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I'm looking at old complex analysis exams and am stuck on the following. Suppose f(z) is holomorphic on $D(0,2)$ and continuous on its closure.

Suppose the $|f(z)|\le 16$ on the closure of $D(0,2)$ and is non-constant and |f(0)|=1. Show $f$ cannot have more than 4 zeros in $D(0,1)$ .

I found this technique Upper bound for zeros of holomorphic function , but it doesn't seem to apply to this problem.

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Mykie
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  • Do you know Rouché's theorem? Compare $f$ with $z^4$. – mrf Oct 28 '12 at 16:25
  • Davide - thanks. – Mykie Oct 28 '12 at 16:25
  • I considered Rouche but it does not seem to be helpful for this problem since there is so little information about $f$ – Mykie Oct 28 '12 at 16:27
  • I added an additional condition |f(0)|=1 – Mykie Oct 28 '12 at 16:28
  • I believe a solution might be to define $g: D(0,4) \rightarrow \mathbb{C}$ as $g(z)=f(\sqrt{z})$. And then apply the technique here

    http://math.stackexchange.com/questions/21437/upper-bound-for-zeros-of-holomorphic-function

    to $g$.

     – Mykie Oct 28 '12 at 17:19

1 Answers1

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If $\alpha_1, \dotsc, \alpha_m$ are zeroes of $f$ on $D(0,1)$ let

$$ p(z) = 2^m\prod_{j=1}^m \frac{z-\alpha_j}{4 - \overline{\alpha}_j z}. $$

Then $p$ is holomorphic on $D(0,2)$ and $p(\alpha_j) = 0$ for $j \in \{1, \dotsc, m\}$ and $|p(z)| = 1$ for all $z$ with $|z|=2$. Since $f(z)/p(z)$ is holomorphic on $D(0,2)$ the maximum principle says that

$$ 2^m\prod_{j=1}^m|\alpha_j|^{-1} = |f(0)/p(0)| \leq \max_{|z|=2} |f(z)/p(z)| = \max_{|z|=2} |f(z)| \leq 16. $$

So $2^m \leq 16 \prod_{j=1}^m |\alpha_m| \leq 16$ and therefore $m \leq 4$.

WimC
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