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: Friday, November 7, 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

(Problem 10)

Solution

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)

(Problem 30)

(Problem 40)

[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)

(Problem 60)

(Problem 70)

(Problem 80)

(Problem 90)

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

(Problem 100)

(Problem 110)

(Problem 120)

(Problem 130)

(Problem 140)

Solution

\(z\bar z\) 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)

(Problem 160)

(Problem 170)

1.2. Real Analysis

(Fact 180)

(Fact 190)

(Bashar, Problem 200)

Solution

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)

(Dibyendu, Problem 220)

Solution

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

(Memory 221)

(Memory 230)

(Memory 240)

(Memory 250)

(Memory 270)

(Memory 280)

1.2. Further Properties of the Complex Numbers

(Hope, Problem 290)

Solution

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)

(Problem 310)

Solution

It converges to \(e^x\).

(Micah, Problem 320)

Solution

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)

[Definition: The complex exponential] If \(x\) is real, we define

(Muhammad, Problem 340)

Solution

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)

Solution

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)

Solution

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)

Solution

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)

Solution

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)

[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)

Solution

\(p\) 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)

Solution

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)

Solution

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)

Solution

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)

Solution

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)

(Memory 450)

(Memory 460)

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)

Solution

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)

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)

(Fact 500)

[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)

(Problem 511)

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

(Muhammad, Problem 520)

(Nisa, Problem 530)

Solution

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

(Robert, Problem 540)

Solution

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)

Solution

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

(Fact 560)

(Fact 570)

Solution

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)

Solution

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)

Solution

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

(Fact 600)

(Wilson, Problem 610)

Solution

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

(Adam, Problem 620)

Solution

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:

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

1.4. Real Analysis

(Memory 630)

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

(Amani, Problem 640)

(Bashar, Problem 650)

Solution

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.

(Problem 651)

Solution

Recall that \(\exp \) is given by \(\exp (x+iy)=e^x(\cos y+i\sin y)=e^x\cos y+ie^x\sin y\).

Writing \(u(x,y)=e^x\cos y\), \(v(x,y)=e^x\sin y\), we have that

\begin{equation*}\frac {\partial u}{\partial x} = e^x\cos y=\frac {\partial v}{\partial y}, \quad \frac {\partial u}{\partial y} = -e^x\sin y=-\frac {\partial v}{\partial x},\end{equation*}

and so \(u\) and \(v\) satisfy the Cauchy-Riemann equations; thus \(f(x,y)=u(x,y)+iv(x,y)=\exp (x+iy)\) is holomorphic.

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

(Dibyendu, Problem 660)

Solution

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)

Solution

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

(Fact 680)

Solution

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)

Solution

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)

Solution

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)

Solution

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)

Solution

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)

Solution

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)

(Memory 750)

(Problem 760)

(Problem 770)

Solution

This will be proven as homework.

1.5. Real and Holomorphic Antiderivatives

(Sam, Problem 780)

(Bonus Problem 790)

(Problem 800)

Solution

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)

(Wilson, Problem 820)

Solution

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

(Adam, Problem 830)

Solution

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)

[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)

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)

(Memory 860)

[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)

(Memory 871)

(Memory 872)

(Problem 873)

Solution

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)

(Memory 890)

(Bashar, Problem 900)

Solution

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)

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)

[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

[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

(Dibyendu, Problem 930)

Solution

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

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)

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)

Solution

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

(Micah, Problem 950)

Solution

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

(Problem 951)

(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)

Show that

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

(Problem 961)

Solution

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)

Solution

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)

Solution

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)

(William, Problem 1000)

(Problem 1010)

(Problem 1011)

(Problem 1012)

(Bonus Problem 1013)

(Proposition 1014)

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)

Solution

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)

Solution

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)

(Amani, Problem 1050)

Solution

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)

Solution

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)

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)

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)

(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)

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

Solution

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)

(Problem 1082)

(Problem 1083)

(Fact 1090)

[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

If in addition \(g(p)\neq 0\) then \(f/g\) has a complex derivative at \(p\) and

(Fact 1100)

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)

Solution

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)

Solution

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)

(Nisa, Problem 1140)

Solution

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)

Solution

Define \(\gamma :[0,1]\to \Omega \) by \(\gamma (t)=p+t(z-p)\). Then by Proposition 2.1.6, 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)

(Problem 1161)

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)

[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)

[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)

[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

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

(Adam, Problem 1180)

[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)

(Amani, Problem 1190)

2.3. Real Analysis

(Memory 1200)

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)

Solution

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)

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)

Solution

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)

Solution

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)

(Memory 1260)

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

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)

(Muhammad, Problem 1280)

(Nisa, Problem 1290)

Solution

\(\oint _\gamma \frac {1}{\zeta -z_0}\,d\zeta =1\) (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)

Solution

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)

Solution

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)

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

(Adam, Problem 1330)

(Bonus Problem 1340)

[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)

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)

[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)

(Bashar, Problem 1380)

(Problem 1381)

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 \([c,d]\).

Solution

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)

Show that

Solution

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)

Solution

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)

(Hope, Problem 1410)

Solution

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)

(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

(Muhammad, Problem 1430)

[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

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)

Solution

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)

Solution

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)

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)

Solution

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)

3.2. Real Analysis

(Memory 1490)

[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)

(Memory 1510)

(Adam, Problem 1520)

Solution

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)

(Bonus Problem 1540)

[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)

(Memory 1560)

(Bashar, Problem 1570)

(Memory 1580)

(Memory 1590)

3.2. Complex Power Series

(Dibyendu, Problem 1600)

Solution

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)

(Micah, Problem 1620)

Solution

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)

Solution

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)

Solution

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)

• 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?

Solution

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)

Solution

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

(Wilson, Problem 1670)

Solution

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)

(Amani, Problem 1690)

[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, \(f^{(n)}\) exists everywhere in \(\Omega \). The Taylor series for \(f\) at \(P\) is the power series \(\sum _{k=0}^\infty \frac {f^{(k)}(P)}{k!}(z-P)^k\).

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 \(k\).

(Bashar, Problem 1700)

(Dibyendu, Problem 1710)

Solution

We will solve the previous two problems using a single calculation. Let \(f(z)=\sum _{k=0}^\infty a_k(z-P)^k\) in \(D(P,r)\). Then by Lemma 3.2.10, we must have that \(f\) is holomorphic in \(D(P,r)\), and if \(n\geq 0\) is an integer then

\begin{equation*}\frac {f^{(n)}(P)}{n!} =\sum _{k=n}^\infty \frac {k!}{n!(k-n)!} a_k (P-P)^{k-n} .\end{equation*}

Recall that in power series we define \(0^0=1\). If \(k>n\) then \((P-P)^{k-n}=0\), and so we have that

\begin{equation*}\frac {f^{(n)}(P)}{n!} =\sum _{k=n}^n \frac {k!}{n!(k-n)!} a_k (P-P)^{k-n} =a_n .\end{equation*}

Thus the Taylor series for \(f\) at \(P\) by definition is

\begin{equation*}\sum _{n=0}^\infty \frac {f^{(n)}(P)}{n!} (z-P)^n = \sum _{n=0}^\infty a_n (z-P)^n.\end{equation*}

This completes Problem Problem 1700.

Furthermore, if \(a_n\) and \(b_n\) are as in Proposition 3.2.11 and \(f(z)=\sum _{k=0}^\infty a_k(z-P)^k=\sum _{k=0}^\infty b_k(z-P)^k\) in \(D(P,r)\) for some \(r>0\), then \(a_n=\frac {f^{(n)}(P)}{n!}=b_n\) and so \(a_n=b_n\) for all \(n\).

[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)

(Problem 1730)

(We will show in Section 3.3 that \(\mathop {\widetilde {\mathrm {exp}}} (z)=\exp (z)\), where \(\exp (z)\) is as defined in Section 1.2.)

