If $f'(z_0)\neq 0$ then $f$ is one to one on some open disk $D_r(z_0)$ This is what I am trying to prove 

Let $D\subset\mathbb{C}$ be open and $f$ be analytic in $D$. If there is $z_0\in D$ such that $f'(z_0)\neq 0$ then there exists $D_r(z_0)\subset D$ and $f$ is one to one in $D_r(z_0)$. 

I am finding it difficult to prove this but on some disk $D_r(z_0)\subset D$ $f'=0$ due to continuity. But I just can't go ahead and get anything out of this. ANy hints will be appreciated. Thanks
 A: This is a special case of the inverse function theorem which tells you something more: the inverse function is itself differentiable.
If you just want to prove that your function is one-to-one in a small circle you might reason as follows.
Consider the incremental ratio:
$$
  g(w,z) = \frac{f(w)-f(z)}{w-z}
$$
and define $g(z,z) = f'(z)$ so that $g$ results in a continuous function defined on $D\times D$. You know that $|g(z_0,z_0)| = |f'(z_0)|=c>0$. Take $r>0$ small enough so that $|g(z,w)| > c/2$ for all $z,w \in D_r(z_0)$. Then:
$$
 |f(z)-f(w)| = |g(z,w)(z-w)| = |g(z,w)|\cdot |z-w| \ge \frac c 2 |z-w| > 0
$$ 
if $z\neq w$.
Hence the function is one-to-one.
A: The Lagrange inversion theorem says that if $f'(z_0)\neq 0$ then $f$ is invertible near $z_0$.
The theorem gives an explicit power series for the inverse function.
A: HINT: An idea, why it should be true. On a small disk the higher derivatives are meaningless, hence the graph of the function is near a plane on this disk.
A: Hint: Let $r>0$ such that $f'(z)\neq 0$ for all $z\in D(z_0,r)$. First prove that $\tilde{f}=f|_{D(z_0,r)}$ is injective. Suppose that $\tilde{f}$ is not injective. By Complex Mean Value Theorem for all $z_1,z_2 \in D(z_0,r) $there are $u,v\in D(z_0,r)$ such that
$$
\mbox{Re}\{\tilde{f}(u)\}=\mbox{Re}\left\{\frac{\tilde{f}(z_2)-\tilde{f}(z_1)}{z_2-z_1}\right\},
\\
\mbox{Im}\{\tilde{f}(v)\}=\mbox{Im}\left\{\frac{\tilde{f}(z_2)-\tilde{f}(z_1)}{z_2-z_1}\right\}.
$$
Use this to get a contradiction and conclude that $\tilde{f}$ is injective. For sufficiently small disk of radius $r>0$ have to $\tilde{f}$ is surjective. Thus $\tilde{f}$ is $1-1$.
