Prove that a analytic function is one-one Is the following statement true?
Suppose, $ f: D\to \mathbb C $ is an analytic function, where $ D $ is the disc of radius 1 around 0(including the circle of radius 1). $ |f(z)|=1\forall |z|=1 $. $ f(z) $ has only one $ 0 $ in $ D $ with multiplicity $ 1 $. Prove that $ f $ is one-one in $ D $.
Sorry, I wrote the problem wrong before.
 A: If $f$ has only one and simple root at $z=0$, then $f(z)=zg(z)$, with $g$ analytic and $g(z)\ne 0$, for all $z\in \overline{D}$. 
$$
\max_{|z|\le 1}\lvert f(z)\rvert=\max_{|z|= 1}\lvert f(z)\rvert=1,
$$ 
and
$$
\max_{|z|\le 1}\lvert g(z)\rvert=\max_{|z|= 1}\lvert g(z)\rvert=\max_{|z|= 1}\left\lvert \frac{f(z)}{z}\right\rvert=1.
$$
In particular $\lvert g(z)\rvert =1$, for $\lvert z\rvert=1$. But as $g(z)\ne 0$, then $h(z)=1/g(z)$ is also analytic and
$$
\max_{|z|\le 1}\lvert h(z)\rvert=\max_{|z|= 1}\lvert h(z)\rvert=1.
$$
But
$$
\max_{|z|\le 1}\lvert h(z)\rvert=\frac{1}{\min_{|z|\le 1}\lvert g(z)\rvert},
$$
which implies that $\lvert g(z)\rvert=1$ for all $z\in D$, and hence $g(z)=a$, for some $a$ with $\lvert a\rvert=1$, and consequently
$$
f(z)=az.
$$
A: Yes, $f$ is one-to-one - because by the Schwarz lemma (case of equality), $f(z)$ must equal $az$ for some $a$ with $|a|=1$.
A: If $f$ is analytic with only one zero at zero, then $\frac{f(z)}{z}$ and $\frac{z}{f(z)}$ are both analytic functions on the unit disc. By the maximum modulus principle, we must have $|\frac{f(z)}{z}|\le 1$ and $|\frac{z}{f(z)}|\le 1$ on the unit disc. Thus there is a constant $c$ such that $\frac{f(z)}{z}=c$, $|c|=1$, hence, $f(z)=c\cdot z$ for every $z$ in the unit disc. We now see that $f$ is certainly one to one. 
