Cauchy's Integral Formula and number of zeros

Question

I'm revising for a complex analysis exam and am a bit stuck on this question.

In the previous part of the question I have stated Cauchy's Integral Formula, and then deduced from it that $$|f^k(z_0)|\leq \frac{k!}{r^k}M$$ where $M:=\max \{|f(z)|:|z-z_0|=r\}$.

Now, assuming that $|f(z_0)|<\min\{|f(z)|:|z-z_0|=r\}$, I want to deduce that $f$ has at least one zero in $B_r(z_0)$. I'm not really sure how to go about this. I don't know if I want to be employing something like Rouche's Theorem here or not? I can see from the deduced result above that $|f(z_0)|\leq M$, but I'm not sure if that's going to help me.

Any help appreciated, thank you.

Answer 1

You're correct in wanting to use Rouche's Theorem. Try comparing f to the function $g(z)=Mz$

Answer 2

The usual proof is that if $$|f(z_0)|<\min\{|f(z)|:|z-z_0|=r\}$$ then the continuous real-valued function $z\mapsto |f(z)|$ has a minimum inside $|z-z_0|\le r$, say at $z_m$.

If $z_m$ is not a root of $f$, then by the open mapping theorem for holomorphic functions, for $\varepsilon>0$ sufficiently small the function $f$ also takes the value $f(z)=(1-ε)f(z_m)$ at some point $z$ close to $z_m$, contradicting $|f(z_m)|$ being the minimum.

Thus $z_m$ can only be a root.