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

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