# Cauchy's Integral Formula and number of zeros

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.

• 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. – LutzL May 31 at 6:17

You're correct in wanting to use Rouche's Theorem. Try comparing f to the function $$g(z)=Mz$$
• 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 $Mz$? – jessg12345 May 29 at 13:34
• 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? – LutzL May 31 at 6:20
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.