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)

Solution

Without loss of generality we may assume \(P=0\). By the Cauchy integral formula (Theorem 2.4.2), if \(z\in D(0,r)=D(P,r)\) and \(|z|<\varrho <r\), then

\begin{equation*}f(z)=\oint _{\partial D(0,\varrho )} \frac {f(\zeta )}{\zeta -z}\,d\zeta .\end{equation*}

We must take \(\varrho <r\) to ensure that \(f\) is continuous up to \(\partial D(0,\varrho )\).

Because \(|z|<|\zeta |\) all \(\zeta \in \partial D(0,\varrho )\), we have that

\begin{equation*}\frac {1}{\zeta -z}=\frac {1/\zeta }{1-z/\zeta } =\sum _{k=0}^\infty \frac {z^k}{\zeta ^{k+1}}.\end{equation*}

Thus

\begin{equation*}f(z)=\oint _{\partial D(0,\varrho )}\sum _{k=0}^\infty \frac {f(\zeta )}{\zeta ^{k+1}}z^k\,d\zeta .\end{equation*}

Because \(f\) is continuous on the compact set \(\partial D(0,\varrho )\), it is bounded there. Let \(m=\sup _{\partial D(0,\varrho )}|f|\). Then \(|\frac {f(\zeta )}{\zeta ^{k+1}}z^k|\leq m|z|^k/\varrho ^{k+1}\). But \(\sum _{k=0}^\infty m|z|^k/\varrho ^{k+1}\) is a convergent geometric series, so by the Weierstrauß \(M\)-test we have that the series converges uniformly on \(\partial D(0,r)\). Thus by Memory 1580 we may interchange the sum with the integral and see that

\begin{equation*}f(z)=\sum _{k=0}^\infty z^k\oint _{\partial D(0,\varrho )} \frac {f(\zeta )}{\zeta ^{k+1}}\,d\zeta \end{equation*}

and the sum converges.

Let \(a_k(\varrho )=\oint _{\partial D(0,\varrho )} \frac {f(\zeta )}{\zeta ^{k+1}}\,d\zeta \) for any \(\varrho \in (0,r)\); we have shown that \(\sum _{k=0}^\infty a_k(\varrho )\, z^k\) converges in \(D(0,\varrho )\) to \(f(z)\). But by Proposition 3.2.11 we have that \(a_k(\varrho )\) is independent of \(\varrho \), and so we must have that \(\sum _{k=0}^\infty a_k(\varrho )\, z^k\) converges in \(D(0,r)\) to \(f(z)\), as desired.

(Micah, Problem 1750)

Solution

By the previous problem, we know that if \(P\in \Omega \) then there are complex numbers \(a_k\) such that \(\sum _{k=0}^\infty a_k(z-P)^k\) has radius of convergence at least \(r\) and \(f(z)=\sum _{k=0}^\infty a_k(z-P)^k\) for all \(z\in D(P,r)\).

But by Problem 1700, we have that the Taylor series for \(f\) is simply \(\sum _{k=0}^\infty a_k(z-P)^k\), and by assumption \(\sum _{k=0}^\infty a_k(z-P)^k\) converges to \(f\) in \(D(P,r)\), and so the Taylor series converges to \(f\) in \(D(P,r)\).

(Muhammad, Problem 1760)

Solution

Define \(F:D(P,R)\to \C \) by \(F(z)=\sum _{k=0}^\infty \frac {f^{(k)}(P)}{k!}(z-P)^k\). By assumption the power series does converge in \(D(P,R)\), and by Lemma 3.2.10 \(F\) is holomorphic in \(D(P,R)\).

Suppose for the sake of contradiction that \(G\) is holomorphic in \(D(P,R)\) and that \(G=f=F\) in \(D(P,r)\). A straightforward induction argument shows that if \(n\in \N \) then \(G^{(n)}=f^{(n)}=F^{(n)}\) in \(D(P,r)\), and so in particular \(G^{(n)}(P)=F^{(n)}(P)\). But by Theorem 3.3.1 we have that

\begin{equation*}F(z)=\sum _{n=0}^\infty \frac {F^{(n)}(P)}{n!}(z-P)^n, \qquad G(z)=\sum _{n=0}^\infty \frac {G^{(n)}(P)}{n!}(z-P)^n\end{equation*}

for all \(z\in D(P,R)\), and so we must have that \(G=F\) in \(D(P,R)\).

(Nisa, Problem 1770)

Solution

Let \(r\) be the radius of convergence of the Taylor series for \(f\) at \(P\). By Lemma 3.2.10 we have that

\begin{equation*}f'(z)=\sum _{k=0}^\infty \frac {f^{(k+1)}(P)}{k!}z^k\end{equation*}

in \(D(P,r)\), and in particular that the series has radius of convergence at least \(r\).

Suppose that the power series for \(f'\) at \(P\) has radius of convergence \(R\); by the above we have that \(R\geq r\). Then

\begin{equation*}g(z)=\sum _{k=0}^\infty \frac {f^{(k+1)}(P)}{k!}z^k\end{equation*}

is holomorphic in \(D(P,R)\) and equals \(f'\) in \(D(P,r)\). By Problem 840 there is a holomorphich function \(G:D(P,R)\to \C \) such that \(G'=g\) in \(D(P,R)\). By adding a constant, we may require that \(G(P)=f(P)\).

By Theorem 3.3.1 we have that the Taylor series for \(G\) at \(P\) has radius of convergence at least \(R\) and converges to \(G\) in \(D(P,R)\).

But the Taylor series for \(G\) is

\begin{equation*}\sum _{n=0}^\infty \frac {G^{(n)}(P)}{n!} (z-P)^n = G(P)+\sum _{n=1}^\infty \frac {g^{(n-1)}(P)}{n!} (z-P)^n = f(P)+\sum _{n=1}^\infty \frac {f^{(n)}(P)}{n!} (z-P)^n \end{equation*}

because \(G'=g\) in \(D(P,R)\) and so \(G^{(n)}=g^{(n-1)}\), \(G(P)=f(P)\), and \(g=f'\) in \(D(P,r)\) and so \(g^{(k)}(P)=(f')^{(k)}(P)=f^{(k+1)}(P)\). This is the Taylor series for \(f\), and so we must have that the Taylor series for \(f\) has radius of convergence at least \(R\) as well. Thus \(R\geq r\) and \(r\geq R\), so \(r=R\), as desired.

(Robert, Problem 1780)

(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)

(Sam, Problem 1800)

Solution

Let \(g:\R \to \R \) be infinitely differentiable and suppose that there exists a function \(f\) that is holomorphic on all of \(\C \) and satisfies \(f(x)=g(x)\) for all \(x\in \R \).

Note that \(g'(x)\) denotes the real derivative \(\lim _{\substack {y\to x\\y\in \R }}\frac {g(y)-g(x)}{y-x}\), while \(f'(z)\) denotes the complex derivative \(\lim _{\substack {w\to z\\w\in \C }}\frac {f(w)-f(z)}{w-z}\).

We then have that \(\frac {\partial ^n }{\partial x^n}f(x)=\frac {d^n}{dx^n} g(x)=g^{(n)}(x)\) for all \(x\in \R \). By the previous fact, \(\frac {\partial ^n }{\partial z^n}f(z)\big \vert _{z=x}=\frac {d^n}{dx^n}g(x)\) for all \(x\in \R \). By Theorem 2.2.1 and Theorem 2.2.2, we have that

\begin{equation*}f^{(n)}(x)=\frac {\partial ^n }{\partial z^n}f(z)\bigg \vert _{z=x}=\frac {d^n}{dx^n}g(x)\end{equation*}

for all \(x\in \R \). By Theorem 3.3.1, we must have that

\begin{equation*}f(z)=\sum _{n=0}^\infty \frac {f^{(n)}(0)}{n!} z^n=\sum _{n=0}^\infty \frac {g^{(n)}(0)}{n!} z^n\end{equation*}

for all \(z\in \C \). In particular, for a given \(g\) there is at most one function \(f\) that satisfies the given conditions, and so the functions given in Problem 1730 by power series are the unique functions that satisfy these conditions with \(g(x)=\exp x\), \(g(x)=\cos x\) or \(g(x)=\sin x\).

In particular, by Problem 651, \(\exp z\) is holomorphic in \(\C \) and satisfies \(\exp x= \mathop {\widetilde {\mathrm {exp}}} x=e^x\) for all \(x\in \R \), and so \(\exp z=\mathop {\widetilde {\mathrm {exp}}} z\) for all \(z\in \C \).

(Wilson, Problem 1810)

Solution

Let \(f(z)=\sum _{n=0}^\infty a_n z^n\) and let \(g(z)=\sum _{n=0}^\infty b_n z^n\); by Problem 1680, \(f\) and \(g\) are both holomorphic in \(D(0,r)\). Thus \(h=fg\) is holomorphic in \(D(0,r)\). By Problem 1740 and Problem 1750, we must have that

\begin{equation*}\sum _{n=0}^\infty \frac {h^{(n)}(0)}{n!} z^n\end{equation*}

has radius of convergence at least \(r\) and must converge to \(h(z)=f(z)g(z)\). A straightforward induction argument and the Leibniz rule (Fact 570) shows that

\begin{equation*}h^{(n)}(z)=\sum _{k=0}^n\frac {n!}{k!(n-k)!} f^{(k)}(z)\, g^{(n-k)}(z) \end{equation*}

and so by Lemma 3.2.10 we have that

\begin{equation*}\frac {h^{(n)}(0)}{n!}=\sum _{k=0}^n a_k\,b_{n-k}.\end{equation*}

3.4. The Cauchy Estimates and Liouville’s Theorem

Theorem 3.4.1. (The Cauchy estimates) Let \(f:\Omega \to \C \) be holomorphic and let \(\overline D(P,r)\subseteq \Omega \). Let \(k\in \N _0\). Then

(Adam, Problem 1820)

Solution

Let \(\gamma :[0,2\pi ]\to \C \) be given by \(\gamma (t)=e^{it}\), so \(\gamma \) is a parameterization of \(\partial D(P,r)\). By Theorem 3.1.1,

\begin{equation*}f^{(k)}(P)=\frac {k!}{2\pi i}\oint _\gamma \frac {f(\zeta )}{(\zeta -P)^{k+1}}d\zeta .\end{equation*}

By Problem 980,

\begin{equation*}\biggl |\oint _\gamma \frac {f(\zeta )}{(\zeta -P)^{k+1}}d\zeta \biggr | \leq \ell (\gamma ) \sup _{\zeta \in \widetilde \gamma } \frac {|f(\zeta )|}{|\zeta -P|^{k+1}}\end{equation*}

where \(\ell (\gamma )=\int _0^{2\pi }|\gamma '(t)|\,dt=2\pi r\). But \(\widetilde \gamma =\partial D(P,r)\), and so if \(\zeta \in \widetilde \gamma \) then \(|\zeta -P|=r\). Thus this reduces to

\begin{equation*}|f^{(k)}(P)|\leq \frac {k!}{2\pi } 2\pi r \sup _{\zeta \in \widetilde \gamma } \frac {|f(\zeta )|}{r^{k+1}} =\frac {k!}{r^k} \sup _{\zeta \in \widetilde \gamma } |f(\zeta )| \end{equation*}

as desired.

(Amani, Problem 1830)

Lemma 3.4.2. If \(f\) is holomorphic on a connected open set \(\Omega \) and \(\p [f]{z}=0\) in \(\Omega \), then \(f\) is constant. (This was proven in Problem 610.)

[Definition: Entire] A function \(f:\C \to \C \) is entire if \(f\) is holomorphic on all of \(\C \).

Theorem 3.4.3. [Liouville’s theorem.] A bounded entire function is constant.

(Bashar, Problem 1840)

Solution

Let \(f:\C \to \C \) be entire and let \(M\in \R \) be such that \(|f(z)|\leq M\) for all \(z\in \C \). By assumption \(M\) exists.

By the Cauchy estimates (Theorem 3.4.1),

\begin{equation*}|f'(z)|\leq \frac {M}{r}\end{equation*}

for all \(r>0\). Taking the limit as \(r\to \infty \) yields that \(f'(z)=0\) for all \(z\) and the result follows from Lemma 3.4.2.

(Dibyendu, Problem 1850)

Solution

By Theorem 3.3.1, we have that

\begin{equation*}f(z)=\sum _{n=0}^\infty \frac {f^{(n)}(P)}{n!}(z-P)^n.\end{equation*}

But \(f^{(k+1)}\equiv 0\) in \(D(P,r)\), and therefore all derivatives of order higher than \(k\) are zero. We then have that

\begin{equation*}f(z)=\sum _{n=0}^k \frac {f^{(n)}(P)}{n!}(z-P)^n\end{equation*}

which is a polynomial of degree at most \(k\).

(Bonus Problem 1860)

[Exercise: Theorem 3.4.4]. If \(f\) is entire and there is a constant \(C\in \R \) and a \(k\in \N _0\) such that \(|f(z)|\leq C+C|z|^k\) for all \(z\in \C \), then \(f\) is a polynomial of degree at most \(k\).

Theorem 3.4.5. (The fundamental theorem of algebra.) Let \(p\) be a nonconstant (holomorphic) polynomial. Prove that \(p\) has a root; that is, prove that there is an \(\alpha \in \C \) with \(p(\alpha )=0\).

[Chapter 3, Problem 36] Let \(p(z)=a_0+a_1z+\dots +a_nz^n\) be a polynomial. Show that there is an \(R\in (0,\infty )\) such that if \(|z|\geq R\), then \(|p(z)|\geq |a_n|\,|z|^n/2\).

(Hope, Problem 1870)

(Micah, Problem 1880)

Solution

Yes. Clearly every linear polynomial (\(k=1\)) can be factored completely.

Suppose that all polynomials of degree at most \(k\) can be factored completely. Let \(p\) be a polynomial of degree \(k+1\).

By the Fundamental Theorem of Algebra, there is a root \(r_1\in \C \) such that \(p(r_1)=0\). By the standard division algorithm, this means that there is a polynomial \(q\) of degree \(k\) such that \(p(z)=(z-r_1)q(z)\). By our induction hypothesis \(q\) may be factored completely, and so \(p\) may be factored completely.

3.5. Uniform Limits of Holomorphic Functions

[Definition: Domain] A domain is a connected open subset of \(\C \).

Theorem 3.5.1. Let \(\Omega \subseteq \C \) be a domain. Let \(f_n\), \(f:\Omega \to \C \). Suppose that each \(f_n\) is holomorphic in \(\Omega \) and that if \(K\subset \Omega \) is compact, then \(f_n\to f\) uniformly on \(K\). Then \(f\) is holomorphic in \(\Omega \).

(Muhammad, Problem 1890)

Solution

Suppose that \(P\in \C \), \(r>0\), and \(\overline D(P,r)\subset \Omega \). Then by the Cauchy integral formula,

\begin{equation*}f_n(z)=\frac {1}{2\pi i} \oint _{\partial D(P,r)} \frac {f_n(\zeta )}{\zeta -z}\,d\zeta \end{equation*}

for all \(z\in D(P,r)\) and all \(n\).

But \(\partial D(P,r)\subset \Omega \) is compact, and so by assumption \(f_n\to f\) uniformly on \(\partial D(P,r)\). For any fixed \(z\in D(P,r)\), \(\frac {1}{\zeta -z}\) is bounded on \(\partial D(P,r)\), and so \(\frac {f_j(\zeta )}{\zeta -z}\to \frac {f(\zeta )}{\zeta -z}\) uniformly on \(\partial D(P,r)\). By assumption, if \(z\in D(P,r)\) then \(f_n(z)\to f(z)\), while by Memory 1580

\begin{equation*}\lim _{n\to \infty }\oint _{\partial D(P,r)} \frac {f_n(\zeta )}{\zeta -z}\,d\zeta =\oint _{\partial D(P,r)} \frac {f(\zeta )}{\zeta -z}\,d\zeta . \end{equation*}

In particular,

\begin{equation*}f(z)=\frac {1}{2\pi i} \oint _{\partial D(P,r)} \frac {f(\zeta )}{\zeta -z}\,d\zeta \end{equation*}

Because each \(f_n\) is continuous on \(\Omega \) and \(f_n\to f\) uniformly on \(\partial D(P,r)\), we have that \(f\) is continuous on \(\partial D(P,r)\). By Theorem 3.1.3, this implies that \(f\) is holomorphic in \(D(P,r)\), as desired.

