If $f$ is analytic at the interior of the circle $C(0;1)$ and if $|f(z)|<1$ for $|z|=1$, show that there exists a only point $z_0$ at the interior of this circle such that $f(z_0)=z_0$
Rouché's theorem : Let $D \subset \mathbb{C}$ a domaine and $h,g: D \to \mathbb{C}$ two holomorphic functions in $D$. Let $C$ a closed path contained in the interior of $D$. If $|h(z)+g(z)| < |h(z)|+|g(z)|$, $\forall z \in \mathbb{C}$, then $h$ and $g$ have the same number of zeroes in the interior of $C$.
I think I have to use Rouché's theorem with $h(z)=f(z)-z$, but it is unclear for me. Does anyone could give me some details?
On the unit circle we have $|f(z)-z+z| =|f(z)|< 1 = |z|$, hence $z \mapsto z$ and $z \mapsto f(z)-z$ have the same number of zeroes in the unit disk. Hence there is exactly one point in the unit disk for which $f(z) = z$.