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

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One Response to Rouche’s Theorem

  1. Pingback: Open mapping theorem « Andy Soffer

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