[Chapter 3, Problem 4] Prove Theorem 3.5.1 using Morera’s theorem.

(Nisa, Problem 1900)

Solution

Let \(f_j(x)=\frac {x}{\sqrt {x^2+1/j^2}}\). \(f_j\) is infinitely differentiable on \(\R \), but \(\{f_j\}_{j=1}^\infty \) converges uniformly to \(|x|\).

Corollary 3.5.2. Let \(\Omega \subseteq \C \) be a domain. Let \(f_n\), \(f:\Omega \to \C \). Suppose that each \(f_n\) is holomorphic in \(\Omega \) and that if \(K\subset \Omega \) is compact, then \(f_n\to f\) uniformly on \(K\). Then \(\frac {\partial }{\partial z} f_n\to \frac {\partial }{\partial z} f\) on \(\Omega \) and the convergence is uniform on all compact subsets \(X\) of \(\Omega \).

(Robert, Problem 1910)

Solution

Let \(X\subset \Omega \) be compact. Because \(\Omega \) is open, a standard real analysis argument yields that there is a \(r>0\) and finitely many points \(z_1\), \(z_2,\dots z_N\) such that \(X\subset \cup _{n=1}^N D(z_n,r/3)\).

Let \(K=\cup _{n=1}^N \overline D(z_n, 2r/3)\). Then \(K\) is the union of finitely many compact sets, and so is compact. Furthermore, if \(z\in X\) then \(z\in D(z_\ell ,r/3)\) for some \(\ell \) and so \(\overline D(z,r/3)\subset D(z_\ell ,2r/3)\subset K\).

By assumption \(f_n\to f\) uniformly on \(K\). That is, for every \(\varepsilon >0\) there is a \(M\in \N \) such that if \(n\geq M\) then \(|f_n-f|<\varepsilon \) on \(K\).

By Theorem 3.5.1, \(f\) and thus \(f_n-f\) is holomorphic, and so by Theorem 3.4.1 we have that if \(n\geq M\) and \(z\in X\) then

\begin{equation*}|f'(z)-f_n'(z)|\leq \frac {3}{r}\sup _{\partial D(z,r/3)} |f-f_n|\leq \frac {3\varepsilon }{r}.\end{equation*}

Thus \(f_n'\to f'\) uniformly on \(X\), as desired.

(Sam, Problem 1920)

Solution

Let \(f_n(x)=\frac {1}{n}\sin (n^2x)\). Then \(f_n\to 0\) uniformly but \(f_n'(x)=n\cos (n^2x)\) does not converge to zero.

3.6. Real Analysis

(Wilson, Problem 1930)

Solution

If \(k\geq 1\), then \(\ln ^{(k)}(x)=(-1)^{k-1} (k-1)! x^{-k}\). So the Taylor series for \(\ln \) expanded about \(c\) is

\begin{equation*}\ln (c)+\sum _{k=1}^\infty \frac {(-1)^{k-1} }{kc^k} (x-c)^k.\end{equation*}

By either the ratio or root test, we have that the series converges whenever \(|x-c|<c\).

It remains to show that the series in fact converges to the function \(\ln x\). We have that \(f(x)=\ln x\) is infinitely differentiable on \((0,\infty )\). Suppose that \(m\geq 2\) is an integer.

Recall from Memory 1500 that if \(x>c\) then

\begin{align*}\ln x &= \ln (c)+\sum _{k=1}^{m-1}\frac {(-1)^{k-1} }{kc^k} (x-c)^k + \frac {(-1)^{m-1}}{m (y_m)^{m} } (x-c)^m\end{align*}

for some \(y_m\) with \(c<y_m<x\). If \(0<x<c\) then we will use a different form of the Taylor remainder, namely

\begin{align*}\ln x&= \ln (c)+\sum _{k=1}^{m-1}\frac {(-1)^{k-1} }{kc^k} (x-c)^k + \frac {(-1)^{m-1}}{m (y_m)^{m} } (x-y_m)^{m-1}(x-c)\end{align*}

for some \(y_m\) with \(x<y_m<c\).

In the first case, \(0<x-c<c<y_m\) and so \(|\frac {x-c}{y_m}|^m\leq 1\) for all \(m\). (Recall that \(y_m\) may depend on \(m\) and so we cannot take a limit as \(m\to \infty \)!) In the second case, \(0<y_m-x<y_m-0=y_m\) and so \(|\frac {(x-y_m)^{m-1}}{(y_m)^{m-1}}|\leq 1\). Thus we have either that

\[\Bigl |\ln x-\ln c-\sum _{k=1}^{m-1}\frac {(-1)^{k-1} }{kc^k} (x-c)^k\Bigr | \leq \frac {1}{m}\]

or

\[\Bigl |\ln x-\ln c-\sum _{k=1}^{m-1}\frac {(-1)^{k-1} }{kc^k} (x-c)^k\Bigr | \leq \frac {c-x}{mx}.\]

In either case the right hand side converges to zero as \(m\to \infty \) and so we must have that the infinite series converges to \(\ln x\), as desired.

[Definition: Relatively open and closed] Let \((X,d)\) be a metric space and let \(Y\subseteq X\). Then \((Y,d)\) is also a metric space. If \(F\subseteq Y\) is closed in \((Y,d)\), then we say that \(F\) is relatively closed. If \(G\subseteq Y\) is open in \((Y,d)\), then we say that \(G\) is relatively open.

(Memory 1940)

(Memory 1950)

(Problem 1960)

(Problem 1970)

3.6. The Zeros of a Holomorphic Function

Corollary 3.6.2. Let \(\Omega \subseteq \C \) be open and connected and let \(f:\Omega \to \C \) be holomorphic. Suppose that there is a \(P\in \Omega \) and an \(r>0\) such that \(D(P,r)\subseteq \Omega \) and \(f=0\) in \(D(P,r)\). Then \(f=0\) everywhere in \(\Omega \).

Corollary 3.6.5. Let \(\Omega \subseteq \C \) be open and connected and let \(f:\Omega \to \C \) be holomorphic. Suppose that there is a \(P\in \Omega \) such that \(f^{(k)}(P)=0\) for all \(k\in \N _0\) (that is, all integers \(k\) such that \(k\geq 0\)). Then \(f=0\) everywhere in \(\Omega \).

(Adam, Problem 1980)

Show that \(E=F\).

Solution

If \(\zeta \in E\), let \(r>0\) be such that \(D(\zeta ,r)\subseteq \Omega \) and \(f=0\) in \(D(\zeta ,r)\). Then all first partial derivatives of \(f\) are zero in \(D(\zeta ,r)\); by induction, all derivatives of \(f\) (of any order) are zero in \(D(\zeta ,r)\), and in particular at \(\zeta \). Thus \(\zeta \in F\) and so \(E\subseteq F\).

Conversely, suppose that \(\zeta \in F\), so \(\zeta \in \Omega \) and \(\frac {\partial ^k f}{\partial z^k}(\zeta )=0\) for all \(k\in \N _0\). Because \(\Omega \) is open there is an \(r>0\) such that \(D(\zeta ,r)\subset \Omega \). By Theorem 3.3.1, if \(z\in D(\zeta ,r)\) then

\begin{equation*}f(z)=\sum _{k=0}^\infty \frac {f^{(k)}{\zeta }}{k!}(z-\zeta )^k.\end{equation*}

But because \(f^{(k)}{\zeta }=0\) for all \(k\), we have that \(f(z)=0\) for all \(z\in D(\zeta ,r)\), as desired.

(Amani, Problem 1990)

Solution

Let \(\{z_n\}_{n=1}^\infty \subseteq F\subseteq \Omega \) and suppose that \(z_n\to z\) for some \(z\in \Omega \).

