A complex map with "bounded" derivative is injective The exercise I try to solve states: "Let $\,f\,$ be analytic in $\,D:=\{z\in\mathbb{C}\;|\;|z|<1\}\,$ , and such that $$|f'(z)-1|<\frac{1}{2}\,\,\,\forall\,z\in D$$
Prove that $\,f\,$ is $\,1-1\,$ in $\,D\,$.
My thoughts: The condition $$|f'(z)-1|<\frac{1}{2}\,\,\,\forall\,z\in D$$
means the range of the analytic function $\,f'\,$ misses lots of points on the complex plane, so applying Picard's Theorem (or some extension of Liouville's) we get that $\,f'(z)=w=\,$ constant, from which it follows that $\,f\,$ is linear on $\,D\,$ and thus $\,1-1\,$ there.
Doubts: $\,\,(i)\,\,$ This exercise is meant to be from an introductory first course in complex functions, so Picard's theorem seems overkill here...yet I can't see how to avoid it.
$\,\,(ii)\,\,$ Even assuming we must use Picard's Theorem, the versions of it I know always talk of "entire functions", yet our function $\,f\,$ above is analytic only in the open unit disk. Is this a problem? Perhaps it is and thus something else must be used...?
Any help will be much appreciated.
 A: Use the fact that the inverse of $f$ is holomorphic in every disk centered at some $w_0 \in f(D)$ since
$$
\lim_{w \to w_0} \frac{f^{-1}(w) - f^{-1}(w_0)}{w-w_0} = \lim_{z \to z_0} \frac{z - z_0}{f(z) - f(z_0)} \to \frac{1}{f'(z_0)},
$$
and the condition $|f'(z)-1| < \frac 12$ ensures that $f'(z_0) \neq 0$.  Therefore $f$ is injective. If you fill in the blanks this is all you need.
EDIT : Yeah... I thought I had it. But as comments noticed I don't think my approach is so straight forward. As requested, I'll fill in some blanks that I thought helped. Recall that 
$$
\frac 1{2\pi i} \int_C \frac{f'(z)}{f(z)} \, dz = Z_f - P_f
$$
for a meromorphic function, $Z_f$ standing for the number of zeros of $f(z)$ in the interior of $C$ and $P_f$ the number of poles (assuming there are no zeros/poles on the contour). Since in our context the function is analytic, it is in particular holomorphic, hence has no poles over $D$, so that this integral counts the number of zeros. Let $z_0 \in D$, and in the above integral, define $N(w)$ by substituting $f(z)$ above by $f(z) - w$ :
$$
N(w) = \frac 1{2 \pi i} \int_{C_r} \frac{f'(z)}{f(z) - w} \, dz 
$$
where $C_r$ stands for a circle of sufficiently small radius around $z_0$ (small enough so that $N(f(z_0)) = 1$), so that the integral counts how many $z$ are such that $f(z) = w$ close enough to $z_0$. You can show that $N(w)$ is continuous, so that since $N(f(z_0)) = 1$, $N(w) = 1$ by continuity for all $w$ such that $|w-f(z_0)| < r$. This means $f$ is locally injective, and furthermore $f(D)$ is open since $|f'(z) - 1| < 1/2$ implies $f'(z) \neq 0$, hence $f(z)$ is not constant (this would be the only case where $f(D)$ would not be open).
Hope that helps,
A: There is a very similar answer here (feel free to ignore the question if it looks intimidating).  The basic idea is to use the triangle inequality applied to $f(z)-z$ to show that $|f(x)-f(y)| > 0$ for any distinct $x,y$.  More precisely your assumption allows us to show $|f(x)-f(y)| \ge \frac12 |x-y|$.
