# Thread: Locally injective holomorphic map

1. ## Locally injective holomorphic map

Hello,

I'm having trouble with part of the following problem:

Suppose $G$ is an open set in the complex plane $\mathbb{C}$, and let $a$ be a point of $G$. Let $f$ be a holomorphic function on $G$, and suppose $f'(a)\neq 0$.

First I'm suppose to show that there exists an $r>0$, such that $|f'(z)-f'(a)|<|f'(a)|$ for all $z\in B_r(a)$, the ball of radius $r$ and center $a$.

This follows from continuity of $f'$ at $a$, with epsilon $=|f'(a)|$.

Then I'm showing that for all $z_1,z_2\in B_r(a)$, it holds that

$|\int_0^1 f'(tz_2+(1-t)z_1)\ dt - f'(a)| < |f'(a)|.$

This follows from the calculation

$|\int_0^1 f'(tz_2+(1-t)z_1)\ dt - f'(a)|$
$= |\int_0^1 f'(tz_2+(1-t)z_1) - f'(a)\ dt|$
$\leq \int_0^1 |f'(tz_2+(1-t)z_1) - f'(a)|\ dt$
$< \int_0^1 |f'(a)|\ dt = |f'(a)|,$

with the last inequality following from the fact that the line segment between $z_1$ and $z_2$ is contained in $B_r(a)$.

Then I'm asked to prove that for all $z_1$ and $z_2$ in $B_r(a)$, it is true that
$f(z_2)-f(z_1) = (z_2-z_1)\int_0^1 f'(tz_2+(1-t)z_1)\ dt,$

which I do by substituting a single variable for the expression $tz_2+(1-t)z_1$ in the integral.

My problem is to show that this implies that $f$ restricted to the ball $B_r(a)$ is injective, which boils down to showing that the integral is non-zero for $z_1\neq z_2$. Any ideas?

2. Well, if $0=\int_0^1 f'(tz_2+(1-t)z_1)\ dt$, then $|\int_0^1 f'(tz_2+(1-t)z_1)\ dt - f'(a)| < |f'(a)|$ implies $|f'(a)|<|f'(a)|$...

3. Ah! I missed this, thanks for pointing it out.