If \(k\in \N _0\), then \(f^{(k)}(z_n)=0\) for all \(n\) because \(z_n\in F\). But each \(f^{(k)}\) is continuous on \(\Omega \) by Theorem 3.1.1, and so we must have that \(f^{(k)}(z)=0\) because \(z_n\to z\) and \(f^{(k)}(z_n)\to 0\). Thus \(z\in F\).

We have showed that \(E\) is open in \(\C \) (and therefore relatively open in \(\Omega \)) and equals \(F\), which relatively closed in \(\Omega \). By definition of connected set, we must have that either \(E=F=\Omega \) or \(E=F=\emptyset \). Recalling the definitions of \(E\) and \(F\) completes the proof of both corollaries.

Theorem 3.6.1. Let \(\Omega \subseteq \C \) be open and connected and let \(f:\Omega \to \C \) be holomorphic.

Suppose that there is a sequence \(\{z_n\}_{n=1}^\infty \) such that

• \(z_n\in \Omega \) for all \(n\in \N \),

• \(f(z_n)=0\) for all \(n\in \N \),

• The sequence \(\{z_n\}_{n=1}^\infty \) is convergent,

• \(z_0=\lim _{n\to \infty } z_n\) is in \(\Omega \),

• \(z_0\neq z_n\) for all \(n\geq 1\).

Then \(f(z)=0\) for all \(z\in \Omega \).

(Bashar, Problem 2000)

Solution

Let \(r>0\) be such that \(D(z_0,r)\subseteq \Omega \); \(r\) must exist because \(\Omega \) is open. By Theorem 3.3.1, there are constants \(a_k\) such that if \(z\in D(z_0,r)\), then \(\sum _{k=0}^\infty a_k(z-z_0)^k\) converges absolutely to \(f(z)\).

We claim that \(a_k=0\) for all \(k\); this immediately implies that \(f=0\) in \(D(z_0,r)\), and so by Corollary 3.6.2 we have that \(f=0\) in \(\Omega \).

We have that, if \(m\geq 0\) is an integer, then \(\sum _{k=m}^\infty a_k(z-z_0)^k\) also converges absolutely. Define \(f_m:D(z_0,r)\to \C \) by

\begin{equation*}f_m(z)=\sum _{k=m}^\infty a_k(z-z_0)^k.\end{equation*}

We observe that for any fixed \(z\in D(z_0,r)\), \(\lim _{m\to \infty } f_m(z)=0\).

If \(z=z_0\) then the series \(\sum _{k=m}^\infty a_k(z-z_0)^{k-m}\) converges trivially. If \(z\in D(z_0,r)\setminus \{z_0\}\), then \((z-z_0)^{-m}\) is independent of \(k\), and so if \(\sum _{k=m}^\infty a_k(z-z_0)^k\) converges absolutely then so does

\[\sum _{k=m}^\infty [a_k(z-z_0)^k](z-z_0)^{-m} =\sum _{k=m}^\infty a_k(z-z_0)^{k-m}.\]

We define \(h_m:D(z_0,r)\to \C \) by \(h_m(z)=\sum _{k=m}^\infty a_k(z-z_0)^{k-m}\). By Lemma 3.2.10, each \(h_m\) is holomorphic and therefore continuous in \(D(z_0,r)\). Furthermore, if \(z\in D(z_0,r)\) then \(\lim _{m\to \infty } h_m(z)=0\).

We have that \(f(z)=f_0(z)=(z-z_0)^0h_0(z)\) for all \(z\in D(z_0,r)\) (with the convention \(0^0=1\) usual in power series).

Suppose that we have established that \(f(z)=f_m(z)=(z-z_0)^m h_m(z)\) for some fixed integer \(m\) and all \(z\in D(z_0,r)\). Then \(a_m=h_m(z_0)=\lim _{n\to \infty } h_m(z_n)\) because \(h_m\) is continuous. But \(h_m(z_n)=f(z_n)/(z_n-z_0)^m=0\) and so \(a_m=0\). Thus \(f(z)=f_m(z)=f_{m+1}(z)=(z-z_0)^{m+1}h_{m+1}(z)\) and \(a_m=0\). By induction we have that \(f(z)=f_m(z)=f_{m+1}(z)=(z-z_0)^{m+1}h_{m+1}(z)\) and \(a_m=0\) for all \(m\in \Z \), and so as noted above the conclusion follows from Corollary 3.6.2.

(Dibyendu, Problem 2010)

Solution

Let \(f(z)=\sin (1/z)\) and let \(z_n=\frac {1}{n\pi }\). Then \(z_n\to 0\) and \(f(z_n)=0\) for all \(n\in \N \), but \(f(z)\neq 0\) for many values of \(z\).

[Definition: Accumulation point] Let \(S\subseteq \C \). Suppose that \(P\in \C \) and that, for every \(r>0\), there is a \(z\in D(P,r)\cap S\) with \(z\neq P\). Then we say that \(P\) is an accumulation point for \(S\).

(Hope, Problem 2020)

Solution

Theorem. Let \(\Omega \subseteq \C \) be open and connected and let \(f:\Omega \to \C \) be holomorphic. Let \(Z=\{z\in \Omega :f(z)=0\}\). If \(Z\) has an accumulation point that lies in \(\Omega \), then \(f(z)=0\) for all \(z\in \Omega \).

(Micah, Problem 2030)

Corollary 3.6.3. Suppose that \(f\) and \(g\) are holomorphic in a connected open set \(\Omega \). If \(\{z\in \Omega :f(z)=g(z)\}\) has an accumulation point in \(\Omega \), then \(f(z)=g(z)\) for all \(z\in \Omega \).

(Muhammad, Problem 2040)

Solution

Let \(h(z)=f(z)-g(z)\). Then \(h\) is holomorphic in \(\Omega \) and \(\{z\in \Omega :f(z)=g(z)\}=\{z\in \Omega :h(z)=0\}\). Thus \(\{z\in \Omega :h(z)=0\}\) has an accumulation point. By Problem 2020, \(h\equiv 0\) in \(\Omega \), and so \(f\equiv g\) in \(\Omega \).

(Nisa, Problem 2050)

Corollary 3.6.4. Suppose that \(f\) and \(g\) are holomorphic in a connected open set \(\Omega \). If \(fg=0\) everywhere in \(\Omega \), then either \(f\equiv 0\) or \(g\equiv 0\) in \(\Omega \).

(Robert, Problem 2060)

Solution

Suppose \(f\not \equiv 0\). Then there is a \(\zeta \in \Omega \) such that \(f(\zeta )\neq 0\). By continuity, there is a \(r>0\) such that \(f\neq 0\) in \(D(\zeta ,r)\), and so we must have that \(g\equiv 0\) in \(D(\zeta ,r)\). The result follows from Corollary 3.6.2.

[Chapter 3, Problem 42] Let \(f\) be holomorphic in the connected open set \(\Omega \) and let \(K\subseteq \Omega \) be compact. Show that if \(f\) has infinitely many zeroes in \(K\) then \(f\equiv 0\) in \(\Omega \).

4.2. Convergence of Laurent series

[Definition: Laurent series] A Laurent series is a formal expression of the form \(\sum _{k=-\infty }^\infty a_k(z-P)^k\), where \(P\in \C \) and each \(a_k\in \C \), with the convention that \(0^0=1\) and \(0\cdot 0^k=0\) even if \(k<0\).

[Definition: Convergence of Laurent series] We say that the Laurent series \(\sum _{k=-\infty }^\infty a_k(z-P)^k\) converges at \(z\) if the two series \(\sum _{k=0}^\infty a_k(z-P)^k\) and \(\sum _{k=1}^\infty a_{-k}(z-P)^{-k}\) both converge, and write

Lemma 4.2.1. Suppose that the doubly infinite series \(\sum _{k=-\infty }^\infty a_k(z-P)^k\) converges at \(z=w_1\) and at \(z=w_2\), where \(0<|w_1-P|<|w_2-P|\). Then the series converges absolutely at \(z\) for all \(z\) such that \(|w_1-P|<|z-P|<|w_2-P|\).

