TITLE: A complex map with "bounded" derivative is injective
QUESTION [6 upvotes]: 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.

REPLY [10 votes]: 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 $|f(x)-f(y)| \ge \frac12 |x-y|$.