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.