If $f:\mathbb{C} \to \mathbb{C}$ is analytic and $|f(z)| \leq 1$ for all $z$ with $|z|=1$. What can be said about the solutions of $f(z)=z$ in the unit disc |z|<1?
Clearly, if $f$ is the constant function $1$, there is no solution. And if you take $f(z)=z$, there are infinite solutions. But except those functions, is there something to be said? I know that if $|f(z)|<1$ on the unit circle, then by Rouché's theorem there is exactly one solution in the unit disc.
In this case we only know that $|f(z)|<1$ (if $f$ is non constant) according to the maximum modulus principle.
Suppose there is at least one solution $z_0$ in the unit disk $D$ such that $f(z_0) = z_0$. Now consider the holomorphic function $h = g \circ f \circ g^{-1}$ where $g$ is the Möbius transformation of $D$ mapping $z_0$ to $0$, namely $$ g :z\mapsto \frac{z-z_0}{1 - \overline{z_0}z}. $$ Since $h(D) \subset D$ and $h(0)=0$, the Schwarz Lemma shows that either $h$ is the identity function or the equation $h(z) = z$ has no other solution than $0$.
In conclusion, there are only three possibilities:
$f(z) = z$ has no solution.
$f$ is the identity function.
$f(z) = z$ has exactly one solution.