(Sam, Problem 2070)

Lemma 4.2.2. Let \(\sum _{k=-\infty }^\infty a_k(z-P)^k\) be a doubly infinite series that converges at \(z=w\) for at least one \(w\in \C \setminus \{P\}\). Then there are extended real numbers \(r\) and \(R\) with \(0\leq r\leq |w-P|\leq R\leq \infty \) such that the series converges absolutely if \(r<|z-P|<R\) and diverges if \(|z-P|<r\) or \(|z-P|>R\).

Furthermore, if \(r<\tau \leq \sigma <R\) then the series converges uniformly on \(\overline D(P,\sigma )\setminus D(P,\tau )\).

(Problem 2080)

(Wilson, Problem 2090)

Solution

Let \(w_1\) and \(w_2\) satisfy \(r<|w_1-P|<\sigma \) and \(\tau <|w_2-P|<R\). Then \(\sum _{k=-\infty }^\infty a_k(w_1-P)^k\) converges, and so in particular \(\sum _{k=0}^\infty a_k(w_1-P)^k\) converges. By Proposition 3.2.9 we have that \(\sum _{k=0}^\infty a_k(z-P)^k\) converges uniformly on \(\overline D(P,\sigma )\).

Similarly, \(\sum _{k=-\infty }^\infty a_k(w_2-P)^k\) converges, and so in particular \(\sum _{k=-\infty }^{-1} a_k(w_2-P)^k\) converges. Thus the power series \(\sum _{n=1}^\infty a_{-n}\bigl (\frac {1}{w_2-P}\bigr )^n\) converges, and so \(\sum _{n=1}^\infty a_{-n}\zeta ^n\) converges uniformly to some \(g(\zeta )\) for \(\zeta \in \overline D(0,\frac {1}{\tau })\). That is, for every \(\varepsilon >0\) there is a \(M>0\) such that if \(m\geq M\) and \(|\zeta |\leq 1/\tau \), then \(|g(\zeta )-\sum _{n=1}^m a_{-n}\zeta ^n|<\varepsilon \). Thus, if \(m\geq M\) and \(|z-P|\geq \tau \), then \(\zeta =\frac {1}{z-P}\) satisfies \(|\zeta |\leq 1/\tau \) and so \(|g(\zeta )-\sum _{k=-m}^{-1} a_{k}(z-P)^k|<\varepsilon \). Thus the series \(\sum _{k=-\infty }^{-1} a_k(z-P)^k\) converges uniformly on \(\{z\in \C :|z-P|\geq \tau \}=\C \setminus D(P,\tau )\).

Thus both series converge uniformly on \(\overline D(P,\sigma )\setminus D(P,\tau )\), and so their sum converges uniformly in this region, as desired.

(Adam, Problem 2100)

Solution

If \(r=R\) or if the series converges nowhere then there is nothing to prove. Otherwise, let \(f_N(z)=\sum _{j=-N}^N a_j(z-P)^j\). Then each \(f_N\) is holomorphic and \(f_N\to f\) uniformly on all compact subsets of \(D(P,R)\setminus \overline D(P,r)\). By Theorem 3.5.1 \(f\) must also be holomorphic.

(Amani, Problem 2110)

• \(r=1\) and \(R=2\).

• \(r=0\), \(R=1\), and \(a_j\neq 0\) for infinitely many values of \(j<0\).

• \(r=1\) and \(R=\infty \) and \(a_j\neq 0\) for infinitely many values of \(j>0\).

• \(r=0\) and \(R=\infty \) and \(a_j\neq 0\) for infinitely many values of \(j>0\) and also for infinitely many values of \(j<0\).

Solution

• One such series is \(\sum _{k=-\infty }^{-1} z^k+\sum _{k=0}^\infty \frac {1}{2^k} z^k\).

• One such series is \(\sum _{k=-\infty }^{-1} \frac {1}{|k|!} z^k+\sum _{k=0}^\infty z^k\).

• One such series is \(\sum _{k=-\infty }^{-1} z^k+\sum _{k=0}^\infty \frac {1}{k!} z^k\).

• One such series is \(\sum _{k=-\infty }^\infty \frac {1}{|k|!}z^k\).

Proposition 4.2.4. Let \(\sum _{j=-\infty }^\infty a_j(z-P)^j\) and \(\sum _{j=-\infty }^\infty b_j(z-P)^j\) be doubly infinite series that both converge to the same value if \(r<|z-P|<R\), for some \(0\leq r<R\leq \infty \). Then \(a_j=b_j\) for all \(j\).

(Bashar, Problem 2120)

Solution

Observe that

\begin{align*} \frac {1}{2\pi i} \oint _{\partial D(P,\tau )} \frac {f(\zeta )}{(\zeta -P)^{n+1}}\,d\zeta &= \frac {1}{2\pi i} \oint _{\partial D(P,\tau )} \frac {1}{(\zeta -P)^{n+1}}\sum _{k=-\infty }^\infty a_k (\zeta -P)^k\,d\zeta = \frac {1}{2\pi i} \oint _{\partial D(P,\tau )} \sum _{k=-\infty }^\infty a_k (\zeta -P)^{k-n-1}\,d\zeta \end{align*}

by definition of \(f\). By Lemma 4.2.2, the series converges uniformly on the compact set \(\partial D(P,\tau )\). Because \(\frac {1}{(\zeta -P)^{n+1}}\) is bounded on \(\partial D(P,\tau )\), we have that \(\sum _{k=-\infty }^\infty a_k (\zeta -P)^{k-n-1}\) also converges uniformly, and so by Memory 1580 we have that

\begin{align*}\frac {1}{2\pi i} &\oint _{\partial D(P,\tau )} \frac {f(\zeta )}{(\zeta -P)^{n+1}}\,d\zeta = \sum _{k=-\infty }^\infty a_k \frac {1}{2\pi i} \oint _{\partial D(P,\tau )} (\zeta -P)^{k-n-1}\,d\zeta .\end{align*}

Using the parameterization \(\gamma (t)=P+\tau e^{it}\), \(0\leq t\leq 2\pi \), of \(\partial D(P,\tau )\), we compute

\begin{equation*}\frac {1}{2\pi i} \oint _{\partial D(P,\tau )} (\zeta -P)^{k-n-1}\,d\zeta =\frac {\tau ^{k-n} }{2\pi }\int _0^{2\pi } e^{it(k-n)} \,dt \end{equation*}

which equals one if \(k=n\) and equals zero otherwise, so

\begin{equation*}\frac {1}{2\pi i} \oint _{\partial D(P,\tau )} \frac {f(\zeta )}{(\zeta -P)^{n+1}}\,d\zeta = a_n.\end{equation*}

Similarly, we have that

\begin{equation*}\frac {1}{2\pi i} \oint _{\partial D(P,\tau )} \frac {f(\zeta )}{(\zeta -P)^{n+1}}\,d\zeta =b_n\end{equation*}

and so \(b_n=a_n\) for all \(n\).

4.3. Existence of Laurent Expansions

[Definition: Extreme discs] If \(z\in \C \), then we take \(D(z,0)=\emptyset \), \(\overline D(z,0)=\{z\}\) (this is not the metric space closure of \(D(z,0)\)), and \(D(z,\infty )=\overline D(z,\infty )=\C \).

Theorem 4.3.2. Let \(0\leq r<R\leq \infty \) and let \(\Omega =D(P,R)\setminus \overline D(P,r)\) for some \(P\in \C \).

Suppose that \(f:\Omega \to \C \) is holomorphic. Then there exist constants \(a_k\) such that the series

Furthermore, if \(r<\sigma <\tau <R\) then the series converges uniformly on \(\overline D(P,\tau )\setminus D(P,\sigma )\).

