Given $f\in Hol(|z|<1)$ and $image(f(z))\subset K\subset D_0(1)$ where K is compact subset Prove that $f$ has one fixed point.

Question

Given $f\in Hol(|z|<1)$ and $image(f(z))\subset K\subset D_0(1)$ where K is compact subset
Prove that $f$ has one fixed point.

My try is:
$g(z)=z-f(z)$ and then we know that $|f(z)|<1$ using Rouche theorem $1=|z|>|f(z)|$ and because $z$ has only 1 zero we get the needed.
I'm not sure if my prove is correct

Answer 1

You cannot apply Rouche's Theorem to the unit circle. But there exist $t\in (0,1)$ such that $|f(z)| <t$ for all $z$. For any $r \in (t,1)$ we can apply Rouche's Theorem to see that $f$ has a unique fixed point in $\{z: |z| <r\}$ for every $r \in (t,1)$ . This implies that $f$ has a unique fixed point in $\{z: |z|<1\}$.