Zeros of Holomorphic (upper bound) 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. 

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 $m \leq 4$.
