Using Rouche's Theorem to find the number of solutions of $f(z)=z$ in the open unit disc [duplicate]

Question

How many roots does the equation $f(z)=z$ have in the circle $|z|<1$ if for $|z|\leq 1$, $f(z)$ is analytic and satisfies $|f(z)|<1$?

My idea: I figured I could do this pretty easily using Rouche: Consider $|z|=1$, and let $g(z)=z$, then $|f(z)-g(z)|=z-z=0<1=|g(z)|$. So, since $g$ has only $1$ root in $|z|<1$, then so does $f$.

I just feel like I am missing something. In particular, can I define $g$, and use it, in the way I did? Any thoughts are greatly appreciated! Thank you.

Answer 1

Let $h(z) = f(z)-z$, $g(z) = z$, then for $|z|=1$ we have $|h(z)+g(z)| = |f(z)| < 1 = |g(z)|$, hence $h,g$ have the same number of zeroes inside the circle. Since $g$ has exactly one zero, so does $h$.

Answer 2

Here is an alternative solution without using Rouché's Theorem. While it is lengthier than cooper.hat's great answer, it determines exactly what the fixed points of $f$ are.

We can consider $f$ as a holomorphic function on $\overline{\mathbb{D}}:=\big\{z\in\mathbb{C}\,\big|\,|z|\leq 1\big\}$. Note that the image of $f$ lies in $\mathbb{D}:=\big\{z\in\mathbb{C}\,\big|\,|z|< 1\big\}$. We first claim that the equation $f(z)=z$ has at most one solution $z\in\mathbb{D}$.

Suppose on the contrary that $f(z)=z$ has two solutions $z=z_i$ for $i\in\{1,2\}$ in $\mathbb{D}$. Then, we can find a Möbius transformation $\mu:\overline{\mathbb{D}}\to\overline{\mathbb{D}}$ that maps $z_1\mapsto 0$. Write $w:=\mu(z_2)$. Let $\phi:=\mu\circ f\circ \mu^{-1}$. Then, $\phi:\overline{\mathbb{D}}\to\overline{\mathbb{D}}$ is such that $\phi(0)=0$ and $\phi(w)=w$. By Schwarz's Lemma, $\phi(z)=z$ for all $z\in\overline{\mathbb{D}}$. Thus, $f(z)=z$ for all $z\in\overline{\mathbb{D}}$. This contradicts the assumption that the image of $f$ lies in $\mathbb{D}\subsetneq \overline{\mathbb{D}}$. Therefore, $f$ has at most one fixed point.

Now, define $f^{\circ 1}:=f$ and $f^{\circ k}:=f\circ f^{\circ (k-1)}$ for $k=2,3,4,\ldots$. Take $I_k$ to be the image of $f^{\circ k}$ for each positive integer $k$. Note that $I_1\supseteq I_2\supseteq I_3\supseteq \ldots$. Because $f$ is a continuous function and $\overline{\mathbb{D}}$ is a compact set, we can easily see that each $I_k$ is a compact set. Due to Cantor's Intersection Theorem, the set $$I:=\bigcap_{k=1}^\infty\,I_k$$ is nonempty. By our previous paragraph, $I=\{\zeta\}$ for some $\zeta\in\mathbb{D}$. Clearly, $f(\zeta)=\zeta$. From this result, it follows that $$\zeta=\lim_{k\to\infty}\,f^{\circ k}(z)$$ for any $z\in\overline{\mathbb{D}}$.