Mimimum of entire function over unit disc I'm trying to solve the following question: 

Let: $f:\mathbb C\to \mathbb C$ be an entire function, $|f(0)|=1$ and $f$ has exactly $n$ roots in $D(0,\frac{1}{2})$. (Not necessarily distinct)
  Prove that: $$2^n\le Max\{|f(z)|\,;|z|=1\}$$

My idea is the following:
Since $f$ has $n$ roots, so we can write: $$f(z)=(z-z_1)(z-z_2)...(z-z_n)g(z)$$ Which: $g(z_i)\neq 0$ when $z_i$ are the roots of $f$ for $i=1,2,...,n$. because of $\,\,|f(0)|=1$, we shoud have: $$|\Pi_{i=1}^nz_i||g(0)|=1$$ By the hypothesis, $|z_i|\lt\frac12\,$, which mean's: $|g(0)|\gt 2^n$
Now define: $$h(z)=(z-z_1)(z-z_2)...(z-z_n)$$ If i prove that: $1\le Max\{|h(z)|\,;|z|=1\}$, I am done?
 A: One neat trick to see $\max \{ \lvert h(z)\rvert : \lvert z\rvert = 1\} \geqslant 1$ is to look at the function
$$k(z) = z^n h(1/z).$$
Then $k(0) = 1$, hence
$$1 \leqslant \max \{ \lvert k(z)\rvert : \lvert z\rvert = 1\}.$$
(Note: since $k$ is not constant, the inequality is in fact strict.)
But we have
$$\max \{ \lvert k(z)\rvert : \lvert z\rvert = 1\} = \max \{ \lvert h(z)\rvert : \lvert z\rvert = 1\}.$$
However, as you noted, the maxima of $\lvert h\rvert$ and $\lvert g\rvert$ on the unit circle need not occur at the same point(s), so the maximum of the product can be smaller than the product of the maxima.
A small modification of the argument fixes that, however. The case $n = 0$ is clear, hence we assume $n > 0$ in the following.
If we replace $h$ with a function with the same zeros, the same value at $0$, and $\lvert h(z)\rvert = 1$ for all $z$ on the unit circle, then the assertion follows from $\lvert g(0)\rvert \geqslant 2^n$.
The automorphisms of the unit disk are precisely the functions of the form
$$T_{a,\lambda}(z) = \lambda\frac{z-a}{1-\overline{a}z},$$
where $\lvert a\rvert < 1$ and $\lvert\lambda\rvert = 1$. In particular, we have $\lvert T_{a,\lambda}(z)\rvert = 1$ for $\lvert z\rvert = 1$.
So instead of $(z-z_1)\cdot \dotsc \cdot (z-z_n)$, we look at
$$\tilde{h}(z) = \prod_{k=1}^n \frac{z-z_k}{1-\overline{z_k}\cdot z}.$$
Then $\tilde{h}$ is holomorphic in a neighbourhood of the closed disk $\{ z : \lvert z\rvert \leqslant 2\}$, $\tilde{h}$ has precisely the $z_k,\, 1 \leqslant k \leqslant n$ as zeros,
$$\lvert\tilde{h}(0)\rvert = \prod_{k=1}^n \lvert z_k\rvert < 2^{-n},$$
and $\lvert \tilde{h}(z)\rvert = 1$ for $\lvert z\rvert = 1$.
Then writing $f(z) = \tilde{h}(z) \cdot \tilde{g}(z)$, we have $\lvert \tilde{g}(0)\rvert > 2^n$ and by the maximum modulus principle it follows that
$$\max \{ \lvert f(z)\rvert : \lvert z\rvert = 1\} = \max \{ \lvert \tilde{g}(z)\rvert : \lvert z\rvert = 1\} > 2^n.$$
