I came across this statement in my lecture:
"Let $U, V \subset \mathbb{C}$ open sets, if $f:U \rightarrow V$ is a $homeomorphism^1$ which is $complex \space differentiable^2$ $\forall z \in U$, then $f'(z)\neq0$ $\space$ $\forall z \in U$."
And I have the followings definitions:
$^1$ $\space$Let $U, V \subset \mathbb{C}$ open sets, a bijection $f:U \rightarrow V$ is a homeomorphism if $f$ and $f^-1$ are continues.
$^2$ $\space$ Let $U \subset \mathbb{C}$ open set, a complex function $f:U \rightarrow \mathbb{C}$ is complexe differentiable at point $z_0 \in \mathbb{C}$ if the limit $ \lim_{z\to\ z_0}\frac{f(z)-f(z_0)}{z-z_0}$ exists.
I know this isn't true for a real function, but i can't find a precise argument for a complex one. Any help or hints will be really appreciated.
This is not an easy thing to prove if you didn't take a course in complex analysis. Basically, the fundamental fact about complex differentiable functions is that they can (locally) be written as entire series: $$f(z)=\sum_{n=0}^\infty a_n (z-z_0)^n$$ From there, one can prove that if $f$ has this form and $f'(z_0)=0$ (ie $a_1=0$) then $f$ is not locally injective near $z_0$, and therefore cannot be a homeomorphism.
These facts can be found in any textbook on complex analysis, but again are not trivial.