Let $\displaystyle f$ be analytic and copies the open unit disc $\displaystyle D$ onto itself. assume $\displaystyle f$ is 1-1 and $\displaystyle f(0)=0$. Show that $\displaystyle f=e^{it}z, t\in R$ for $\displaystyle |z|<1$.

OK, so according to Schwartz's lemma all I need to show is that there exists $\displaystyle 0\neq z\in D$ such that $\displaystyle |f(z)|=|z|$ or that $\displaystyle f'(0)=1$, how do I use the fact that $\displaystyle f$ is 1-1 to show this? I was trying to define a new function such as $\displaystyle \frac{f(z)}{a}$ where $\displaystyle a$ is a point in the unit disc, but didnt get anywhere.