I'm currently trying to work through the following problem:
Let $\mathbb{D}=\{z\in\mathbb{C}~:~|z|<1\}$ denote the open unit disk. Suppose $f:\mathbb{D}\to\mathbb{D}$ is a holomorphic function whose range is contained in a compact subset of $\mathbb{D}$. Show that $f$ has a unique fixed point.
So, we must show that there exists a point $z^{*}\in\mathbb{D}$ with $f(z^{*})=z^{*}$, and that $f(z)\neq z$ for all $z\in\mathbb{D}$ with $z\neq z^{*}$. I'm not even sure where to start on this problem. I'm not sure if I should expand $f$ in a power series near the origin (or if that does nothing). Seeing the fact that $f(\mathbb{D})$ lies in a compact subset of $\mathbb{D}$ suggests to me that I should be using Rouche's Theorem somehow, but this could also be completely useless, I'm not sure. I'm also not sure how I would apply it, since it deals with zeros of functions and not fixed points.
Thanks in advance for any suggestions.
Rouche's theorem can be applied in the following form: if $g$ and $h$ are holomorphic in a region containing a circle $C$, and if $|g|>|h|$ on $C$, then $g$ and $g+h$ have the same number of zeroes inside $C$.
Now since the image of $f$ is a compact subset of the open unit disk, it is contained in a closed disk of radius $r<1$ centered at the origin, and so we can apply Rouche's theorem with $g(z)=z$, $h(z)=-f(z)$, and $C$ any circle of radius $s$ centered at the origin with $r<s<1$ to conclude that $f(z)-z$ has exactly one zero in the unit disk.