The argument principle and residues a higher order poles.

The argument principle and residues at higher order poles.

These notes are set up so that you get to prove the main results by solving smaller problems that when put together give the big result. The answers to the problems are in the videos. You will get the most out of these notes if you do (or try) the problems before looking at the videos.

$ \newcommand{\Res}{\operatorname{Res}} $ We have seen that if $f(z)$ has a simple pole at $z=a$, say $$ f(z)= \frac{g(z)}{h(z)} $$ where $h(a)=0$ and $h'(a)\ne 0$ then the residue of $f$ at $z=a$ $$ \Res(f,a) = \frac{g(a)}{h'(a)} $$
We have also seen that if $f(z)$ has a root of order $m$ at $z=a$, then it can be written as $$ f(z) = (z-a)^mf_0(z) $$ where $f_0(z)$ is analytic near $a$ and $f_0(a)\ne 0$. Cauchy noticed that these two facts could be put together to give a nice result.

Proposition. Let $f(z)$ have a zero of order $m$ at $z=a$. Then the function $$ \frac{f'(z)}{f(z)} $$ has a simple pole at $z=a$ and the residue is $$ \Res\Big( \frac{f'}{f} , a\Big) = m. $$ That is the residue of $ \dfrac{f'}{f}$ at $z=a$ is the order of the zero of $f$ at $z=a$. Problem 1. Prove this by writing $f(z)$ as $$ f(z) = (z-a)^m f_0(z) $$ with $f_0(a)\ne 0$ and computing $\dfrac{f'}{f}$ and seeing that it has simple pole with the residue of $m$.
Solution:

Now let $D$ be a bounded domain with nice boundary and let $f(z)$ be analytic on the $D$ and boundary and assume that $f(z)\ne 0$ for $z$ on $\partial D$. Then $$ \frac{f'(z)}{f(z)} $$ will be analytic on $D$ expect at with isolated singularities at the zeros of $f$. If we now apply the residue theorem to this function the result is Theorem (The argument principle) Let $f(z)$ be analytic on the closure of the bounded domain $D$ and assume that $f(z)$ has no zeros on the boundary $\partial D$. Then $$ \int_{\partial D} \frac{f'(z)}{f(z)} dz = 2\pi i (\text{the number of zeros of $f(z)$ in $D$}) $$ where the zeros are counted with multiplicity. (That is a zero of order $m$ is counted $m$ times.)
Problem 2. Write out the details of this to make sure you understand them.

Solution:

The zeros of $f(z)$ are the solutions to $f(z)=0$. There is nothing special about solving for $0$, we could have just as easily looked for solutions to $f(z)=b$ for for some constant $b$. So a slightly more general version of the last theorem is

Theorem Let $f(z)$ be analytic on the closure of the bounded domain $D$ and assume that $f(z)=b$ has no solutions on the boundary $\partial D$. Then $$ \int_{\partial D} \frac{f'(z)}{f(z)} dz = 2\pi i (\text{number of solution of $f(z)=b$}) $$ where the solutions are counted with multiplicity. (That is the if $f(a)=b$ then $a$ is counted $m$ times if $f(z)-b$ has a zero of order $m$.)
Problem 3. Prove this.

Solution:

So far we only have an easy to use formula for finding the residue of functions at simple poles. So let take a small step toward messier and look at a pole of order $2$. So we have a function whose Laurent expansion looks like $$ f(z)= \frac{c_{-2}}{(z-a)^2} + \frac{c_{-1}}{(z-a)} + c_0 + c_1(z-a) + c_2(z-a)^2 + \dots $$ and our job is to compute the residue at $z=a$, that is the coefficient $c_{-1}$. The idea to to clear of fractions by multiplying by $(z-a)^2$ and then use the same argument that we have used to find the coefficients of a Taylor's expansion. Proposition. With the setup we have just given, the residue of $f$ at $z=a$ is $$ \Res(f,a) = c_{-1} = \lim_{z\to a} \frac{d}{dz} \Big( (z-a)^2f(z)\big). $$ Problem 4. Prove this.
Solution: