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!