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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.

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  • $\begingroup$ Do you know the open mapping theorem for holomorphic functions? Or the maximum principle for harmonic functions? I do not see how you can apply Rouché directly. $\endgroup$ – LutzL May 31 at 6:17
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You're correct in wanting to use Rouche's Theorem. Try comparing f to the function $g(z)=Mz$

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  • $\begingroup$ Okay, I'm still a bit stuck, because you know that $|f(z_0)|\le M$ by Cauchy's integral formula, but then how do you use the information given about the minimum, and also change $M$ to $Mz$? $\endgroup$ – jessg12345 May 29 at 13:34
  • $\begingroup$ Could you elaborate what the comparison function with a root in $B_r(z_0)$ is? How does it account for the possibility of multiple roots of $f$ in that disk? $\endgroup$ – LutzL May 31 at 6:20
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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.

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