Theorem 4.3.1. Let \(f\), \(r\), \(R\) be as in Theorem 4.3.2. If \(r<\sigma <|z-P|<\tau <R\), then

(Dibyendu, Problem 2130)

Solution

Because \(z\) is outside \(D(P,\sigma )\), we have that \(\oint _{\partial D(P,\sigma )} \frac {1}{\zeta -z}\,d\zeta =0\) by Theorem 2.4.3. Because \(z\in D(P,\tau )\), we have that \(\oint _{\partial D(P,\tau )} \frac {1}{\zeta -z}\,d\zeta =2\pi i\) by the special case of the Cauchy integral formula (Lemma 2.4.1). Thus

\begin{equation*}\frac {1}{2\pi i} \oint _{\partial D(0,\tau )} \frac {f(z)}{\zeta -z} \,d\zeta - \frac {1}{2\pi i} \oint _{\partial D(0,\sigma )} \frac {f(z)}{\zeta -z} \,d\zeta =f(z).\end{equation*}

(Hope, Problem 2140)

Solution

Define

\[g(\zeta )=\begin {cases}\frac {f(\zeta )-f(z)}{\zeta -z}, &\zeta \neq z,\\f'(z),&\zeta =z.\end {cases}\]

Clearly \(g\) is holomorphic in \(\Omega \setminus \{z\}\). Furthermore, \(\lim _{\zeta \to z} g(\zeta )=g(z)\) and so \(g\) is continuous at \(z\). By Theorem 2.3.3 and Corollary 3.1.2, we have that \(g\) is holomorphic in \(\Omega \).

Thus by Proposition 2.6.6 we have that

\begin{equation*}\oint _{\partial D(P,\sigma )} g=\oint _{\partial D(P,\tau )} g\end{equation*}

and so by definition of \(g\) we have that

\begin{equation*} \oint _{\partial D(P,\tau )} \frac {f(\zeta )-f(z)}{\zeta -z} \,d\zeta = \oint _{\partial D(P,\sigma )} \frac {f(\zeta )-f(z)}{\zeta -z} \,d\zeta .\end{equation*}

Straighforward algebra completes the proof.

(Micah, Problem 2150)

(Muhammad, Problem 2160)

Solution

Define

\begin{equation*}g(z)=\sum _{k=0}^\infty a_k(z-P)^k,\qquad h(z)=\frac {1}{2\pi i}\oint _{\partial D(P,\tau )} \frac {f(\zeta )}{\zeta -z}\,d\zeta .\end{equation*}

By the previous problem and Lemma 3.2.10, \(g\) is holomorphic in \(D(P,\tau )\), while by Theorem 3.1.3 and because holomorphic functions are continuous, we have that \(h\) is also holomorphic in \(D(P,\tau )\).

Furthermore, by our generalization of Theorem 3.1.3, we have that if \(n\geq 0\) is an integer then

\begin{equation*}h^{(n)}(z)=\frac {n!}{2\pi i} \oint _{\partial D(P,\tau )} \frac {f(\zeta )}{(\zeta -z)^{n+1}}\,d\zeta \end{equation*}

while by Lemma 3.2.10

\begin{equation*}g^{(n)}(z)=\sum _{k=n}^\infty \frac {k!}{(k-n)!}a_k(z-P)^{k-n}.\end{equation*}

In particular,

\begin{equation*}h^{(n)}(P)=\frac {n!}{2\pi i} \oint _{\partial D(P,\tau )} \frac {f(\zeta )}{(\zeta -P)^{n+1}}\,d\zeta , \quad g^{(n)}(z)=n! a_n.\end{equation*}

By definition of \(a_n\), we have that \(g^{(n)}(P)-h^{(n)}(P)=0\) for all \(n\geq 0\), and so by Corollary 3.6.5, \(g=h\) everywhere in \(D(P,\tau )\).

(Nisa, Problem 2170)

Solution

We need only show that if \(r<\sigma <|z-P|<\tau <R\), then

\begin{equation*}\sum _{k=-\infty }^{-1} a_k(z-P)^k = -\frac {1}{2\pi i} \oint _{\partial D(P,\sigma )} \frac {f(\zeta )}{\zeta -z}\,d\zeta .\end{equation*}

Recall that

\[a_k=\frac {1}{2\pi i} \oint _{\partial D(P,\sigma )} \frac {f(\zeta )}{(\zeta -P)^{k+1}}\,d\zeta .\]

Then

\begin{equation*}\sum _{k=-\infty }^{-1} a_k (z-P)^k =\sum _{k=-\infty }^{-1} \frac {1}{2\pi i} \oint _{\partial D(P,\sigma )} \frac {f(\zeta )\,(z-P)^k}{(\zeta -P)^{k+1}}\,d\zeta .\end{equation*}

Let \(M=\sup _{|\zeta -P|=\sigma } |f(\zeta )|\); because \(f\) is continuous on \(\Omega \) and \(\partial D(P,\sigma )\subset \Omega \) is compact, \(M\) is finite. Then

\begin{equation*}\left |\frac {f(\zeta )}{(\zeta -P)^{k+1}}(z-P)^k\right |\leq M\sigma ^{-1}\left (\frac {|z-P|}{\sigma }\right )^k\end{equation*}

for all \(\zeta \in \partial D(P,\sigma )\), and \(\sum _{k=-\infty }^{-1} M\sigma ^{-1}\left (\frac {|z-P|}{\sigma }\right )^k\) converges because \(|z-P|>\sigma \). Thus by Memory 1590 we have that

\[\sum _{k=-\infty }^{-1} \frac {f(\zeta )}{(\zeta -P)^{k+1}}\ (z-P)^k\]

converges absolutely for all \(\zeta \in \partial D(P,\sigma )\), uniformly for \(\zeta \in \partial D(P,\sigma )\). Thus by Memory 1580, we have that

\begin{equation*}\sum _{k=-\infty }^{-1} a_k (z-P)^k =\frac {1}{2\pi i} \oint _{\partial D(P,\sigma )} \sum _{k=-\infty }^{-1} \frac {f(\zeta )\,(z-P)^k}{(\zeta -P)^{k+1}} \,d\zeta .\end{equation*}

Applying the formula for the sum of a geometric series, we see that

\begin{align*} \sum _{k=-\infty }^{-1} a_k (z-P)^k &= \frac {1}{2\pi i} \oint _{\partial D(P,\sigma )} f(\zeta )\frac {1}{z-P} \sum _{k=-\infty }^{-1} \biggl (\frac {z-P}{\zeta -P}\biggr )^{k+1} \,d\zeta \\&= \frac {1}{2\pi i} \oint _{\partial D(P,\sigma )} f(\zeta )\frac {1}{z-P} \sum _{n=0}^{\infty } \biggl (\frac {\zeta -P}{z-P}\biggr )^{n} \,d\zeta \\&= \frac {1}{2\pi i} \oint _{\partial D(P,\sigma )} f(\zeta )\frac {1}{z-P} \frac {1}{1-(\zeta -P)/(z-P)} \,d\zeta \\&= \frac {1}{2\pi i} \oint _{\partial D(P,\sigma )} \frac {f(\zeta )}{z-\zeta } \,d\zeta \end{align*}

as desired.

4.1. The Behavior of a Holomorphic Function Near an Isolated Singularity

[Definition: Isolated singularity] If \(\Omega \subseteq \C \) is open and \(P\in \C \), and if \(f\) is a function defined and holomorphic in \(\Omega \setminus \{P\}\), then we say that \(f\) has an isolated singularity at \(P\).

[Definition: Removable singularity] If \(f\) has an isolated singularity at \(P\) and if \(f\) is defined and bounded on \(D(P,r)\setminus \{P\}\) for some \(r>0\), then we say that \(f\) has a removable singularity at \(P\).

Theorem 4.1.1. [The Riemann removable singularities theorem.] Suppose that \(f\) has a removable singularity at \(P\). Then \(\lim _{z\to P}f(z)\) exists (and is a finite complex number), and the function

(Observe that if the limit exists, then \(\widehat f\) is continuous on \(\Omega \) and holomorphic on \(\Omega \setminus \{P\}\), so the fact that \(\widehat f\) is holomorphic is simply Problem 1450.)

[Chapter 4, Problem 8a] Suppose that \(P\in \Omega \subseteq \C \) for some open set \(\Omega \). Suppose that \(f:\Omega \setminus \{P\}\to \C \) is holomorphic and that \(\lim _{z\to P} (z-P)f(z)=0\). Then \(\lim _{z\to P}f(z)\) exists.

