {equation*} \frac {\partial f}{\partial x}(x,y) =h(x,y_0)+\int _{y_0}^y \frac {\partial g}{\partial x}(x,t)\,dt. \end {equation*} Again because \(x\neq x_0\), we have that \(\p [g]{x}(x,t)=\p [h]{t}(x,t)\) and so by the fundamental theorem of calculus \(\p [f]{x}=h\) provided \(x\neq x_0\).

We need only show that \(\p [f]{x}=h\) even if \(x=x_0\). Fix a \(y\) and let \(F_y(x)=f(x,y)\). Then \(I=\{x\in \R :(x,y)\in \Omega \}\) is an open interval. We need only consider the case where \(I\neq \emptyset \). Then \(F_y\) is continuous on \(I\), \(F_y'(x)=h(x,y)\) for all \(x\neq x_0\), and \(\lim _{x\to x_0} F_y'(x)=h(x_0,y)\) because \(h\) is continuous. Thus by Lemma 2.3.1, \(F_y'(x_0)=h(x_0,y)\) and so \(\p [f]{x}=h\) even if \(x=x_0\).

(Bonus Problem 1221) Must the function \(f\) in Theorem 2.3.2 be \(C^2\) (that is, must \(g\) and \(h\) be continously differentiable at \(P\) as well as elsewhere in \(\Omega \)? If so, prove it; if not, give a counterexample.

Theorem 2.3.3. Let \(P\in \Omega \), where \(\Omega \) is an open rectangle or disc. Suppose that \(f\) is continuous on \(\Omega \) and holomorphic on \(\Omega \setminus \{P\}\). Then there is a function \(F\) that is holomorphic on all of \(\Omega \) (including \(P\)) such that \(\frac {\partial F}{\partial z} = f\).

(Dibyendu, Problem 1230) Prove Theorem 2.3.3.

Let \(f=u+iv\) where \(u\) and \(v\) are real valued; then by definition \(u\), \(v\) are continuous on \(\Omega \) and \(C^1\) on \(\Omega \setminus \{P\}\).

Then by the Cauchy-Riemann equations, we have that \(\p [u]{x}=\p [v]{y}\). Therefore, by Theorem 2.3.2, there is a \(V\in C^1(\Omega )\) such that \(\p [V]{y}=u\) and \(\p [V]{x}=v\) in all of \(\Omega \). Similarly, \(\p [u]{y}=\p [(-v)]{x}\), and so there is a \(U\in C^1(\Omega )\) such that \(\p [U]{x}=u\) and \(\p [U]{y}=-v\).

Let \(F=U+iV\). Then \begin {equation*} \p [F]{z} = \frac {1}{2}\biggl (\p [F]{x}+\frac {1}{i}\p [F]{y}\biggr ) =\frac {1}{2}\biggl (u+iv+\frac {1}{i}(-v+iu)\biggr )=f \end {equation*} and \begin {equation*} \p [F]{\bar z} = \frac {1}{2}\biggl (\p [F]{x}-\frac {1}{i}\p [F]{y}\biggr ) =\frac {1}{2}\biggl (u+iv-\frac {1}{i}(-v+iu)\biggr )=0 \end {equation*} in \(\Omega \), as desired.

(Hope, Problem 1240) Is the previous problem true if we relax the assumption that \(f\) is continuous at \(P\) (and that \(\p [F]{z}=f\) at \(P\))?

No. Let \(\Omega =D(0,2)\) and let \(P=0\). Then \(f(z)=1/z\) is holomorphic on \(\Omega \setminus \{P\}\) (see Problem 620) and by Problem 2.4a in your book, if \(\gamma (t)=e^{it}\), \(0\leq t\leq 2\pi \), then \(\oint _{\gamma } f(z)\,dz=2\pi i\).

But if \(F\) is holomorphic in \(\Omega \setminus \{P\}\) and \(F'=f\), then by Problem 970 \begin {align*} \oint _\gamma f(z)\,dz &= \oint _\gamma \frac {\partial F}{\partial z} \,dz=F(\gamma (2\pi ))-F(\gamma (0))=F(1)-F(1)=0. \end {align*}

This is a contradiction; therefore, no such \(F\) can exist.

2.4. Real Analysis

(Memory 1250) In \(\C \), \(\overline D(P,r)\) is the closure of \(D(P,r)\).

(Memory 1260) In \(\C \), \(\partial D(P,r)=\partial \overline D(P,r)=\{z\in \C :|z-P|=r\}\).

2.4. The Cauchy Integral Formula and the Cauchy Integral Theorem

Theorem 2.4.3. [The Cauchy integral theorem.] Let \(f\) be holomorphic in \(D(P,R)\). Let \(\gamma :[a,b]\to D(P,R)\) be a closed curve. Then \(\oint _\gamma f(z)\,dz=0\).

[Chapter 2, Problem 1] Prove the Cauchy integral theorem.

Theorem 2.4.2. [The Cauchy integral formula.] Let \(\Omega \subseteq \C \) be open and let \(\overline D(z_0,r)\subset \Omega \). Let \(f\) be holomorphic in \(\Omega \) and let \(z\in D(z_0,r)\). Then \begin {equation*} f(z)=\frac {1}{2\pi i}\oint _\gamma \frac {f(\zeta )}{\zeta -z}\,d\zeta \end {equation*} where \(\gamma (t)=z_0+re^{it}\), \(0\leq t\leq 2\pi \), is the parametrization of \(\partial D(z_0,r)\) traversed once counterclockwise.

Lemma 2.4.1. The Cauchy integral formula is true in the special case where \(f(\zeta )=1\) for all \(\zeta \in \C \).

(Micah, Problem 1270) Let \(\gamma :[a,b]\to K\) be a \(C^1\) curve for some set \(K\subseteq \C \) (not necessarily open). Let \(V\subseteq \C \) be either open or compact and let \(f: K\times V\to \C \) be continuous. Let \(F:V\to \C \) be defined by \begin {gather*} F(z)=\oint _\gamma f(\zeta ,z)\,d\zeta . \end {gather*} Show that \(F\) is continuous on \(V\).

(Muhammad, Problem 1280) In this problem we will begin the proof of Lemma 2.4.1 (and thus ultimately of Theorem 2.4.2). Let \(\gamma :[a,b]\to K\) be a \(C^1\) curve for some set \(K\subseteq \C \) (not necessarily open). Let \(W\subseteq \C \) be open and let \(f: K\times W\to \C \) be continuous. Suppose that the functions \(\frac {\partial f}{\partial x}\) and \(\frac {\partial f}{\partial y}\) given by \(\frac {\partial f}{\partial x}(\zeta ,x+iy)=\frac {\partial }{\partial x} f(\zeta ,x+iy)\) and \(\frac {\partial f}{\partial y}(\zeta ,x+iy)=\frac {\partial }{\partial y} f(\zeta ,x+iy)\) are continuous on \(K\times W\). Show that \begin {gather*} \frac {\partial }{\partial x} \oint _\gamma f(\zeta ,x+iy)\,d\zeta = \oint _\gamma \frac {\partial }{\partial x} f(\zeta ,x+iy)\,d\zeta ,\\ \frac {\partial }{\partial y} \oint _\gamma f(\zeta ,x+iy)\,d\zeta = \oint _\gamma \frac {\partial }{\partial y} f(\zeta ,x+iy)\,d\zeta ,\\ \frac {\partial }{\partial z} \oint _\gamma f(\zeta ,z)\,d\zeta = \oint _\gamma \frac {\partial }{\partial z} f(\zeta ,z)\,d\zeta ,\\ \frac {\partial }{\partial \bar z} \oint _\gamma f(\zeta ,z)\,d\zeta = \oint _\gamma \frac {\partial }{\partial \bar z} f(\zeta ,z)\,d\zeta \end {gather*} for all \(z=x+iy\in W\).

