Show $f$ can be extended to be analytic in $\mathbb{C}$ except at finitely many poles.

Question

I have attempted the following problem but I am stuck on one part:

Suppose $f$ is analytic on the unit disk and continuous on the boundary of the disk. Also, suppose $|f(z)|=1$ for $|z|=1$. Show that $f$ can be extended to be analytic in $\mathbb{C}$, except for finitely many poles, by defining $F(z) = \left(\overline{f(\frac{1}{\overline{z}})}\right)^{-1}$.

My attempt: Let $z, z_0$ be in $\mathbb{C}\setminus\mathbb{D}$. Then $\frac{1}{\overline{z}}$ and $\frac{1}{\overline{z_0}}$ are in $\mathbb{D}$ so there is a power series centered at $\frac{1}{\overline{z_0}}$ convergent in a neighborhood contained in $\mathbb{D}$. Then we can write $f(\frac{1}{\overline{z}}) = \sum_{n=0}^{\infty}a_n(\frac{1}{\overline{z}} - \frac{1}{\overline{z_0}})^n$ so that $\overline{f(\frac{1}{\overline{z}})}= \sum_{n=0}^{\infty}\overline{a_n}(\frac{1}{{z}} - \frac{1}{z_0})^n$.

Now I am stuck on writing $F$ as a power series in $z$. Once this is done, I know that I can use that fact that $|f(z)| = 1$ on the boundary of $\mathbb{D}$ to show that $F$ and $f$ agree on the boundary, and therefore invoke the symmetry principle to get my result. Regarding the finiteness of the poles, I know this follows from the fact that $f$ must have finitely many zeros in the disk.

Any help is appreciated!

Answer 1

Hint: Use Cauchy-Riemann equations to show that if $g$ is analytic on $U$, $\overline{g(\overline{z})}$ is analytic on $\{z: \overline{z} \in U\}$. Using the fact that $z \to 1/z$ is analytic on $\mathbb C \backslash \{0\}$, $F(z)$ is analytic on $\{z:|z|>1, f(1/\overline{z}) \ne 0\}$.

Answer 2

With $$ \frac1{\overline{z+h}}=\frac1{\overline z}-\frac1{\overline z^2}\overline h+O(h^2) $$ we have $$f\left(\frac1{\overline{ z+ h}}\right)=f\left(\frac1{\overline z}\right)-\frac1{\overline z^2}f'\left(\frac1{\overline z}\right)\cdot \overline h+O(h^2)$$ and so $$\frac1{F(z+h)}=\overline{f\left(\frac1{\overline{ z+ h}}\right)}=\overline{f\left(\frac1{\overline z}\right)}-\frac1{ z^2}\overline{f'\left(\frac1{\overline z}\right)}\cdot h+O(h^2)$$ showing that $\frac1{F(z)}$ is complex differentiable and hence analytic.