“Geometric” proof of Rouche's theorem on the number of zeros?

I understand the analytic proof of Rouche's theorem as presented in Stein and Shakarchi's complex analysis - $|f(z)| > |g(z)|$ on the boundary circle C ensures that the argument principle can be applied, and an application of that principle shows that function counting zeros along homotopy $f + gt$ between $f$ and $g$ is continuous function, so is necessarily constant as $Z$ is discrete.

Thinking about this homotopy, I think that there is a geometric way in which the theorem is almost obvious - in particular, the hypothesis ensure that the zeros do not cross the boundary circle. These zeros move continuously with $t$, though I might have to use the open mapping theorem to prove that given a small pertubation of the image there is a new zero near each old zero (using the open mapping theorem would be circular reasoning, at least in the development of complex analysis that I'm reading). It should also be the case that if two zeros overlap, their orders add when considered as zeros of the holomorphic function $f+tg$. However, I'm not sure how to convince myself of this last idea (other than by applying Rouche's theorem...).

Any thoughts? I'm not looking for anything particularly formal, though that would be nice too.

(The geometric explanation on the Wikipedia page is different, though also compelling.)

• The condition that you write as $f(z) > g(z)$ is bad for complex numbers, for which there is no good notion of inequalities. – KCd Jun 20 '14 at 20:24
• @KCd - typo - thanks, I fixed it. – Lorenzo Najt Jun 20 '14 at 20:40

The condition $$|1-z| < 1+|z|$$ is equivalent to the condition that $z$ is never on the negative real axis bcause it's the same as $-\Re z < |z|$, which is satisfied if $\Im z \ne 0$, or if $z=\Re z > 0$. So the above inequality characterizes the slitted plane $\{ re^{i\theta} : 0 < r,\; -\pi < \theta < \pi\}$, a nice domain for a logarithm. Any curve which winds around inside this region must have winding number $0$.
If $f$ and $g$ satisfy $$|f-g|<|f|+|g|$$ on a simple closed curve $C$ on and inside of which $f$, $g$ are holomorphic, then neither $f$ nor $g$ is $0$ on $C$, and the image of $C$ under $f/g$ must have winding number $0$. So the above condition forces the winding number of $f$ on $C$ to equal the winding number of $g$ on $C$. This proof has a straightforward geometric interpretation, and the condition is less restrictive than $|1-z|<1$.