(Nisa, Problem 1290) Prove Lemma 2.4.1. Hint: Start by proving Lemma 2.4.1 in the special case \(z=z_0\). Computing \(\re \oint _\gamma \frac {1}{\zeta -z}\,d\zeta \) and \(\im \oint _\gamma \frac {1}{\zeta -z}\,d\zeta \) directly from the definition of line integral is very difficult if \(z\neq z_0\). Instead compute the derivatives of \(G(z)= \oint _\gamma \frac {1}{\zeta -z}\,d\zeta \) and use the known value of \(\oint _\gamma \frac {1}{\zeta -z_0}\,d\zeta \).

\(\oint _\gamma \frac {1}{\zeta -z_0}\,d\zeta =1\) (this is a routine computation).

If \(\zeta \neq x+iy\), then \begin {align*} \p {x} \frac {1}{\zeta -(x+iy)} &=\p {x}\frac {\overline \zeta -x+iy}{(\re \zeta -x)^2+(\im \zeta -y)^2} \\&=\frac {-|\zeta -(x+iy)|^2+(\overline \zeta -x+iy)(2(\re \zeta -x))}{|\zeta -(x+iy)|^4} \\&=\frac {-|\zeta -(x+iy)|^2+(\overline \zeta -x+iy)((\overline \zeta -x+iy)+(\zeta -x-iy))}{(\zeta -x-iy)^2(\overline \zeta -x+iy)^2} \\&= \frac {1}{(\zeta -(x+iy))^2} . \end {align*}

Now, by the previous problem, if \(x+iy\in D(P,r)\), then \begin {align*} \frac {\partial }{\partial x} \oint _\gamma \frac {1}{\zeta -(x+iy)}\,d\zeta &= \oint _\gamma \p {x}\frac {1}{\zeta -(x+iy)}\,d\zeta \\&= \oint _\gamma \frac {1}{(\zeta -(x+iy))^2}\,d\zeta . \end {align*}

Let \(F(\zeta )=\frac {-1}{\zeta -(x+iy)}\). By Problem 620 and the chain rule we have that \(\p {\zeta } F(\zeta )=\frac {1}{(\zeta -(x+iy))^2}\). Thus \begin {align*} \frac {\partial }{\partial x} \oint _\gamma \frac {1}{\zeta -(x+iy)}\,d\zeta &= \oint _\gamma \p {\zeta }\frac {-1}{\zeta -(x+iy)}\,d\zeta \end {align*}

which by Problem 970 is zero because \(\gamma \) is a closed curve.

Similarly \begin {align*} \frac {\partial }{\partial y} \oint _\gamma \frac {1}{\zeta -(x+iy)}\,d\zeta &= 0 \end {align*}

for all \(x+iy\in D(z_0,r)\). Thus \(\oint _\gamma \frac {1}{\zeta -(x+iy)}\,d\zeta \) (regarded as a function of \(x+iy\)) must be a constant; since it equals \(1\) at \(x+iy=z_0\), it must be \(1\) everywhere.

2.5. The Cauchy Integral Formula: Some Examples

(Robert, Problem 1300) Let \(\gamma (t)=z_0+re^{i t}\), \(0\leq t\leq 2\pi \). Let \(n\) be an integer. Let \(z\in D(z_0,r)\). Show that \(\oint _\gamma (\zeta -z)^n\,d\zeta =0\) if \(n\neq -1\).

By Problem 580 and Problem 620, and by the chain rule (Problem 1.49), if \(n\neq -1\) then \begin {equation*} (\zeta -z)^n=\frac {1}{n+1} \frac {\partial }{\partial \zeta }(\zeta -z)^{n+1}. \end {equation*} Thus \begin {align*} \oint _\gamma (\zeta -z)^n\,d\zeta &= \oint _\gamma \frac {1}{n+1} \frac {\partial }{\partial \zeta }(\zeta -z)^{n+1}\,d\zeta =\frac {(\gamma (2\pi )-z)^{n+1}}{n+1} -\frac {(\gamma (0)-z)^{n+1}}{n+1} =0 \end {align*}

by Proposition 2.1.6.

(Sam, Problem 1310) Show that if \(n\geq 0\) and \(\gamma \) is as in the previous problem, then \begin {equation*} \frac {1}{2\pi i}\oint _\gamma \frac {\zeta ^n}{\zeta -z}\,d\zeta =z^n. \end {equation*} (As we have not yet proven the Cauchy integral formula, do not cite the Cauchy integral formula to perform this computation.)

If \(n=0\) then the result follows from Lemma 2.4.1. Otherwise, by the binomial theorem \begin {equation*} \zeta ^n=[(\zeta -z)+z]^n = \sum _{k=0}^n \binom {n}{k} z^{k}(\zeta -z)^{n-k}. \end {equation*}

Thus \begin {equation*} \oint _\gamma \frac {\zeta ^n}{\zeta -z}\,d\zeta =\sum _{k=0}^n \binom {n}{k}z^{k}\oint _\gamma (\zeta -z)^{n-k-1}\,d\zeta . \end {equation*} If \(k<n\) then the integral is zero, while if \(k=n\) then the integral is \(2\pi i\) by Lemma 2.4.1, as desired.

(Wilson, Problem 1320) Let \(p\) be a holomorphic polynomial. Find \(\frac {1}{2\pi i}\oint _\gamma \frac {p(\zeta )}{\zeta -z}\,d\zeta \). (As we have not yet proven the Cauchy integral formula, do not cite the Cauchy integral formula to perform this computation.)

2.4. The Cauchy Integral Formula and the Cauchy Integral Theorem

Theorem 2.4.2. [The Cauchy integral formula.] Let \(\Omega \subseteq \C \) be open and let \(\overline D(z_0,r)\subset \Omega \). Let \(f\) be holomorphic in \(\Omega \) and let \(z\in D(z_0,r)\). Then \begin {equation*} f(z)=\frac {1}{2\pi i}\oint _\gamma \frac {f(\zeta )}{\zeta -z}\,d\zeta . \end {equation*}

(Adam, Problem 1330) Prove Theorem 2.4.2. Hint: Let \(h(\zeta )=\frac {f(\zeta )-f(z)}{\zeta -z}\) if \(\zeta \neq z\). How should you define \(h(z)\)? What can you say about the behavior of \(h\) at \(z\) and in \(\Omega \setminus \{z\}\)?

(Bonus Problem 1340) Prove the previous result without using Problem 1230.

[Definition: Integral over a circle] We define \(\oint _{\partial D(P,r)} f(z)\,dz=\oint _\gamma f(z)\,dz\), where \(\gamma \) is a counterclockwise simple parameterization of \(\partial D(P,r)\).

[Chapter 2, Problem 20] Let \(f\) be continuous on \(\overline D(P,r)\) and holomorphic in \(D(P,r)\). Show that \(f(z)=\oint _{\partial D(P,r)} \frac {f(\zeta )}{\zeta -z}\,d\zeta \) for all \(z\in D(P,r)\).

