## Maximum modulus principle

Here’s a fact you probably never noticed: Holomoprhic functions have no local maxima. Okay, constant functions do, but those are lame.

Theorem 1 (Maximum modulus principle) Let ${f\in{\mathcal H}(U)}$. Then if ${z_0\in U}$ has ${|f(z_0)|\ge|f(z)|}$ for every ${z\in U}$, then ${f}$ is constant.

Proof: If ${|f|}$ has a local maximum at ${z_0}$, then in a small ball around ${z_0}$, consider the image of ${|f|}$. It looks something like ${(f(z_0)-\varepsilon, f(z_0)]}$. We don’t really care about the lower bound. The important point is that the upper bound is attained. This tells us that, the image of the small ball ${B_r(z_0)}$ under ${f}$ is completely contained in ${\overline{B_{f(z_0)}(0)}}$, and that it touches the boundary. Thus, the image of this ball has a boundary and thus cannot be an open set. But by the open mapping theorem, this is only possible if ${f}$ were in fact constant. $\Box$

## Open mapping theorem

Today we’ll prove the open mapping theorem:

Theorem 1 (Open mapping Theorem) Let ${f\in{\mathcal H}(U)}$, for some open set ${U}$ in ${{\mathbb C}}$. Then ${f(U)}$, the set of all possible images of ${f}$ is either constant, or is open in ${{\mathbb C}}$.

Let ${z_0\in f(U)}$, and let ${w_0}$ be a preimage. That is, ${f(w_0)=z_0}$. Since ${U}$ is open, there must be a small closed ball around ${w_0}$ completely contained in ${U}$. Let ${r}$ be it’s radius. Then ${\overline{B_r(w_0)}\subseteq U}$. Let ${g(z)=f(z)-z_0}$. What do we know about ${g}$?

We know that the roots of ${g}$ are isolated points. After all, if not, we would have a convergent sequence of roots, contradiction one of the theorems we proved here. So we can choose ${r}$ to be even smaller so that the only root of ${g}$ in ${\overline{B_r(w_0)}}$ is ${w_0}$. Since the boundary of ${\overline{B_r(w_0)}}$ is a circle, and hence compact, and ${|g(z)|}$ is continuous it attains its minimum when restricted to the boundary. Let ${m}$ be the minimum value of ${|g(z)|}$ for ${z}$ on the boundary of ${\overline{B_r(w_0)}}$.

Now, by Rouché’s theorem, ${g}$ will have the same number of roots in in ${B_r(w_0)}$ as ${f(z)-z_1}$ for any ${z_1}$ in ${B_m(z_0)}$. (Application of Rouché’s theorem is left as an exercise.)

In particular, this means that for every complex number in ${z_1\in B_m(z_0)}$, the function ${f(z)-z_1}$ has at least one root in ${B_r(w_0)}$, and so ${B_m(z_0)}$ is contained in ${f(U)}$, meaning our arbitrarily chosen point ${z_0}$ is in the interior of ${f(U)}$. Thus ${f(U)}$ is open.

## Rouche’s Theorem

Today we prove Rouché’s theorem. The gist is that it helps us count the number of roots of a holomorphic function, given some bounds on its values.

Theorem 1 Suppose ${f}$ and ${g}$ are holomorphic functions inside and on the boundary of some closed contour ${\gamma}$. If

$\displaystyle |g(z)|<|f(z)|$

on ${\gamma}$, then ${f}$ and ${f+g}$ have the same number of zeros on the interior of ${\gamma}$.

Before we begin proving this, it should be emphasized that we count with multiplicity. We would count the number 1 as a root of ${x^2-2x+1}$ twice. Most root counting we every do will be done this way. I feel confident in saying that it is the correct way to count roots, even if at first it is unintuitive.

Proof: By hypothesis, ${f}$ has no roots on the boundary ${\gamma}$. Define ${F(z)=\frac{f(z)+g(z)}{f(z)}}$. The roots of ${F}$ are the roots of ${f+g}$. The poles of ${F}$ are the roots of ${f}$. So it suffices to use the argument principal to show that ${N(F)=P(F)}$. That is, we need to show that

$\displaystyle \displaystyle\frac1{2\pi i}\int_\gamma\frac{F'}{F}=N(F)-P(F)$

is zero.

