This is the problem I'm trying to solve (I'm not sure it is solvable):
Suppose $f$ is an analytic function on a neighborhood of $\overline{B(0,1)}$, and that $|f(z)|=1$ whenever $|z|=1$, and also that $f$ has at least one zero. Show that $f$ restricted to $B(0;1)$ is 1-1 and onto itself.
I already know by the Maximum Modulus Theorem that $f(B(0;1))\subset B(0;1)$.
I have showed that $f$ is onto by contradiction, so suppose $\alpha \in B(0;1)$, $\alpha\neq 0$, is not in $f(B(0;1))$, then define the function $g$ on a neighborhood of $\overline{B(0,1)}$, given by
$$ g(z) = \frac{1}{f(z)-\alpha} $$
Since $f(z)\neq\alpha$ for all $z$ in that neighborhood, g is analytic, so the maximum value of $g$ on $\overline{B(0,1)}$ occurs on $\partial B(0;1)$, hence there is $c\in\mathbb{C}$ such that $|c|=1$ and
$$ \frac{1}{|f(z)-\alpha|} < \frac{1}{|c-\alpha|}\ \ \ \ \forall\ z\in B(0;1) $$
Which means that
$$ |f(z)-\alpha| > |c-\alpha|\ \ \ \ \forall\ z\in B(0;1) $$
And since $f$ has at least one zero in $B(0;1)$, we have
$$ |\alpha| > |c-\alpha| \geq |c| - |\alpha| = 1 - |\alpha| $$
Thus,
$$ |\alpha| > \frac{1}{2} $$
So we've concluded that $\overline{B(0;1/2)}\subset f(B(0;1))$, and if you define
$$ r_1 = \frac{1}{2};\ \ r_2 = \frac{1}{2} + \frac{1}{4}; \ \ \dots \ \ r_n = \sum_{k=1}^n \frac{1}{2^n} $$
it is not difficult to see that the facts that $\overline{B(0;r_1)}\subset f(B(0;1))$ together with $|f(z)-\alpha| > |c-\alpha|\ \ \ \ \forall\ z\in B(0;1)$, implies that $\overline{B(0;r_2)}\subset f(B(0;1))$, which implies that $\overline{B(0;r_3)}\subset f(B(0;1))$, ans so on.
Since $\lim r_n = 1$ we have that $B(0;1)\subset f(B(0;1))$, thus proving that $f$ is onto.
The fact that $f$ is 1-1 I'm not able to prove, I'm thinking it may be false, but I can't think of an example, can anyone help me on this one?