(Problem 1350) Let \(f\) and \(g\) be holomorphic in \(D(P,r)\) and continuous on \(\overline D(P,r)\). Suppose that \(f(\zeta )=g(\zeta )\) for all \(\zeta \in \partial D(P,r)\). Show that \(f(z)=g(z)\) for all \(z\in D(P,r)\).

2.6. An Introduction to the Cauchy Integral Theorem and the Cauchy Integral Formula for More General Curves

Proposition 2.6.6. Let \(\Omega =D(P,\tau )\setminus \overline D(P,\sigma )\) for some \(P\in \C \) and some \(0<\sigma <\tau \). Let \(\sigma <r<R<\tau \) and let \(\gamma _r\), \(\gamma _R\) be the counterclockwise parameterizations of \(\partial D(P,r)\), \(\partial D(P,R)\). Suppose that \(f\) is holomorphic in \(\Omega \). Then \(\oint _{\gamma _r} f=\oint _{\gamma _R} f\).

(Amani, Problem 1360) Prove Proposition 2.6.6. Hint: Define \(\gamma _s\) in the natural way and find a function \(h\) such that \(\frac {d}{ds}\oint _{\gamma _s} f=\oint _{\gamma _s} h\).

[Definition: Homotopic curves] Let \(a<b\), \(c<d\). Let \(\Omega \subseteq \C \) be open. Let \(\gamma _c\), \(\gamma _d:[a,b]\to \Omega \) be two \(C^1\) curves with the same endpoints (so \(\gamma _c(a)=\gamma _d(a)\), \(\gamma _c(b)=\gamma _d(b)\)).

We say that \(\gamma _c\) and \(\gamma _d\) are \(C^1\)-homotopic in \(\Omega \) if there is a function \(\Gamma \) such that:

• \(\Gamma :[a,b]\times [c,d]\to \Omega \),

• \(\Gamma (t,c)=\gamma _c(t)\), \(\Gamma (t,d)=\gamma _d(t)\) for all \(t\in [a,b]\),

• \(\Gamma (a,s)=\gamma _c(a)=\gamma _d(a)\), \(\Gamma (b,s)=\gamma _c(b)=\gamma _d(b)\) for all \(s\in [c,d]\),

• \(\Gamma \) is continuous on \([a,b]\times [c,d]\),

• \(\Gamma \) is \(C^1\) in the first variable in the sense that if \(\gamma _s(t)=\Gamma (t,s)\), then \(\gamma _s\in C^1[a,b]\) for all \(c\leq s\leq d\).

[Definition: Homotopic closed curves] Let \(a<b\), \(c<d\). Let \(\Omega \subseteq \C \) be open. Let \(\gamma _c\), \(\gamma _d:[a,b]\to \Omega \) be two closed \(C^1\) curves.

We say that \(\gamma _c\) and \(\gamma _d\) are \(C^1\)-homotopic in \(\Omega \) if there is a function \(\Gamma \) such that:

• \(\Gamma :[a,b]\times [c,d]\to \Omega \),

• \(\Gamma (t,c)=\gamma _c(t)\), \(\Gamma (t,d)=\gamma _d(t)\) for all \(t\in [a,b]\),

• \(\Gamma (a,s)=\Gamma (b,s)\) for all \(s\in [c,d]\),

• \(\Gamma \) is continuous on \([a,b]\times [c,d]\),

• \(\Gamma \) is \(C^1\) in the first variable on \([a,b]\times [c,d]\).

(Bonus Problem 1370) Show that the assumption that \(\Gamma \) be \(C^1\) in the first variable is unnecessary: if \(\gamma _c\) and \(\gamma _d\) are \(C^1\) and there is a function \(\Gamma \) satisfying all of the above conditions except that \(\Gamma \) is not \(C^1\) in the first variable, then \(\Gamma \) may be perturbed slightly to yield a \(C^1\) function.

(Bashar, Problem 1380) Let \(\Omega =D(P,\tau )\setminus \overline D(P,\sigma )\) for some \(P\in \C \) and some \(0<\sigma <\tau \). Let \(\sigma <r<R<\tau \) and let \(\gamma _r\), \(\gamma _R\) be the counterclockwise parameterizations of \(\partial D(P,r)\), \(\partial D(P,R)\). Show that \(\gamma _r\) and \(\gamma _R\) are homotopic in \(\Omega \).

(Problem 1381) Let \(\Omega \subseteq \C \) be an open set and let \(f:\Omega \to \C \) be holomorphic.

Let \(a<b\), \(c<d\), and let \(\Gamma :[a,b]\times [c,d]\to \Omega \) be continuous.

Assume that, if \(s\in [c,d]\), then \(\gamma _s:[a,b]\to \Omega \) given by \(\gamma _s(t)=\Gamma (t,s)\) is a \(C^1\) curve.

Further assume that the function \(\frac {\partial \Gamma }{\partial t}\) (with \(\frac {\partial \Gamma }{\partial t}(a,s)\) and \(\frac {\partial \Gamma }{\partial t}(b,s)\) given by one-sided derivatives) is also continuous on \([a,b]\times [c,d]\).

Show that the function \(I(s)=\oint _{\gamma _s} f\) is continuous on \([c,d]\).

By definition of line integral, \begin {equation*} \oint _{\gamma _s} f = \int _a^b f(\Gamma (t,s))\,\p {t}\Gamma (t,s)\,dt. \end {equation*} By Problem 760, if \(c\leq s\leq d\) then \(\oint _{\gamma _s} f\) is continuous (as a function of \(s\)) at \(s\).

(Problem 1382) Let \(\Omega \), \(f\), and \(\Gamma \) be as in the previous problem. Assume that \(\frac {\partial \Gamma }{\partial s}\) and \(\frac {\partial ^2 \Gamma }{\partial s\,\partial t}\) exist and are continuous on \([a,b]\times (c,d)\).

Show that \begin {align*} \frac {d}{ds}\oint _{\gamma _s} f &= \biggl (f(\Gamma (b,s))\,\p {s}\Gamma (b,s)\biggr ) - \biggl (f(\Gamma (a,s))\,\p {s}\Gamma (a,s)\biggr ) . \end {align*}

By definition of line integral, \begin {equation*} \oint _{\gamma _s} f = \int _a^b f(\Gamma (t,s))\,\p {t}\Gamma (t,s)\,dt. \end {equation*} By Problem 770, \begin {equation*} \frac {d}{ds}\oint _{\gamma _s} f = \int _a^b \p {s}\biggl (f(\Gamma (t,s))\,\p {t}\Gamma (t,s)\biggr )\,dt. \end {equation*} By the product rule and problem Problem 1120, \begin {align*} \frac {d}{ds}\oint _{\gamma _s} f &= \int _a^b f'(\Gamma (t,s))\,\p {s}\Gamma (t,s)\,\p {t}\Gamma (t,s) +f(\Gamma (t,s))\,\frac {\partial ^2}{\partial s\,\partial t} \Gamma (t,s) \,dt. \end {align*}

By Clairaut’s theorem and the product rule, \begin {equation*} \frac {d}{ds}\oint _{\gamma _s} f = \int _a^b\p {t}\biggl (f(\Gamma (t,s))\,\p {s}\Gamma (t,s)\biggr )\,dt. \end {equation*} By the fundamental theorem of calculus, \begin {align*} \frac {d}{ds}\oint _{\gamma _s} f &= \biggl (f(\Gamma (b,s))\,\p {s}\Gamma (b,s)\biggr ) - \biggl (f(\Gamma (a,s))\,\p {s}\Gamma (a,s)\biggr ) \end {align*}

as desired.

