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

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!

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 $$\{z:|z|>1, f(1/\overline{z}) \ne 0\}$$.
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.