Homeomorphism from interior of unit disk and 0 as a fixed point $f$ is analytic in $|z|<1$, $|f(z)|<1$ and $f(0)=0$, $f'(0)=\alpha \ne 0$
Find a sharp estimate of the radius $r$ of a disc such that for: 
$z1$,$z2$ in the disc: $|z1| \lt r$ and $|z2| \lt r$
$$ f(z1)=f(z2) \Rightarrow z1=z2$$
Cauchy's integral theorem can help me to prove that f is injective:
\begin{align}
f(z1) - f(z2) &= \frac{1}{2\pi i} \int_{C_r} (\frac{1}{t-z1}-\frac{1}{t-z2}) f(t) dt \\
&= \frac{z1-z2}{2\pi i}  \int_{C_r} \frac{f(t)} {(t-z1)(t-z2)}dt
\end{align}
And then use the fact $f'(0)$ being non zero then $z1=z2$ in the neighborhood of $0$.
But I don't think this is rigorous and also it does not give me a method to find the estimate.
 A: Theorem (Landau-Dieudonne):
Let $f$ be analytic in $|z|<1,$ $|f(z)|<1$ and $f(0)=0,$ $f^\prime(0)=\alpha \ne 0.$
Then $f(z)$ is univalent in $|z|<r=\frac{|\alpha |}{1+\sqrt{1-|\alpha |^2}}$.
The estimate $r=\frac{|\alpha |}{1+\sqrt{1-|\alpha |^2}}$ is sharp. 
Addendum:
Lemma 1. Suppose that $g(z)$ is analytic in $|z|<1$, $|g(z)|<1$. Then $$
\frac{|g(0)|-|z|}{1-|g(0)||z|}\le |g(z)|\quad (|z|<1).$$
This is frequently asked in this site, see   Schwarz lemma problem
for instance.  
Lemma 2. Suppose that $f(z)$ is analytic in $|z|<1$, $|f(z)|<1$, $f(0)=0$. Then $$
\frac{\left(|f^\prime(0)|-|z|\right)|z|}{1-|f^\prime(0)||z|}\le |f(z)|\quad (|z|<1).$$
Proof. Apply Lemma 1 to $g(z)=\frac{f(z)}{z}.$ Note that $g(0)=f^\prime (0).$  
Lemma 3.  Suppose that $f(z)$ is analytic in $|z|<1$, $|f(z)|<1$, $f(0)=0$. If there exist $z_1,z_2$ $(z_1\ne z_2)$ with $|z_1|=|z_2|=r$, $f(z_1)=f(z_2)=\omega$, then $|\omega|\le r^2$.
Proof. Let $$
F(z)=\frac{f(z)-\omega}{1-\bar{\omega}f(z)}\cdot \frac{1-\bar{z_1}z}{z-z_1}\cdot\frac{1-\bar{z_2}z}{z-z_2}.$$
Note that $F(z)$ is analytic in $|z|<1$ and $|F(z)|\le 1$. Especially we have $$|F(0)|=\left|\frac{\omega}{z_1z_2}\right|\le 1.$$ So we have $|\omega|\le |z_1z_2|=r^2.$
Now we prove Landau-Dieudonne's Theorem.
Suppose that $f(z)$ is not univalent in $|z|< \rho $. Then there are two points $z_1,z_2$ ( $|z_1|<\rho ,\,|z_2|<\rho )$ such that $f(z_1)=f(z_2)=\omega .$
Let $C_\rho =\{z:\, |z|=\rho \}$ and $\Gamma _\rho =f(C_\rho ),$ the image of $C_\rho $ under $f$. 
Since $f(z)$ has (at least) two $\omega $-points $z_1$ and $ z_2,$ $$
\frac{1}{2\pi} \int_{C_\rho } d\arg (f(z)-\omega )=\frac{1}{2\pi i}\int_{C_\rho }\frac{f^\prime(z)}{f(z)-\omega }dz\ge 2.\tag{1}$$
Suppose that $\Gamma _\rho $ is a Jordan curve. Then 
$$
\frac{1}{2\pi} \int_{C_\rho } d\arg (f(z)-\omega )=1,$$
since $\omega $ lies inside $\Gamma _\rho$. This is contrary to $(1)$. Therefore $\Gamma _\rho $ is not a Jordan curve, and hence there are two points $\zeta _1,\,\zeta _2$ on $ C_\rho $ such that $f(\zeta _1)=f(\zeta _2)$. By lemma 3, we have $$
|f(\zeta_1)|\le \rho ^2.$$
Lemma 2 says that 
$$
\frac{\left(|\alpha |-|\zeta _1|\right)|\zeta _1|}{1-|\alpha ||\zeta _1|}\le |f(\zeta _1)|,$$
and hence we have $$
\frac{\left(|\alpha |-\rho \right)\rho }{1-|\alpha |\rho }\le \rho ^2,$$
which leads to $$
\rho \ge \frac{|\alpha |}{1+\sqrt{1-|\alpha |^2}}.$$
Thus we have proven that $f(z)$ is univalent in $|z|<r=\frac{|\alpha |}{1+\sqrt{1-|\alpha |^2}}.$
About sharpness:
Consider $f(z)=\frac{z(\alpha -z)}{1-\alpha z}$ $(0<\alpha <1)$ with $f(0)=0,\, f^\prime(0)=\alpha $. This function satisfies $f^\prime(z_0)=0$ at $z_0=\frac{\alpha }{1+\sqrt{1-\alpha ^2}}$. Therefore $f(z)$ is univalent only in $|z|<\frac{\alpha }{1+\sqrt{1-\alpha^2}}$, in other words, the estimate is sharp. 
