$f$ is meromorphic on $\mathbb{C}$ and $|f(z)|=1$ on $|z|=1$, show $f$ is rational.

Question

I believe $f$ should be in the form of finite product of Blacshke factors and reciprocals. I have no idea how to show number of zeros or poles are finite.

Answer 1

Hint: since $f(\mathbb{T})\subseteq\mathbb{T}$, if it has infinitely many zeroes or poles there would be an accumulation point somewhere inside the unit disc by reflection principle. Now what happens at that accumulation point?

Alternatively, immediately appeal to reflection principle to extend $f$ as a holomorphic map $S^2\to S^2$, so $f$ must be rational.

Answer 2

Let $g\left(z\right)=f\left(\frac{1}{z}\right)$. Then, $g$ has exactly the same properties as $f$. The same is true for the reciprocals of $f$ and $g$.

With that in mind, we turn our attention to $f$ as a function on the Riemann sphere, $\hat{\mathbb{C}}$. Now, suppose $f$ has infinitely many poles. Thus, the set $P_{f}$ of poles of $f$ in $\hat{\mathbb{C}}$ is infinite. Since the Riemann sphere is a compact Riemann surface, compactness guarantees that $P_{f}$ has an accumulation point $w$ in $\hat{\mathbb{C}}$. If $w$ is not in the open unit disk, then its reciprocal $v$ will be in the open unit disk, and vice versa. Thus, we have that one of $f$ or $g$ has an accumulation point of poles in $\mathbb{D}$, and thus, that one of the reciprocals of $f$ and $g$ must have an accumulation point of zeroes in the closure of $\mathbb{D}$. However, since both of these functions have unit magnitude on the unit circle, the accumulation point must lie inside $\mathbb{D}$. Thus, by the identity theorem, since one of $f$ and $g$'s reciprocals has an accumulation point of zeroes in its domain, it must be a constant. By construction, this then forces both $f$ and $g$ to be constant on the Riemann circle, which contradicts the assumption that $f$ had infinitely many poles.

Repeating this argument with the reciprocals of $f$ and $g$ in place of $f$ and $g$ then gives the corresponding result about the zeroes of $f$.

Q.E.D.