(Dibyendu, Problem 1390) Let \(\Omega \) be an open set, let \(f:\Omega \to \C \) be holomorphic, and let \(\gamma _c\), \(\gamma _d:[a,b]\to \Omega \) be two curves with the same endpoints that are homotopic in \(\Omega \). Suppose that the homotopy \(\Gamma \) between \(\gamma _c\) and \(\gamma _d\) satisfies the conditions of Problem 1381 and Problem 1382. Show that \[\oint _{\gamma _c} f=\oint _{\gamma _d} f.\]

By Problem 1382 \begin {equation*} \frac {d}{ds}\oint _{\gamma _s} f = \biggl (f(\Gamma (b,s))\,\p {s}\Gamma (b,s)\biggr )- \biggl (f(\Gamma (a,s))\,\p {s}\Gamma (a,s)\biggr ) . \end {equation*} By definition of homotopy between curves with the same endpoints, \(\Gamma (b,s)=\gamma _c(b)\) for all \(s\), and so \(\p {s}\Gamma (b,s)=0\). Similarly \(\p {s}\Gamma (a,s)=0\) and so \begin {equation*} \frac {d}{ds}\oint _{\gamma _s} f = 0 \end {equation*} for all \(c<s<d\). Thus by the mean value theorem and Problem 1381, \(\oint _{\gamma _c}f=\oint _{\gamma _d} f\).

(Bonus Problem 1400) Let \(\Omega \subseteq \C \) be an open set, let \(\gamma _c\) and \(\gamma _d\) be two curves homotopic in \(\Omega \) with the same endpoints, and let \(f:\Omega \to \C \) be holomorphic. Show that \(\oint _{\gamma _c} f=\oint _{\gamma _d}f\) even if the homotopy is merely continuous and \(C^1\) in the first variable.

(Hope, Problem 1410) Let \(\Omega \subseteq \C \) be any open set, let \(\gamma _c\) and \(\gamma _d\) be any two closed curves that are homotopic in \(\Omega \), and let \(f:\Omega \to \C \) be holomorphic. Assume the homotopy \(\Gamma \) satisfies the conditions of Problem 1381 and Problem 1382. Show that \(\oint _{\gamma _c} f=\oint _{\gamma _d}f\).

By Problem 1381 and Problem 1382, \(\oint _{\gamma _s} f\) is a continuous function of \(s\) and \begin {align*} \frac {d}{ds}\oint _{\gamma _s} f &= \biggl (f(\Gamma (b,s))\,\p {s}\Gamma (b,s)\biggr ) - \biggl (f(\Gamma (a,s))\,\p {s}\Gamma (a,s)\biggr ) . \end {align*}

By definition of homotopy between closed curves, \(\Gamma (b,s)=\Gamma (a,s)\) for all \(s\), and so in particular \(\p {s}\Gamma (b,s)=\p {s}\Gamma (a,s)\). Thus \begin {align*} \frac {d}{ds}\oint _{\gamma _s} f &= \biggl (f(\Gamma (a,s))\,\p {s}\Gamma (a,s)\biggr ) - \biggl (f(\Gamma (a,s))\,\p {s}\Gamma (a,s)\biggr ) =0 . \end {align*}

By the mean value theorem, \(\oint _{\gamma _c} f=\oint _{\gamma _d}f\).

(Micah, Problem 1420) Let \(\Omega \) be open and let \(f:\Omega \to \C \) be holomorphic.

(a)

Let \(\gamma :[0,1]\to \Omega \) be homotopic in \(\Omega \) to a point (constant function). Show that \(\oint _\gamma f=0\).

(b)

Suppose that \(\overline D(z,r)\subset \Omega \) and that \(\gamma \) is homotopic in \(\Omega \setminus \{z\}\) to \(\partial D(z,r)\) (traversed once with counterclockwise orientation). Show that \(\frac {1}{2\pi i}\oint _\gamma \frac {f(\zeta )}{\zeta -z}\,d\zeta =f(z)\).

3.1. Differentiability Properties of Holomorphic Functions

Theorem 3.1.3. [Generalization.] Let \(D(P,r)\subset \C \). Let \(\varphi :\partial D(P,r)\to \C \) be continuous. Let \(k\) be a nonnegative integer. Define \(f:D(P,r)\to \C \) by \begin {equation*} f(z)=\frac {k!}{2\pi i} \oint _{\partial D(P,r)} \frac {\varphi (\zeta )}{(\zeta -z)^{k+1}}\,d\zeta . \end {equation*} Then \(f\) is \(C^1\) and holomorphic in \(D(P,r)\), and \begin {equation*} \p [f]{z}=\frac {(k+1)!}{2\pi i} \oint _{\partial D(P,r)} \frac {\varphi (\zeta )}{(\zeta -z)^{k+2}}\,d\zeta . \end {equation*}

(Muhammad, Problem 1430) Prove Theorem 3.1.3.

[Chapter 2, Problem 21] If \(z\in \partial D(P,r)\), is it necessarily true that \(\lim _{w\to z} f(w)=\varphi (z)\)?

Theorem 3.1.1. Let \(\Omega \subseteq \C \) be open and let \(f:\Omega \to \C \) be holomorphic. Then \(f\in C^\infty (\Omega )\). Moreover, if \(\overline D(P,r)\subset \Omega \), then \begin {equation*} \frac {\partial ^k f}{\partial z^k}(z) = \frac {k!}{2\pi i} \oint _{\partial D(P,r)} \frac {f(\zeta )}{(\zeta -z)^{k+1}}\,d\zeta . \end {equation*}

Corollary 3.1.2. Let \(\Omega \subseteq \C \) be open and let \(f:\Omega \to \C \) be holomorphic. Then \(\frac {\partial ^k f}{\partial z^k}\) is holomorphic in \(\Omega \) for all \(k\in \N \).

(Nisa, Problem 1440) Prove Theorem 3.1.1 and Corollary 3.1.2.

Let \(\overline D(P,r)\subset \Omega \). By the Cauchy integral formula, if \(z\in D(P,r)\) then \begin {equation*} f(z)=\frac {1}{2\pi i} \oint _{\partial D(P,r)} \frac {f(\zeta )}{\zeta -z}\,d\zeta . \end {equation*} Because \(f\) is holomorphic, it must be continuous, and so we may apply Theorem 3.1.3.

We perform an induction argument. Suppose that \(f\), \(f'=\p [f]{z},\dots ,f^{(k-1)}=\frac {\partial ^{k-1}f}{\partial z^{k-1}}\) exist and are \(C^1\) and holomorphic in \(D(P,r)\), and that \(f^{(k)}=\frac {\partial ^{k}f}{\partial z^{k}}\) exists and satisfies \begin {equation*} f^{(k)}(z)=\frac {k!}{2\pi i} \oint _{\partial D(P,r)} \frac {f(\zeta )}{(\zeta -z)^{k+1}}\,d\zeta \end {equation*} in \(D(P,r)\). We have shown that this is true for \(k=0\). By Theorem 3.1.3, we have that \(f^{(k)}\) is also \(C^1\) and holomorphic in \(D(P,r)\), and that \begin {equation*} f^{(k+1)}=\frac {(k+1)!}{2\pi i} \oint _{\partial D(P,r)} \frac {f(\zeta )}{(\zeta -z)^{k+2}}\,d\zeta . \end {equation*} Because \(f^{(k)}\) is holomorphic, \(f^{(k+1)}=\frac {\partial ^{k+1}f}{\partial z^{k+1}}\). Thus by induction this must be true for all nonnegative integers \(k\).

