Show that for this holomorphic function $ f$, $|f(z)| \le 4|$ if $|z|=2$

Question

Let $f$ be a holomorphic function on some open neighbourhood $U$ of the ring $ D = \{ z: 1 \le |z| \le 3 \} $ such that $ |f(z)| \le 1 $ for $ |z| = 1 $ and $ |f(z)| \le 9 $ for $ |z| = 3 $. Prove that $ |f(z)| \le 4 $ for $ |z| = 2 $.

This looks like something that should be a pretty simple exercise, but for some reason I can't see any way to use the basic complex analysis theorems/inequalities to get closer to solving this.

In what context did you find this exercise? – Nov 14 '16 at 00:57 — , Nov 14 '16 at 00:57

score 4 · Accepted Answer · answered Nov 14 '16 at 01:03

4

Hint: apply the Maximum Modulus Principle to the function $$ g(z)=\frac{f(z)}{z^2}. $$

answered Nov 14 '16 at 01:03

A.Γ.

29,518

+1Good answer. And good remainder to me that the MMP also applies to non-simply connected domains. :) – Nov 14 '16 at 01:10

Show that for this holomorphic function $ f$, $|f(z)| \le 4|$ if $|z|=2$

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