2
$\begingroup$

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.

  • 0
    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 $g$.2012-10-28

1 Answers 1

4

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 $m \leq 4$.

  • 0
    @jojo A famous result of Jensen (http://en.wikipedia.org/wiki/Jensen%27s_formula) that relates the distribution of zeroes to the growth of a function.2012-10-28