This proves Corollary 3.1.2 and part of Theorem 3.1.1. To prove that \(f\in C^\infty (\Omega )\), observe that \begin {equation*} \frac {\partial ^{j+\ell } f}{\partial x^j\partial y^\ell } = \biggl (\p {z}+\p {\bar z}\biggr )^j \biggl (\p {z}-\p {\bar z}\biggr )^\ell f \end {equation*}

We can write all partial derivatives of \(f\) (in terms of \(x\) and \(y\)) as linear combinations of \(\frac {\partial ^j f}{\partial z^j}\) and \(\frac {\partial ^{m+n+\ell } f}{\partial x^m\partial y^n\partial \bar z^\ell }\) for various values of \(j\), \(\ell \), \(m\), and \(n\), and so

(Robert, Problem 1450) Suppose that \(P\in \Omega \subseteq \C \) for some open set \(\Omega \). Suppose that \(f\) is continuous on \(\Omega \) and holomorphic on \(\Omega \setminus \{P\}\). Show that \(f\) is holomorphic on \(\Omega \).

We must show that \(\nabla f(P)\) exists, that \(\nabla f\) is continuous at \(P\), and that \(\p [f]{\bar z}(P)=0\). That is, we only need to work at \(P\). Let \(\Omega \) be an open disc centered at \(P\) and contained in \(\Omega \); by definition of open set \(\Omega \) must exist. Then by Theorem 2.3.3 (Problem 1230) there is a function \(F:\Omega \to \C \) that is holomorphic in \(\Omega \) such that \(f=\p [F]{z}\) in \(\Omega \) (including at \(P\)). By Theorem 3.1.1 and Corollary 3.1.2 (Problem 1440), \(f=\p [F]{z}\) is \(C^1\) and holomorphic in \(\Omega \), and in particular at \(P\).

3.1. Real Analysis

(Memory 1460) Let \(\Omega \subseteq \C \) be open and connected. Show that \(\Omega \) is path connected and that the paths may be taken to be \(C^1\); that is, if \(z\), \(w\in \Omega \) then there is a \(\gamma :[0,1]\to \Omega \) with \(\gamma \) a \(C^1\) function such that \(\gamma (0)=z\) and \(\gamma (1)=w\).

3.1. Morera’s theorem

Theorem 3.1.4. (Morera’s theorem.) Let \(\Omega \subseteq \C \) be open and connected. Let \(f\in C(\Omega )\) be such that \(\oint _\gamma f=0\) for all closed curves \(\gamma \). Then \(f\) is holomorphic in \(\Omega \).

(Sam, Problem 1470) Prove Morera’s theorem. Furthermore, show that there is a function \(F\) holomorphic in \(\Omega \) such that \(F'=f\).

Fix some \(z_0\in \Omega \). Suppose that \(z\in \Omega \). By Memory 1460, there is a \(C^1\) curve \(\psi =\psi _z:[0,1]\to \Omega \) such that \(\psi (0)=z_0\) and \(\psi (1)=z\).

Suppose that \(\tau \) is another such curve, that is, a \(C^1\) function \(\tau :[0,1]\to \Omega \) such that \(\tau (0)=z_0\) and \(\tau (1)=z\). Let \(\tau _{-1}(t)=\tau (1-t)\). Then by Problem 1010, we have that \(\oint _{\tau _{-1}} f=-\oint _{\tau } f\). Furthermore, by Problem 1050, there is a \(C^1\) curve \(\gamma :[0,1]\to \C \) such that \(\gamma (0)=\psi (0)=z_0\), \(\gamma (1)=\tau _{-1}(1)=\tau (0)=z_0\) and such that \begin {equation*} \oint _\gamma f=\oint _\psi f+\oint _{\tau _{-1}}f=\oint _\psi f-\oint _\tau f. \end {equation*} But \(\gamma \) is closed and so \(\oint _\gamma f=0\), and so \(\oint _\psi f=\oint _\tau f\).

Thus, if we define \(F(z)=\oint _{\psi _z} f\), then \(F\) is well defined, as \(F(z)\) is independent of our choice of path \(\psi _z\) from \(z_0\) to \(z\).

Now, let \(z\in \Omega \) and let \(r>0\) be such that \(D(z,r)\subseteq \Omega \); such an \(r\) must exist by definition of \(\Omega \). If \(w\in D(z,r)\setminus \{z\}\), then \begin {equation*} F(w)-F(z)=\oint _{\psi _w} f-\oint _{\psi _z} f. \end {equation*} Let \(\varphi (t)=z+t(w-z)\), so \(\varphi :[0,1]\to D(z,r)\) is a \(C^1\) path from \(z\) to \(w\). We may assume without loss of generality that \(\psi _w\) is generated from \(\psi _z\) and \(\varphi \) by Problem 1050; thus, \begin {equation*} \frac {F(w)-F(z)}{w-z}=\frac {1}{w-z}\oint _{\varphi } f =\int _0^1 f(z+t(w-z))\,dt . \end {equation*} A straightforward \(\varepsilon \)-\(\delta \) argument yields that \begin {equation*} \lim _{w\to z}\frac {F(w)-F(z)}{w-z}=f(z) \end {equation*} so \(F\) possesses a complex derivative at \(z\). Furthermore, \(F'=f\) is continuous on \(\Omega \). Thus \(F\in C^1(\Omega )\) and is holomorphic on \(\Omega \) by Problem 1130, and so by Theorem 3.1.1 and Corollary 3.1.2 \(f=F'\) is \(C^\infty \) (in particular \(C^1\)) and holomorphic in \(\Omega \).

(Wilson, Problem 1480) Can you rewrite Morera’s theorem to involve a statement true for all holomorphic functions (can you write it with the phrase “if and only if”)?

3.2. Real Analysis

(Memory 1490) State the Root Test and Ratio Test from undergraduate real analysis.

[Definition: Taylor series] Let \(f\in C^\infty (a,b)\) and let \(a<c<b\). The Taylor series for \(f\) at \(c\) is \(\sum _{n=0}^\infty \frac {f^{(n)}(c)}{n!}(x-c)^n\) (with the convention \(0^0=1\)).

(Memory 1500) Let \(P_{m,c}(x)=\sum _{n=0}^m \frac {f^{(n)}(c)}{n!}(x-c)^n\) be the \(m\)th partial sum of the Taylor series at \(c\). Suppose that \(x\in (a,b)\), \(x\neq c\), \(m\in \N \). Show that there is a \(y_{m}\in (a,b)\) with \(|y_{m}-c|<|x-c|\) such that \begin {equation*} f(x)=P_{m-1,c}(x)+\frac {1}{m!}f^{(m)}(y_m) \,(x-c)^{m}. \end {equation*}

(Memory 1510) The Taylor series for \(\sin \), \(\cos \), and \(\exp \) converge to the parent function on all of \(\R \).

(Adam, Problem 1520) Give an example of a function \(f\in C^\infty (\R )\) such that the Taylor series for \(f\) converges for all \(x\in \R \) but such that \(f(x)\neq \sum _{n=0}^\infty \frac {f^{(n)}(c)}{n!}(x-c)^n\) for all \(x\neq c\).

The function \begin {equation*} f(x)=\begin {cases} \exp (-1/x^2),&x\neq 0,\\0, &x=0\end {cases} \end {equation*} satisfies \(f^{(n)}(0)=0\) for all nonnegative integers \(n\), and thus the Taylor series is zero everywhere; however, \(f(x)\neq 0\) if \(x\neq 0\) and so the Taylor series never converges to the function.

