Prove that if derivative of a holomorphic function is non-zero, then it is locally univalent

Question

Let $f$ be a non-constant entire function such that $f'(z)≠ 0$. Then $f$ is locally univalent.

Try relating the holomorphic derivative to the usual derivative (when viewing $f$ as a map of $\mathbb{R}^2$ into itself). — WoolierThanThou, Jul 20 '19 at 08:43
I got it what you are saying. But I am looking for conventional proof by not viewing f as subset of $\mathbb{R^2}$ — Nik, Jul 20 '19 at 08:55
Appealing to the Inverse Function Theorem is very much the conventional way of proving the statement. — WoolierThanThou, Jul 20 '19 at 09:30
That means whenever a non constant holomorphic function has non vanishing derivative, then that function is locally invertible? — Nik, Jul 20 '19 at 10:48
It means if you have a $C^1$ map $f: U\to \mathbb{R}^n$ with $U\subseteq \mathbb{R}^n$ open, then for any point $x\in U$ such that $Df(x)$ is non-singular (i.e. invertible), then there exists some $\varepsilon$ such that $f$ restricted to $B(x,\varepsilon)$ is an invertible function onto its image. — WoolierThanThou, Jul 20 '19 at 10:54

score 1 · Accepted Answer · answered Jul 20 '19 at 11:19

A proof "by complex analysis", no IFT needed:

Normalize $f$ so $f(0)=0$, $f'(0)=1$: $$f(z)=z+O(|z|^2).$$You can use Rouche to prove

There exist $r,\delta>0$ such that if $|\alpha|<\delta$ then $f-\alpha$ has exactly one zero in $\{|z|<r\}$.

Now choose $\rho\in(0,r)$ so that $|z|<\rho$ implies $|f(z)|<\delta$ and it follows that $f$ is univalent in $\{|z|<\rho\}$.

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