(Robert, Problem 2180)

(a)

Give an example of a \(C^\infty \) function \(f:\R ^2\setminus \{(0,0)\}\to \R \) that is bounded but such that \(\lim _{(x,y)\to (0,0)} f(x,y)\) does not exist.

(b)

Give an example of a \(C^\infty \) function \(f:\R \setminus \{0\}\to \R \) such that \(f\) and \(f'\) are both bounded but such that \(\lim _{x\to 0} f(x)\) does not exist.

Solution

(a)

Let \(f(x,y)=\sin (1/(x^2+y^2))\). Then \(f\) is bounded and is \(C^\infty \) on \(\R ^2\setminus \{(0,0)\}\) but has no limits at the origin.

(b)

Let \(f(x)=\sgn (x)\) (that is, \(f(x)=1\) if \(x>1\) and \(f(x)=0\) if \(x<0\)). Then \(f\) is bounded and \(f'=0\) is bounded but \(\lim _{x\to 0} f(x)\) does not exist.

[Definition: Pole] If \(f\) has an isolated singularity at \(P\), and if \(\lim _{z\to P}|f(z)|=\infty \), then we say that \(f\) has a pole at \(P\).

(Sam, Problem 2200)

• Let \(\Omega \subseteq \C \) be open and let \(f:\Omega \to \C \) be holomorphic. Show that \(\Psi =\Omega \setminus \{z\in \Omega :f(z)=0\}\) is open.

• Suppose that \(P\in \Omega \subseteq \C \) for some open set \(\Omega \). Suppose that \(f:\Omega \setminus \{P\}\to \C \) is holomorphic and that \(\lim _{z\to P} |f(z)|=\infty \). Let \(W=\Omega \setminus \{z\in \Omega \setminus \{P\}:f(z)=0\}\). Observe that \(P\in W\). Show that \(W\) is open.

Solution

For part (a), let \(z\in \Psi \). Then \(f(z)\neq 0\). By Problem 2030 there is a \(r>0\) with \(D(z,r)\subseteq \Omega \) and such that \(f\neq 0\) on \(D(z,r)\setminus \{z\}\). Thus \(f\neq 0\) on \(D(z,r)\) and so \(D(z,r)\subseteq \Psi \). Thus \(\Psi \) is open.

For part (b), observe that \(\Omega \setminus \{P\}\) is open and \(f\) is holomorphic on \(\Omega \setminus \{P\}\). Thus \(\Psi =\{z\in \Omega \setminus \{P\}:f(z)\neq 0\}\) is open. Because \(W=\Psi \cup \{P\}\), we need only consider the point \(P\).

Because \(\Omega \) is open we have that there is a \(\varrho >0\) such that \(D(P,\varrho )\subseteq \Omega \). By definition of limit there is a \(\delta >0\) such that, if \(0<|z-P|<\delta \), then \(|f(z)|>7\) and so in particular \(f(z)\neq 0\). Let \(r=\min (\delta ,\varrho )\); then \(P\in W\) by definiton and, if \(z\in D(P,r)\) with \(z\neq P\), then \(z\in \Omega \) and \(f(z)\neq 0\), and so \(z\in W\). Thus \(D(P,r)\subseteq W\) and so \(W=\Psi \cup D(P,r)\) is the union of two open sets, and so is open.

(Problem 2190)

[Chapter 4, Problem 15a] Suppose that \(P\in \Omega \subseteq \C \) for some open set \(\Omega \). Suppose that \(f:\Omega \setminus \{P\}\to \C \) is holomorphic and that \(\lim _{z\to P} |f(z)|=\infty \). Let \(g:W\setminus \{P\}\to \C \) be given by \(g(z)=1/f(z)\). Then \(g\) has a removable singularity at \(P\) and \(\lim _{z\to P} g(z)=0\).

[Definition: Essential singularity] If \(f\) has an isolated singularity at \(P\), and if \(f\) has neither a pole nor a removable singularity at \(P\), then we say that \(f\) has an essential singularity at \(P\).

(Wilson, Problem 2210)

Solution

If it is false that \(\lim _{z\to P} |f(z)|=\infty \) (either the limit does not exists, or it exists but is finite), then there exists a \(N\in \R \) such that for every \(\delta >0\) there is a \(z\) with \(0<|z-P|<\delta \) and such that \(|f(z)|\leq N\).

(Adam, Problem 2220)

Solution

Recall that there exist real numbers \(\rho \) and \(\theta \) such that \(w=\rho e^{i\theta }\). Furthermore \(\rho >0\) and we may require \(0\leq \theta <2\pi \).

Define \(z_k\) by \(1/z_k=\ln \rho +i\theta +2k\pi i\). Then \(f(z_k)=w\) for all \(k\in \Z \). But \(\lim _{k\to \infty }|1/z_k|=\infty \), and so we may find a \(k\) such that \(|z_k|<r\).

(Amani, Problem 2230)

Theorem 4.1.4. Suppose that \(f\) has an essential singularity at \(P\). Let \(r>0\) be such that \(D(P,r)\subseteq \Omega \). Then \(f(D(P,r)\setminus \{P\})\) is dense in \(\C \).

(Bashar, Problem 2240)

(Dibyendu, Problem 2250)

(Hope, Problem 2260)

(Micah, Problem 2270)

4.3. The Laurent series near an isolated singularity

Proposition 4.3.3. Suppose that \(f\) is holomorphic in the punctured disc \(\Omega =D(P,R)\setminus \{P\}\) for some \(P\in \C \) and some \(0<R\leq \infty \). Then \(f\) has a unique Laurent series

(Muhammad, Problem 2280)

(Nisa, Problem 2290)

[Definition: Order of a zero] Suppose that \(f\) is holomorphic in \(D(P,r)\) for some \(r>0\) and \(f(P)=0\). Then \(f(z)=\sum _{k=0}^\infty a_k (z-P)^k\) for all \(z\in D(P,r)\). The order of the zero of \(f\) at \(P\) is the smallest \(n\) such that \(a_n\neq 0\); note that the order is at least \(1\).

[Chapter 4, Problem 15b] Suppose that \(f\) is holomorphic in \(D(P,R)\setminus \{P\}\) and that \(f\) has a pole at \(P\). Then \(1/f\) is holomorphic in \(D(P,r)\setminus \{P\}\) for some \(r>0\), has a removable singularity at \(P\), and the holomorphic extension \(g\) of \(1/f\) to \(D(P,r)\) satisfies \(g(P)=0\). Let \(n\) be the order of the zero of \(g\) at \(P\). Then \((z-P)^n f(z)\) has a removable singularity at \(P\).

(Robert, Problem 2300)

[Chapter 4, Problem 15c] Suppose that \(f\) is holomorphic in \(D(P,R)\setminus \{P\}\), that \(f\) is not bounded in \(D(P,R/2)\setminus \{P\}\), and that there is a \(m\in \N \) such that the function \(g\) given by \(g(z)=(z-P)^m f(z)\) is bounded in \(D(P,R/2)\). Show that \(f\) has a pole at \(P\).

(Sam, Problem 2310)

[Definition: Order of a pole] Suppose that \(f\) has a pole at \(P\). Then there is some \(N>0\) such that \(a_{-N}\neq 0\) and such that \(f(z)=\sum _{k=-N}^\infty a_k (z-P)^k\) for all \(z\) in a punctured neighborhood of \(P\) (that is, for all \(z\) in in \(D(P,r)\setminus \{P\}\) for some \(r>0\)). We call \(N\) the order of the pole at \(P\). If \(N=1\) we say that \(f\) has a simple pole at \(P\).

[Definition: Pole/zero of nonpositive order] If you write “pole of order 0”, I will assume that you mean “removable singularity”. If you write “zero of order 0 at \(P\)”, I will assume that you mean “holomorphic near \(P\) and nonzero at \(P\).” If you write “zero/pole of order \(-n\)”, for \(n\in \N \), I will assume that you mean a pole/zero of order \(n\) (possibly after the additional step of applying the Riemann removable singularities theorem).

(Wilson, Problem 2320)

4.4. Examples of Laurent Expansions

(Adam, Problem 2330)

(Amani, Problem 2340)