(Amani, Problem 1530) Give an example of a function \(f\in C^\infty (0,\infty )\) such that the Taylor series for \(f\) at \(2\) diverges for all \(|x-2|>2\). Can we do this for a function \(f\in C^\infty (\R )\)?

(Bonus Problem 1540) Give an example of a function \(f\in C^\infty (\R )\) such that the Taylor series for \(f\) at \(0\) diverges for all \(x\neq 0\).

[Definition: Absolute convergence] Let \(\sum _{n=0}^\infty a_n\) be a series of real numbers. If \(\sum _{n=0}^\infty |a_n|\) converges, then we say \(\sum _{n=0}^\infty a_n\) converges absolutely.

[Definition: Uniform convergence] Let \(E\) be a set, let \((X,d)\) be a metric space, and let \(f_k\), \(f:E\to X\). We say that \(f_k\to f\) uniformly on \(E\) if for every \(\varepsilon >0\) there is a \(N\in \N \) such that if \(k\geq N\), then \(d(f_k(z),f(z))<\varepsilon \) for all \(z\in E\).

[Definition: Uniformly Cauchy] Let \(E\) be a set, let \((X,d)\) be a metric space, and let \(f_k:E\to X\). We say that \(\{f_k\}_{k=1}^\infty \) is uniformly Cauchy on \(E\) if for every \(\varepsilon >0\) there is a \(N\in \N \) such that if \(n>m\geq N\), then \(d(f_n(z),f-m(z))<\varepsilon \) for all \(z\in E\).

[Definition: Uniform convergence and Cauchy for series] If \(E\) is a set, \(V\) is a vector space, and \(f_k:E\to V\) for each \(k\in \N \), then the series \(\sum _{k=1}^\infty f_k\) converges uniformly to \(f:E\to V\) or is uniformly Cauchy, respectively, if the sequences of partial sums \(\bigl \{\sum _{k=1}^n f_k\bigr \}_{n=1}^\infty \) converge uniformly or are uniformly Cauchy.

(Memory 1550) Suppose that \((X,d)\) is a complete metric space. Then any uniformly Cauchy sequence is uniformly convergent.

(Memory 1560) Suppose that \((E,\rho )\) and \((X,d)\) are two metric spaces. Let \(f_k\), \(f:E\to X\). Suppose \(f_k\to f\) uniformly on \(E\) and that each \(f_k\) is continuous. Then \(f\) is also continuous.

(Bashar, Problem 1570) Give an example of a compact metric space \((X,d)\) and a sequence of continuous functions from \(X\) to \(\R \) that converge pointwise, but not uniformly, to a continuous function.

(Memory 1580) Let \(f_k\), \(f:[a,b]\to \R \). Suppose that each \(f_k\) is Riemann integrable and that \(f_k\to f\) uniformly on \([a,b]\). Then \(f\) is Riemann integrable, \(\lim _{k\to \infty }\int _a^b f_k\) exists, and \(\lim _{k\to \infty }\int _a^b f_k=\int _a^b f\).

(Memory 1590) (The Weierstrauß \(M\)-test.) Suppose that \(A\) is a set and that for each \(n\), \(f_n:A\to \C \) is a bounded function. Suppose that there is a sequence \(\{M_n\}_{n=0}^\infty \subset [0,\infty )\) such that \(|f_n(x)|\leq M_n\) for all \(x\in A\) and \(\sum _{n=0}^\infty M_n<\infty \). Then the series \(\sum _{n=0}^\infty f_n(x)\) converges absolutely and uniformly on \(A\).

3.2. Complex Power Series

(Dibyendu, Problem 1600) Let \(\sum _{n=0}^\infty a_n\) be a series of complex numbers. Show that if \(\sum _{n=0}^\infty |a_n|\) converges then \(\sum _{n=0}^\infty a_n\) converges (that is, that in the complex numbers, we still have that absolute convergence implies convergence).

Let \(a_n=x_n+iy_n\) where \(x_n\), \(y_n\in \R \). Then \(|x_n|\leq |a_n|\) and \(|y_n|\leq |a_n|\). We recall from real analysis that a nondecreasing sequence converges if and only if it is bounded. Therefore \(\{\sum _{n=0}^m |a_n|\}_{m\in \N }\) is bounded. Because \(|x_n|\leq |a_n|\), we have that \(\sum _{n=0}^m |x_n|\leq \sum _{n=0}^m|a_n|\leq \sup _{m\in \N } \sum _{n=0}^m |a_n|\) and so \(\{\sum _{n=0}^m |x_n|\}_{m\in \N }\) is bounded. Thus \(\sum _{n=0}^\infty x_n\) converges absolutely, and therefore \(\sum _{n=0}^\infty x_n\) converges. Similarly, \(\sum _{n=0}^\infty y_n\) converges. By Problem 220, \(\sum _{n=0}^\infty a_n\) converges.

Definition 3.2.2. (Complex power series.) A complex power series is a formal sum \(\sum _{k=0}^\infty a_k (z-P)^k\) for some \(\{a_k\}_{k=1}^\infty \subseteq \C \). The series converges at \(z\) if \(\lim _{n\to \infty } \sum _{k=0}^n a_k (z-P)^k\) exists.

Lemma 3.2.3. Suppose that the series \(\sum _{k=0}^\infty a_k (z-P)^k\) converges at \(z=w\) for some \(w\in \C \). Then the series converges absolutely at \(z\) for all \(z\) with \(|z-P|<|w-P|\).

Proposition 3.2.9. Suppose that the series \(\sum _{k=0}^\infty a_k (z-P)^k\) converges at \(z=w\) for some \(w\in \C \). If \(0<r<|w-P|\), then the series converges uniformly on \(\overline D(P,r)\).

(Hope, Problem 1610) Prove Lemma 3.2.3.

(Micah, Problem 1620) Prove Proposition 3.2.9.

Because \(\sum _{k=0} a_k(w-P)^k\) converges, we have that \(\lim _{k\to \infty } a_k(w-P)^k=0\). In particular, \(\{a_k(w-P)^k\}_{k=0}^\infty \) is bounded. Let \(A=\sup _{k\geq 0} |a_k(w-P)^k|\).

Because \(0<r/|w-P|<1\), the geometric series \(\sum _{k=0}^\infty A(r/|w-P|)^k\) converges. Thus for every \(\varepsilon >0\) there is an \(N>0\) such that \(\sum _{k=N}^\infty A(r/|w-P|)^k<\varepsilon \).

If \(z\in \overline D(P,r)\), then \(|z-P|\leq r\) and so \(|a_k(z-P)^k| \leq |a_k(w-P)^k | (r/|w-P|)^k\leq A(r/|w-P|)^k\). We then have that \(\sum _{k=N}^\infty |a_k(z-P)^k|\leq \sum _{k=N}^\infty A(r/|w-P|)^k<\varepsilon \) for all \(z\in \overline D(P,r)\). Furthermore, by Lemma 3.2.3 and Problem 1600 \(\sum _{k=0}^\infty a_k(z-P)^k\) exists, and if \(m\geq N\) then \begin {align*} \Bigl |\sum _{k=0}^\infty a_k(z-P)^k-\sum _{k=0}^m a_k(z-P)^k\Bigr | &=\Bigl |\sum _{k=m+1}^\infty a_k(z-P)^k\Bigr | \leq \sum _{k=m+1}^\infty |a_k(z-P)^k| <\varepsilon . \end {align*}

Thus the series converges uniformly on \(\overline D(P,r)\).

