i would really appreciate if you can help me with the following demonstration please:
Let $C$ be a closed simple contour such that $| f (z) | = cte$ ($f$ is analytic inside $C$ and is not an identical constant). Show that $f$ has at least one zero inside $C$.
Assuming $f$ is analytic inside $C$ and continuous on $\operatorname{Int}C\cup C$ (here $\operatorname{Int}C$ denotes the interior of the simple closed curve $C$) we'll make use of the maximum principle which states that an analytic function attains its maximum values on the boundary of its domain given that such domain is bounded, which is the case here.
Assume $f$ has no zeros inside $C$ (i.e. in $\operatorname{Int}C$). Apply the maximum principle to $1/f$ to deduce that $f$ possesses a minimum value at some point $z_0\in C$. Then $$ M=|f(z_0)|\leq |f(z)|\leq M\quad\forall z\in\operatorname{Int}C , $$ where $M=|f|$ on $C$. We have now proved that $|f|=M$ on the whole of $\operatorname{Int}C\cup C$. What's left here is to prove that this condition implies that $f$ is constant, contrary to our hypothesis. I'll leave this to you: Consider the open mapping theorem for analytic functions.