Let $B$ be the unit open disk in $\mathbb{C}$ and $f:B\rightarrow \mathbb{C}$ be a holomorphic function satisfying $|f'(z)-f'(0)|<|f'(0)|$ on $B$.
Then, how do I prove that $f$ is injective?
By the Rouché theorem, $f'$ is never zero on $B$. How do I proceed next?
$\lvert f'(z) - f'(0)\rvert < \lvert f'(0)\rvert$ means that $f'(z)$ lies in the open disk with radius $\lvert f'(0)\rvert$ and centre $f'(0)$. This disk is contained in an open half-plane, hence there is an $\alpha \in \mathbb{C}$ with $\lvert\alpha\rvert = 1$ such that
$$\operatorname{Re} \bigl(\alpha\cdot f'(z)\bigr) > 0$$
for all $z \in B$. It follows that
$$\operatorname{Re} \alpha\int_0^1 f'(z + t(w-z))\,dt > 0$$
for all $z,w \in B$. In particular the integral doesn't vanish, and hence neither does
$$f(w) - f(z) = (w-z) \int_0^1 f'(z + t(w-z))\,dt$$
unless $w = z$.
Expressing $$ f(z)=f(0)+f'(0)z+g(z) $$ with $g$ holomorphic in $B$, we have, for $z\ne w$, and $z,w\in B$, $$ f(z)-f(w)=(z-w)\,f'(0)+\int_w^z g'(\zeta)\,d\zeta. $$ In order to show that $f$ is injective, it suffices to show that $f(z)=f(w)$ if and only if $z=w$, and hence the above vanishes only for $z=w$, which in turn is equivalent to $$ f'(0)+\frac{1}{z-w}\int_w^z g'(\zeta)\,d\zeta\ne 0. $$ We do know that $\,f'(z)=f'(0)+g'(z)$, and hence $$ \lvert\,f'(0)\rvert>\lvert\,f'(z)-f'(0)\rvert=|g'(z)|. $$ Hence $$ \left|\frac{1}{z-w}\int_w^z g'(\zeta)\,d\zeta\,\right|\le \max_{\zeta\in[z,w]}|g'(\zeta)|<|\,f'(0)|. $$
Note. Observe that this proof work in the case even when $\,f:U\to\mathbb R^n$, is a $C^1$ mapping and $U\subset\mathbb R^n$ is a convex neighbourhood of the origin. In fact, satisfaction of the condition $\,\|\,f'(0)-f'(x)\|<\|\,f'(0)\|,\,$ guarantees the existence of a smooth inverse.