(Muhammad, Problem 1630) Suppose that the series diverges at \(w\) for some \(w\in \C \). Show that the series diverges at \(z\) for all \(z\) with \(|z-P|>|w-P|\).

Suppose for the sake of contradiction that the series converges at \(z\). By Lemma 3.2.3 with \(z\) and \(w\) interchanged, we know that the series converges at \(w\). But we assumed that the series diverged at \(w\), a contradiction. Therefore the series must diverge at \(z\).

Definition 3.2.4. (Radius of convergence.) The radius of convergence of \(\sum _{k=0}^\infty a_k (z-P)^k\) is \(\sup \{|w-P|:\sum _{k=0}^\infty a_k (w-P)^k\) converges\(\}\).

(Nisa, Problem 1640) Show that the radius of convergence is also \(\inf \{|w-P|:\sum _{k=0}^\infty a_k (w-P)^k\) diverges\(\}\).

Let \(R_1=\sup \{|w-P|:\sum _{k=0}^\infty a_k (w-P)^k\) converges\(\}\), \(R_2=\inf \{|\zeta -P|:\sum _{k=0}^\infty a_k (\zeta -P)^k\) diverges\(\}\).

If \(r\in \{|w-P|:\sum _{k=0}^\infty a_k (w-P)^k\) converges\(\}\), then \(r=|w-P|\) for some \(w\) such that the series converges. If the series diverges at \(\zeta \), then \(|\zeta -P|\geq |w-P|\) by Lemma 3.2.3, and so \(r\) is a lower bound on \(\{|\zeta -P|:\sum _{k=0}^\infty a_k (\zeta -P)^k\) diverges\(\}\). Thus \(r\leq R_2\). So \(R_2\) is an upper bound on \(\{|w-P|:\sum _{k=0}^\infty a_k (w-P)^k\) converges\(\}\), and so \(R_2\geq R_1\).

If \(R_2>R_1\), let \(z\in \C \) be such that \(R_1<|z-P|<R_2\). Then the series either converges or diverges at \(z\). If it converges, then \(|z-P|\in \{|w-P|:\sum _{k=0}^\infty a_k (w-P)^k\) converges\(\}\), and so \(|z-P|\leq R_1\), a contradiction. We similarly derive a contradiction if the series diverges at \(z\), and so we must have that \(R_2=R_1\), as desired.

(Robert, Problem 1650) The root test for real numbers states that if \(\{b_n\}_{n=0}^\infty \subset \R \), then

• If \(\limsup _{n\to \infty } \sqrt [n]{|b_n|} <1\), then \(\sum _{n=0}^\infty b_n\) converges absolutely.

• If \(\limsup _{n\to \infty } \sqrt [n]{|b_n|}>1\), then the sequence \(\{b_n\}_{n=0}^\infty \) is unbounded (and in particular the series \(\sum _{n=0}^\infty b_n\) diverges).

What does the root test tell us about complex power series?

The root test for real numbers states that if \(\{b_n\}_{n=0}^\infty \subset \R \), then

• If \(\limsup _{n\to \infty } \sqrt [n]{|b_n|} <1\), then \(\sum _{n=0}^\infty b_n\) converges absolutely.

• If \(\limsup _{n\to \infty } \sqrt [n]{|b_n|}>1\), then the sequence \(\{b_n\}_{n=0}^\infty \) is unbounded (and in particular the series \(\sum _{n=0}^\infty b_n\) diverges).

Let \(\sum _{k=0}^\infty a_k(z-P)^k\) be a complex power series. Fix a \(z\in \C \) and observe that \begin {equation*} \limsup _{k\to \infty }\sqrt [k]{|a_k(z-P)^k|} =|z-P|\limsup _{k\to \infty }\sqrt [k]{|a_k|}. \end {equation*}

Thus the series \(\sum _{k=0}^\infty a_k(z-P)^k\) converges if \(|z-P|\limsup _{k\to \infty }\sqrt [k]{|a_k|}<1\) and diverges if \(|z-P|\limsup _{k\to \infty }\sqrt [k]{|a_k|}>1\). Thus the radius of convergence must be \begin {equation*} \frac {1}{\limsup _{k\to \infty }\sqrt [k]{|a_k|}} \end {equation*} with the convention that \(\frac {1}{0}=\infty \) and \(\frac {1}{\infty }=0\); that is, if \(\limsup _{k\to \infty }\sqrt [k]{|a_k|}=0\) then the series converges everywhere and if \(\limsup _{k\to \infty }\sqrt [k]{|a_k|}=\infty \) then the series diverges unless \(z=P\).

(Sam, Problem 1660) State the ratio test from undergraduate real analysis. What does the ratio test say about power series?

The ratio test for real numbers states that if \(\{b_n\}_{n=0}^\infty \subset \R \), then

• If \(\lim _{n\to \infty } \frac {|b_{n+1}|}{|b_n|}\) exists and is less than \(1\), then \(\sum _{n=0}^\infty b_n\) converges absolutely.

• If \(\lim _{n\to \infty } \frac {|b_{n+1}|}{|b_n|}\) exists and is greater than \(1\), then the sequence \(\{b_n\}_{n=0}^\infty \) is unbounded (and in particular the series \(\sum _{n=0}^\infty b_n\) diverges).

Let \(\sum _{k=0}^\infty a_k(z-P)^k\) be a complex power series. Fix a \(z\in \C \setminus \{P\}\) and observe that if either \(\lim _{k\to \infty } \frac {|a_{k+1}(z-P)^{k+1}|}{|a_k(z-P)^k|}\) or \(\lim _{k\to \infty }\frac {|a_{k+1}|}{|a_k|}\) exists, then the other must exist and \begin {equation*} \lim _{k\to \infty } \frac {|a_{k+1}(z-P)^{k+1}|}{|a_k(z-P)^k|} = |z-P|\lim _{k\to \infty }\frac {|a_{k+1}|}{|a_k|}. \end {equation*} Thus, the series converges absolutely if \(|z-P|<\lim _{k\to \infty }\frac {|a_k|}{|a_{k+1}|}\) and diverges if \(|z-P|>\lim _{k\to \infty }\frac {|a_k|}{|a_{k+1}|}\), so the radius of convergence must be \(\lim _{k\to \infty }\frac {|a_k|}{|a_{k+1}|}\).

Lemma 3.2.10. Let \(\sum _{k=0}^\infty a_k (z-P)^k\) be a power series with radius of convergence \(R>0\). Define \(f:D(P,R)\to \C \) by \(f(z)=\sum _{k=0}^\infty a_k (z-P)^k\).

Then \(f\) is \(C^\infty \) and holomorphic in \(D(P,R)\), and if \(n\in \N \) then the series \begin {equation*} \sum _{k=n}^\infty \frac {k!}{(k-n)!}a_k (z-P)^{k-n} \end {equation*} has radius of convergence at least \(R\) and converges to \(f^{(n)}(z)=\frac {\partial ^n f}{\partial z^n}\).

(Wilson, Problem 1670) Begin the proof of Lemma 3.2.10 by showing that \(f\) is continuous on \(D(P,R)\).

If \(m\in \N \), define \(f_m(z)=\sum _{k=0}^m a_k(z-P)^k\). Then each \(f_m\) is continuous. By Proposition 3.2.9, if \(0<r<R\) then \(f_m\to f\) uniformly on \(\overline D(P,r)\).