But from our hypotheses, we can conclude that

$\displaystyle |F(z)-1|=\left|\frac{f(z)+g(z)}{f(z)}-1\right|=\left|\frac{f(z)}{g(z)}\right|<1.$

That is, ${F}$ never takes on values more than ${1}$ away from ${1}$. Imagine a dog tethered by a leash of length less than ${1}$ to the point ${1\in{\mathbb C}}$. That dog can’s reach the origin, but more importantly, he can’t walk a loop around the origin. So computing the winding number about the origin must give us zero. This proves the theorem. $\Box$

## Argument Principle

Let ${z_0}$ be a zero of a meromorphic ${f}$ with multiplicity ${m}$. Then we can write ${f(z)=(z-z_0)^m\cdot g(z)}$ where ${g(z_0)\ne0}$. Taking derivatives yields

$\displaystyle f'(z)=m(z-z_0)^{m-1}\cdot g(z)+(z-z_0)^m\cdot g'(z).$

Hence, for ${f'/f}$, we get

$\displaystyle \displaystyle\frac{f'(z)}{f(z)}=\frac{m}{z-z_0}+\frac{g'(z)}{g(z)}.$

The residue of this sum is simply the sum of the residues of the two parts. The first term has residue ${m}$ at ${z_0}$. The second part has no pole at ${z_0}$, and hence zero residue. Thus, the residue of ${f'/f}$ at ${z_0}$ is ${m}$.

What if we pick a pole of ${z_p}$ of ${f}$? Then by a similar construction, if ${z_p}$ is an order ${q}$ pole, we can write ${f(z)=(z-z_p)^{-q}\cdot h(z)}$ and compute

$\displaystyle \displaystyle\frac{f'(z)}{f(z)}=\frac{-q}{z-z_p}+\frac{h'(z)}{h(z)},$

yielding a total residue of ${-q}$.

If a point ${z}$ is neither a pole nor a zero, then ${f'/f}$ is holomorphic at ${z}$, and has residue zero at ${z}$.

If we take a big contour around all of these points, then the integral will be the sum of the residues inside the contour, which we have just shown is the number of roots minus the number of poles (counted with multiplicity and order, respectively).

That is, if ${N_\gamma(f)}$ denotes the number of zeros counted with multiplicity inside a contour ${\gamma}$, and ${P_\gamma(f)}$ denotes the number of poles counted with order, then

$\displaystyle \displaystyle\frac{1}{2\pi i}\int_\gamma \frac{f'(z)}{f(z)}dz=N_\gamma(f)-P_\gamma(f)$

This result is known as the argument principle.

## A joke

Today, a definition, then a joke.

I defined poles of meromorphic functions, but we can be a bit more descriptive. Suppose we have a meromorphic function $f$ which is is undefined at some point $z_0$. We can expand it as a Laurent series, and get something like:

$\displaystyle\sum_{-\infty}^\infty a_n(z-z_0)^n$

It may be that we can make $n$ arbitrarily negative and still have $a_n$ be nonzero. This is basically the worst situation possible. We don’t in fact call it a pole. We say that it is an essential singularity at $z_0$

In a better situation, it might be that $a_{-17}$ is nonzero, but with any smaller (more negative) index, $a_n$ is zero. Then, we would say that $f$ has a pole  at $z_0$ of order $17$. Of course there’s nothing special about seventeen.

If a pole has order 1, we say that it is a simple pole.

Now for a joke.

An airplane is on its way out of Warsaw, and the pilot suffers a heart attack and dies. A passenger is asked to navigate the plane to safety. He looks worried, so the stewardess asks “what’s wrong?” He responds “I’m just a simple Pole in a complex plane!”

Laugh, damn it!

## Residue calculus

There’s a neat trick we can use to integrate real integrals using complex analysis. These can be made arbitrarily complicated, but I’ll give you a simple example. Compute the integral:

$\displaystyle \displaystyle\int_{-\infty}^\infty\frac1{(1+x^2)^3}dx$

This is another way to write down

$\displaystyle \displaystyle\lim_{a\rightarrow\infty}\int_{-a}^a\frac1{(1+x^2)^2}dx.$

Let’s change all the ${x}$s to ${z}$s, and pretend we’re integrating along the path ${[-a,a]}$ in the complex plane. Our notion of integrating in ${{\mathbb C}}$ is defined in such a way that this makes sense. Okay, but we normally integrate loops, not paths, so let’s complete a full loop ${\gamma_a}$ is the semi-circle in the upper half-plane with base ${[-a,a]}$ and radius ${a}$. Then we can break the path into two pieces:

$\displaystyle\int_{\gamma_a}\frac1{(1+z^2)^{2}}dz=\int_{-a}^a\frac1{(1+x^2)^{2}}dx+\int_0^\pi\frac1{(1+(ae^{i\theta})^2)^{2}}d\theta.$

The first piece is the integral we want to compute, and the second is the curved part of the semi-circle. Together they make a closed loop whose integral we can actually calculate using the residue theorem. Since our function is meromorphic on all of ${{\mathbb C}}$, we need to simply figure out which poles are inside the semicircle ${\gamma_a}$, and find their residues. We can tell that the only poles are at ${i}$ and ${-i}$, of which only ${i}$ is in the semicircle ${\gamma_a}$ (for every ${a>1}$). It’s residue we compute as $-i/4$.

Thus, the left hand side, via the residue theorem, must be ${2\pi i\cdot (-i/4)=\pi/2}$.

Now we just need to show that the semi-circular arc’s contiribution is neglibible for large ${a}$. Then, taking ${a\rightarrow\infty}$ yields

$\displaystyle \displaystyle\lim_{a\rightarrow\infty}\int_{-a}^a\frac1{(1+x^2)^2}dx=\pi/2.$

Indeed,

$\displaystyle \begin{array}{rcl} \left|\displaystyle\int_0^\pi\frac1{(1+(ae^{i\theta})^2)^{2}}d\theta\right| &\le& \displaystyle\int_0^\pi\left|\frac1{(1+(ae^{i\theta})^2)^{2}}\right|d\theta\\ &=& \displaystyle\int_0^\pi\frac1{(1+a^2)^{2}}d\theta\\ &=& \frac{\pi}{(1+a^2)^2} \end{array}$

which decreases to zero as ${a\rightarrow\infty}$.

I’m awful at integration in general, so I’m not sure if this integral can be done using integration by parts, or a tricky ${u}$-substitution. There are definitely some integrals though, for which those standard methods just don’t cut it.

## Meromorphic functions and residues

Last time, we discussed Laurent series, which are essentially two-way power series. They are almost as nice as holomorphic functions, but not quite. Maybe we can recoup some of the lost beauty of holomorphicity by imposing a reasonable condition.

We want to allow ourselves to have points where the function isn’t defined, but let’s limit these points. Let’s require them to be isolated points. We say that a function ${f}$ is meromorphic on an open set ${U\subseteq {\mathbb C}}$ if ${f}$ is holomorphic on ${U}$, except at some number of isolated points. We write ${f\in{\mathcal M}(U)}$. These isolated points are called poles. We obviously shouldn’t expect ${f}$ to have a power series centered at one of these poles, but it does have a Laurent series.

Let ${f\in{\mathcal M}(U)}$ and let ${z_0\in U}$ be a pole of ${f}$. Then as we saw last time, ${f}$ has a Laurent series centered at ${z_0}$:

$\displaystyle f(z)=\cdots+a_{-2}(z-z_0)^{-2}+a_{-1}(z-z_0)^{-1}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots$

Moreover, the series has inner radius of convergence 0, so this representation is valid for all ${z}$ close enough to ${z_0}$.

Now if we take a loop ${\gamma}$ in ${U}$, and integrate, if ${f}$ has no poles on the interior of ${\gamma}$, then the integral is zero. This is the Cauchy integral theorem. What if it does have a pole? We can use the generalized Cauchy integral formula we saw last time:

Theorem 1 (Residue theorem) Let ${f\in{\mathcal M}(U)}$ and let ${z_0}$ be a pole of ${f}$ in ${U}$. Expand ${f}$ as a Laurent series as above. Let ${\gamma}$ be a small counter-clockwise circle about ${z_0}$ such that the only pole in its interior is ${z_0}$. Then

$\displaystyle \displaystyle\frac{1}{2\pi i}\int_\gamma f(z)dz=a_{-1}.$

Proof:The generalized Cauchy integral formula we saw last time said that

$\displaystyle a_n=\displaystyle\frac{1}{2\pi i}\int_\gamma\frac{f(z)}{(z-z_0)^{n+1}}dz.$

Let ${n=-1}$.

