# Analytic function mapping disk into disk, Does $f(1)=1$ implies $f$ is locally one to one at $1$?

Let $f$ be analytic on an open set containing unit disc. If $|f(z)|\le 1$ for $z$ in open disc and $f(1)=1$, prove that $f'(1)\ne 0$. On similar lines is a problem I have been trying since several days: suppose you have a polynomial $p(z)= z^n+ q_{n-1}(z)$ such that $|p(z)|\le 1$ for all $|z|\le 1$, then actually $|p(z)|<1$ for all $|z|\le 1$ or $q=0$. Any help will be greatly appreciated.

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The down vote is probably because you phrased your post as a command. Also, it would be helpful to tell us what you have tried so far. –  Trevor Wilson Dec 22 '12 at 4:37
The second part does not look right: take $p(z)=(z+1)/2$. –  user53153 Dec 22 '12 at 5:23

Hint: Suppose that $f'(1) = 0$. Then the power series of $f$ around $1$ is of the form $$f(z) = 1 + a_n(z-1)^n + a_{n+1}(z-1)^{n+1} + \cdots,$$ where $a_n\neq 0$ and $n\geq 2$. When $z$ is very close to $1$, it follows that $f(z)\approx 1 + a_n(z-1)^n$. Using such an approximation, can you show that $f$ maps some points within the unit disk outside the unit disk?