# 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. $\Box$