$\Box$

If we take a loop that goes around several poles, the integral must be the sum of the integrals, as we can build a homotopy as shown in the images below (click to blow up).

The value ${a_{-1}}$ (for an expansion about ${z_0}$) is called the residue of ${f}$ at the pole ${z_0}$. In other words, the theorem states that the value of an integral about a contour is the sum of the residues of the poles inside.

## Laurent series

I must confess, this post will be hand-waivy. My goal in these posts has always been to understand mathematics, and this mostly coincides with careful proofs and decent intuitive explanations. The contents of this post I think are best explained intuitively, leaving the proofs to you, the reader.

Rest assured, you can prove everything in this post. As you will see, this post generalizes much of what we have done thus far. Rather than reprove everything in greater generality, I invite you to generate the new proofs. The proofs we have seen thus far can be altered slightly to get the results presented in this post.

Let’s start with a motivating example. Consider the function ${f(z)=1/z}$. It’s holomorphic on ${{\mathbb C}-\{0\}}$, so for any ${z_0}$ other than zero, we can expand it as a power series centered at ${z_0}$. The expansion is valid in the open ball of radius ${|z_0|}$; that is, right up to, but not including the origin. Well, that’s fine, but a bit annoying for two reasons. For one, we have to pick some ${z_0}$, and different choices of $z_0$ gives us different power series coefficients. Further, no choice is valid everywhere. In fact, we need infinitely many of these choices to cover all of ${{\mathbb C}-\{0\}}$.

So here’s an idea. What if we allow negative terms in a power series expansion? Then, in this “super-expansion,” we could just write ${1/z}$ as ${z^{-1}}$.

Definition 1 A Laurent series centered at ${z_0}$ is a series of the form:

$\displaystyle \cdots+a_{-2}(z-z_0)^{-2}+a_{-1}(z-z_0)^{-1}+\cdots+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots$

As for the radius of convergence of such an expansion, now there are two radii to consider. How far out can we go and have it be valid, and how close to ${z_0}$ can we get and have it be valid. The first has the same formula. The outer radius of convergence is still given by

$\displaystyle R=\left(\displaystyle\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}\right)^{-1}.$

The inner radius of convergence, ${r}$, has a formula which is not terribly surprising:

$\displaystyle r=\displaystyle\limsup_{n\rightarrow\infty}\sqrt[n]{|a_{-n}|}.$

Recognize that power series are just special cases of Laurent series, and the inner radius of convergence just evaluates to zero (since all the terms are zero). As for the proofs of these facts, they contain no new ideas, so I’ll omit them.

You might not expect the Cauchy integral formula to still hold, but it does. The proof is pretty much the same:

Theorem 2 (Cauchy integral formula) Let ${f(z)=\sum_{-\infty}^\infty a_n(z-z_0)^n}$. Let ${\gamma}$ denote a counter-clockwise path around ${z_0}$ on which we can integrate ${f}$. Then

$\displaystyle a_n=\displaystyle\frac{1}{2\pi i}\int_\gamma\frac{f(z)}{(z-z_0)^{n+1}}dz.$

Yes, this is true, even for ${n<0}$. Quite remarkable.

## Two fun facts

Well, it seems that my 15 minutes are up, so let’s get back to some complex analysis. Here are two fun facts about holomorphic functions:

Theorem 1 Suppose ${f\in {\mathcal H}(U)}$, and ${U}$ is connected. Then if ${f}$ constant in a neighborhood of some point ${z_0\in U}$. Then ${f}$ is constant on all of ${U}$.

Proof: Since the set ${\{f(z_0)\}}$ is closed, so is it’s preimage. Let ${V}$ denote this preimage. It suffices to show that ${V}$ has no boundary. Then ${V}$ is open, closed, and non-empty. Since ${U}$ is connected, it follows that ${U=V}$.

Suppose ${V}$ has a boundary, and let ${v}$ be some point on the boundary. Expand a power series about ${v}$:

$\displaystyle f(z)=a_0+a_1(z-v)+a_2(z-v)^2+\cdots.$

Since ${f}$ is holomorphic at ${v}$, this power series must have a non-zero radius of convergence so it is valid on some part of the interior of ${V}$. As we know, the only power series with constant image is the constant series ${f(z)=a_0}$. But now this series is valid in a small ball about ${v}$. This contradicts the assumption that ${v}$ was on the boundary of ${V}$, so ${V}$ must be open, proving the theorem.

$\Box$

Theorem 2 Suppose ${f\in{\mathcal H}(U)}$, and ${U}$ is connected. Suppose we have a sequence of points ${z_1,z_2,\dots}$ converging to ${z_\infty\in U}$ such that ${f(z_1)=f(z_2)=\cdots}$. Then ${f}$ is constant on ${U}$

Proof: By continuity of ${f}$, ${f(z_1)=f(z_2)=\cdots=f(z_\infty)}$. Expanding ${f}$ as a power series about ${z_\infty}$, write

$\displaystyle f(z)=a_0+a_1(z-z_\infty)+a_2(z-z_\infty)^2+\cdots$

Since we have a sequence of points converging to ${z_\infty}$ whose ${f}$-values all agree with ${z_\infty}$, we see that the derivative at ${z_\infty}$ must be

$\displaystyle f'(z_\infty)=\lim_{n\rightarrow\infty}\frac{f(z_n)-f(z_\infty)}{z_n-z_\infty}=\lim_{n\rightarrow\infty}\frac{0}{z_n-z_\infty}=0$

Thus, ${f'(z)=0}$ in a neighborhood of ${z_\infty}$, meaning that the power series: ${f'(z)=a_1+2a_2(z-z_\infty)+3a_3(z-z_\infty)^2+\cdots}$ must in fact be constant. Thus, ${a_n=0}$ for ${n>0}$, meaning that ${f(z)=a_0}$ in a neighborhood of ${z_\infty}$.

Now we can apply the previous theorem to conclude that ${f}$ is in fact constant of all of ${U}$.

$\Box$

## Fundamental theorem of algebra

I really think this is a misnomer. The theorem has very little to do with algebra. The theorem, is an assertion about the complex numbers which are very much an analytic object. Anyway, the proof is pretty short, so here goes.

Theorem 1 Every polynomial with coefficients in ${{\mathbb C}}$ has at least one root in ${{\mathbb C}}$.

Proof: Let ${P:{\mathbb C}\rightarrow{\mathbb C}}$ be a non-constant polynomial. Assume for the sake of contradiction that ${P}$ has no roots in ${{\mathbb C}}$. Then indeed, there is some small open set in ${{\mathbb C}}$ that must not be in the image of ${P}$. Why? If ${P(z)=a_n\cdot z^n+\cdots a_1\cdot z+a_0}$ assuming ${a_n\ne0}$, then

$\displaystyle \lim_{z\rightarrow\infty} \frac{P(z)}{z^n}=a_n,$

so for ${z}$ far enough away from zero, ${P(z)/z^n}$ must be close to ${a_n}$. In particular, for large enough ${|z|}$ we can make ${|P(z)|/|z|^n}$ at least ${|a_n|/2}$. Hence, ${|P(z)|}$ must be, (for large ${|z|}$), at least ${|a_n|\cdot|z|^n}$, and hence, far away from zero. Let ${R}$ be any such value which is large enough to guarantee that for ${|z|>R}$, we have the result just described.

But what about the small complex numbers? Couldn’t those get close to zero? Alas, no. the set ${\{z\mid |z|\le R\}}$ is compact, and so if ${P(z)}$ gets arbitrarily close to zero on this set, it in fact hits zero. By assumption, ${P(z)}$ is never zero, and so it must be bounded away from zero. So let’s say that if ${|w|<\varepsilon}$, then ${w}$ is not in the image of ${P}$. In other words, for each ${z\in{\mathbb C}}$, ${|P(z)|\ge\varepsilon}$.

Now if ${P}$ is never zero, then ${f=1/P}$ is defined on all of ${{\mathbb C}}$. But remember that ${|P(z)|\ge\varepsilon}$. This means that ${|f(z)|\le 1/\varepsilon}$. But now ${f}$ is bounded and entire (holomorphic on all of ${{\mathbb C}}$), so ${f}$ must be a constant, implying that ${P}$ must be a constant polynomial. This contradicts our assumption that ${P}$ was non-constant, so it must be that every polynomial has a root. $\Box$

This is just one of many proofs. You can read a bunch of them on Adam Azzam’s blog (which you should already be reading).