Suppose $f: \overline{\mathbb{D}} \to \mathbb{C}$ is continuous, analytic in $\mathbb{D}$ and satisfies $|f(z)|<1$ for $|z|=1$. Find the number of solutions to the equation $f(z) = z^n$ where $n$ is a positive integer. ($\mathbb{D}$ is the unit disk).
Some of my own attempts: Using the Maximum/Minimum Principle, we can split up the the problem in $2$ parts: Suppose $f$ is constant. It then holds that $z^n = c$ has exactly $n$ zeros because of the fundamental theorem of algebra.
Suppose $f$ is not constant. It holds that $f$ has at least one zero in $\mathbb{D}$. It of course also holds that $z^n$ has a zero of multiplicity $n$ at the origin. I think I'm supposed to hit with stuff like Rouché's theorem, the Argument Principle or Schwartz lemma, but nothing seems to fit.
Thanks in advance.
$g_t(z)=z^n-tf(z)$ has $n$ solutions inside the unit disk for $t=0$. Due to the continuity of the roots of holomorphic functions, one can continue these $n$ solutions for values of $t\in[0,1]$. Since $|g_t(z)|=1-t|f(z)|>0$ on the unit circle for all these $t$, no root path can cross the unit circle, so no root can escape and no root can enter the unit disk. Thus the number of roots inside the unit disk stays at $n$ until $t=1$.
For a more exact and in its entirety more elementary argument, use the Rouché theorem.
Set $a(z)=z^n$ and $b(z)=z^n-f(z)$, then $$ |a(z)-b(z)|=|f(z)|<1=|a(z)|\le|a(z)|+|b(z)|\quad\text{for}\quad |z|=1 $$ so the assumptions of the Rouché theorem are satisfied in all of its forms and $a(z)$ has the same number of roots inside the unit disk as $b(z)$.