With hypotheses of Schwarz's lemma, estimate the radius around zero where $f$ must be one-to-one 
Suppose $f(z)$ is analytic in the open unit disc and $|f(z)|<1$ there.
  Suppose further that $f(0) =0$ and $f'(0) = a \neq 0$. Show that there
  is a disc of positive radius $|z|<\rho$ such that for $z_1$ and $z_2$
  in the disc, $$f(z_1)=f(z_2) \Longrightarrow z_1=z_2.\tag{1}$$ Find an
  estimate for $\rho$. Try to make the estimate as sharp as you can.
  Hint: $$f(z_2)-f(z_1) = \int_{z_1}^{z_2}f'(z)dz = a(z_2-z_1)+ \dotsb .$$

My answer so far: I can show (1) in a neighborhood of zero as follows: $z=0$ is an isolated zero of $f$ by the isolated zero theorem (note $f$ is non-constant since $f'(0)\neq 0$). Therefore there is a closed neighborhood $\overline{B_{\rho}(0)}$ of zero such that $f\neq 0$ on the punctured disc $\overline{B_{\rho}(0)} \setminus \{0\}$. Let $M = \max_{z\in \partial{B_{\rho}(0)}}|f(z)|$. Now for $w\in B_{M}(0)$, we have that $|f(z)-0|> |w-0|$ for $z\in \partial B_\rho(0)$, so $f(z)-0$ and $f(z)-0 + (w-0)=f(z)-w$ have the same number of zeros in $\overline{B_{\rho}(0)}$, meaning that $f$ is one-to-one there.
For the estimate, I'm having more trouble. I know by Schwarz's theorem that $|f(z)|<|z|$ in the disc, and $a<1$, unless $f(z)=\lambda z$ for $\lambda \in S^1$, in which case both are equalities. But how can I use this?
If I try to use the hint, I get (I think) $$f(z_2)-f(z_1) = a(z_2-z_1)+ \frac12 f''(0)(z_2^2 - z_1^2) + \dotsb$$ but I'm not sure what to do with this. 
 A: Hint. Try to find a maximum disk $D(0,r)$ where the inverse of $f$ is definable as a holomorphic function.
A: I interpret the hint as the suggestion to apply the  Mean Value Inequality to $f(z)-az$: 
$$|f(z_2)-f(z_1) - a(z_2-z_1)| \le |f'(\xi)-a| |z_2-z_1|$$ 
where $\xi$ is on the line segment between $z_1$ and $z_2$. So, you have injectivity in $|z|<\rho$ if you can show that $|f'-a|<|a|$ there. 
One way to go from here is to use Cauchy's integral formula: 
$$
f'(z)-f'(0) = \frac{1}{2\pi i} \int_{|\zeta|=r}
\left(\frac{1}{(\zeta-z)^2} - \frac{1}{\zeta^2}\right)f(\zeta)\,d\zeta 
$$
This way, the given bound for $|f|$ fits right, and the rest is algebra around 
$$
\frac{|z| \, |2\zeta-z|}{|\zeta-z|^2 |\zeta|^2}
$$
Pushing $r\to 1^-$, you can get this bounded by
$$
\frac{\rho (2+\rho) }{(1-\rho)^2 }
$$
As long as this  thing is $<|a|$, injectivity holds.

The beginning of the solution can be worded differently. Write $f(z)=a(z+g(z))$. If you can find a region where $g$ is a strict contraction, then $z+g(z)$ is injective there (easy to see). The rest is about getting $|g'|<1$. 
A: Alternatively,try to find the minimum $ρ>0$ such that if $f(z_1)=f(z_2)$ then $z_1\neq z_2$.
This has logic because the $z_1,z_2$ that have the property $(1)$ lie in a disk centred at $0$. So all the others will be outside.
Now let $z_1,z_2$ such that $f(z_1)=f(z_2)$ .Then from the hint we have $a(z_2-z_1)+c_2(z_2^2-z_1^2)+...+c_n(z_2^n-z_1^n)+...=0$ $(*)$ with $$c_n=\frac {f^{(n_)}(0)}{n!}$$.
Now for these $z_2\neq z_1$ we can divide with $z_2-z_1$ the $(*)$.Until when you can divide with $z_2-z_1$? I mean which is the biggest $k$ such that if you divide with $(z_2-z_1)^k$ everything is OK?
A: We have f(0) = 0 and f'(0) $\ne$ 0.  Let c > 0 be such that f has no zeros other than 0 in |z| < c, and for some $z_1, z_2$ in this domain we have $f(z_1) = f(z_2)$ .  We have
$2 \pi if(z_1) = \oint_{|z| = c} f(z)/(z- z_1)dz = 2 \pi if(z_2) = \oint_{|z| = c} f(z)/(z- z_2)dz.$  So these two integrals are equal.
This gives $\oint_{|z| = c} [f(z)/(z- z_1)] -[f(z)/(z- z_2)] dz = 0$ and
$\oint_{|z| = c} f(z)(z_1- z_2)/[(z-z_1)(z-z_2)] dz  = 0.$
If f(z) $\ne$ 0 in this domain (except at zero) there are two cases:
1.  f(z) does not have period 2$\pi$. Then the only term in this last integral that can be 0 is $(z_1- z_2)$ so we conclude $z_1= z_2$ in this domain.
2.  f(z) is periodic in $2\pi$.  Then f($2\pi$) = 0.  Then c must be less than
the smallest divisor of $2\pi$ in which f is periodic.
Suppose $f(z_0) = 0$ where $z_0 \ne$ 0 and |$z_0$| is the smallest modulus of the zeros of f (outside of 0).   Then $f(z_0) = f(0)$, so we have a duplicate value. 
Thus c must be less than the modulus of the nearest zero or singularity of f. 
If there is a smaller bound on c this argument does not show it.
