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Let $D\in \mathbb{C}$ be an open set and $f:D\to \mathbb{C}$ be holomorphic. Suppose we fix $z_0\in D$ then we know that there is a local Taylor's expansion in $B(z_0,r)$ for some $r>0.$ (the corresponding closed ball also lies in $D$.)

The question asks if $|f(z_0)|<\min_{z\in\partial B(z_0,r)}|f(z)|$ then it must contain a zero in the aforementioned open ball.

I thought to prove by contradiction by saying that $\frac{1}{f}$ must be holomorphic on this open ball then I could also deduce the inequality $$|\frac{1}{f(z_0)}|<\max_{z\in\partial B(z_0,r)}|\frac{1}{f(z)}|.$$

However I am not too sure how to proceed further. This kind of reminds me of Rouches Theorem but I am not sure how to apply it.

Many thanks in advance!

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2 Answers 2

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I think you have the inequality backward! Your idea of using contradiction is great, now google the maximum modulus principle.

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  • $\begingroup$ OP is using MMP to $\frac 1 f$ so there is no mistake in the stated inequality. $\endgroup$ Aug 12, 2020 at 23:44
  • $\begingroup$ My answer is correct, there is a mistake in the inequality since he is starting from $f(z_0) < \min f(z)$. $\endgroup$ Aug 13, 2020 at 13:10
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Maximum Modulus Principle shows that if there is no zero then $\frac 1 {|f(z_0)|} \leq \max_{|z-z_0| \leq r} \frac 1 {|f(z)|}=\max_{|z-z_0| = r} \frac 1 {|f(z)|}$ But this is equivalent $|f(z_0| \geq \min \{|f(z)|: |z|= r\}$. This is a contradiction and hence $f$ must have a zero.

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