By Memory 1560, we have that \(f\) must be continuous on \(\overline D(P,r)\). But if \(z\in D(P,R)\), then \(|z-P|<R\) and so there is a \(\varepsilon >0\) and an \(r<R\) such that \(D(z,\varepsilon )\subset \overline D(P,r)\), and so \(f\) is continuous at \(z\) for all \(z\in D(P,R)\). Thus \(f\) is continuous on \(D(P,R)\).

(Adam, Problem 1680) Continue the proof of Lemma 3.2.10 by showing that \(f\) is holomorphic in \(D(P,R)\). You may either follow the proof given in the book, use Morera’s theorem, or use another method of proof of your choice (I find Morera’s theorem to be the simplest method of proof).

(Amani, Problem 1690) Complete the proof of Lemma 3.2.10 by showing that \begin {equation*} \sum _{k=n}^\infty \frac {k!}{(k-n)!}a_k (z-P)^{k-n} \end {equation*} indeed converges to \(f^{(n)}(z)\). Hint: Use Theorem 3.1.1 and Memory 1580.

[Definition: Taylor series] Let \(P\in \Omega \subseteq \C \) where \(\Omega \) is open, and let \(f\) be holomorphic in \(\Omega \). By Theorem 3.1.1 (Problem 1440), \(f^{(n)}\) exists everywhere in \(\Omega \). The Taylor series for \(f\) at \(P\) is the power series \(\sum _{k=0}^\infty \frac {f^{(k)}(P)}{k!}(z-P)^k\).

(Bashar, Problem 1700) Let \(f\) be as in Lemma 3.2.10. Show that the Taylor series for \(f\) at \(P\) is simply \(\sum _{k=0}^\infty a_k (z-P)^k\).

Proposition 3.2.11. Suppose that the two power series \(\sum _{k=0}^\infty a_k (z-P)^k\) and \(\sum _{k=0}^\infty b_k (z-P)^k\) both have positive radius of convergence and that there is some \(r>0\) such that \(\sum _{k=0}^\infty a_k (z-P)^k=\sum _{n=0}^\infty b_k (z-P)^k\) (and both sums converge) whenever \(|z-P|<r\). Then \(a_k=b_k\) for all \(k\).

(Dibyendu, Problem 1710) Prove Proposition 3.2.11.

[Definition: Analytic function] Let \(\Omega \subseteq \C \) be open and let \(f:\Omega \to \C \) be a function. If for every \(P\in \Omega \) there is a \(r>0\) with \(D(P,r)\subseteq \Omega \) and a sequence \(\{a_n\}_{n=1}^\infty \subset \C \) such that \(f(z)=\sum _{n=0}^\infty a_n (z-P)^n\) for all \(z\in D(P,r)\), we say that \(f\) is analytic.

(Problem 1720) Show that analytic functions are holomorphic.

(Problem 1730) Recall that if \(x\in \R \), then \(\exp x=\sum _{n=0}^\infty \frac {x^n}{n!}\), \(\sin x=\sum _{n=0}^\infty \frac {(-1)^n}{(2n+1)!}x^{2n+1}\), \(\cos x=\sum _{n=0}^\infty \frac {(-1)^n}{(2n)!}x^{2n}\). Show that the functions \(\exp z=\sum _{n=0}^\infty \frac {z^n}{n!}\), \(\sin z=\sum _{n=0}^\infty \frac {(-1)^n}{(2n+1)!}z^{2n+1}\), and \(\cos z=\sum _{n=0}^\infty \frac {(-1)^n}{(2n)!}z^{2n}\) are holomorphic on \(\C \) and take the correct values at all real numbers.

3.3. The Power Series Expansion for a Holomorphic Function

Theorem 3.3.1. Let \(\Omega \subseteq \C \) be an open set and let \(f\) be holomorphic in \(\Omega \). Let \(D(P,r)\subseteq \Omega \) for some \(r>0\).

Then the Taylor series for \(f\) at \(P\) has radius of convergence at least \(r\) and converges to \(f(z)\) for all \(z\in D(P,r)\).

(Hope, Problem 1740) Let \(f\) be holomorphic in \(D(P,R)\) and let \(0<r<R\). Begin the proof of Theorem 3.3.1 by showing that there is a power series with radius of convergence at least \(r\) that converges to \(f\) in \(D(P,r)\).

(Micah, Problem 1750) Complete the proof of Theorem 3.3.1 by showing that the power series for \(f\) in \(D(P,r)\) must be the Taylor series for \(f\) at \(P\) and that the radius of convergence of the Taylor series for \(f\) at \(P\) must be at least \(R\).

(Muhammad, Problem 1760) Let \(f\) be holomorphic in \(D(P,r)\). Let \(R\) be the radius of convergence of the Taylor series for \(f\) at \(P\). Observe that \(R\geq r\). Suppose \(R>r\). Show that there is a unique function \(F\) that is holomorphic in \(D(P,R)\) with \(F=f\) in \(D(P,r)\).

(Nisa, Problem 1770) Let \(f\) be an analytic function in a neighborhood of \(P\). Show that the Taylor series for \(f'\) at \(P\) has the same radius of convergence as the Taylor series for \(f\) at \(P\).

(Robert, Problem 1780) Let \(\Omega =\{re^{i\theta }:0<r<\infty ,\>-\pi <\theta <\pi \}=\C \setminus (-\infty ,0]\). Define \(F:\Omega \to \C \) by \(F(re^{i\theta })=\ln r+i\theta \) whenever \(-\pi <\theta <\pi \). Recall that \(F\) is holomorphic and that \(F'(z)=\frac {1}{z}\) for all \(z\in \Omega \).

(a)

Show that the Taylor series for \(F\) at \(-3+4i\) has radius of convergence \(5\).

(b)

Can we extend \(F\) to a function that is holomorphic on \(\Omega \cup D(-3+4i,5)\)?

(Fact 1790) Recall from Proposition 1.4.3 (Problem 660) that if \(f\) is holomorphic in \(\Omega \) then \(\p [f]{z}=\p [f]{x}\) in \(\Omega \). By Corollary 3.1.2 (Problem 1440), \(\p [f]{z}=\p [f]{x}\) is holomorphic in \(\Omega \). A straightforward induction argument yields that \(\frac {\partial ^n f}{\partial z^n}=\frac {\partial ^n f}{\partial x^n}\) in \(\Omega \) for all \(n\in \N \).

(Sam, Problem 1800) Show that the functions \(\exp \), \(\sin \), and \(\cos \) in Problem 1730 are the only functions that are holomorphic on all of \(\C \) and take the correct values for all real numbers.

(Wilson, Problem 1810) Suppose that \(\sum _{n=0}^\infty a_n z^n\) and \(\sum _{n=0}^\infty b_n z^n\) are two power series with radius of convergence at least \(r\). Show that \begin {equation*} \sum _{n=0}^\infty \Bigl (\sum _{k=0}^n a_k\, b_{n-k}\Bigr )z^n \end {equation*} has radius of convergence at least \(r\) and that \begin {equation*} \sum _{n=0}^\infty \Bigl (\sum _{k=0}^n a_k\, b_{n-k}\Bigr )z^n = \biggl (\sum _{n=0}^\infty a_n z^n\biggr )\biggl (\sum _{n=0}^\infty b_n z^n\biggr ) \end {equation*} for all \(|z|<r\).