Meromorphic on unit disc with absolute value 1 on the circle is a rational function. Let $f$ be a meromorphic function on the open unit disk such that $f$ has a continuous extension to the boundary circle.Suppose $f$ has only poles in the open unit disc and suppose $|f(z)|=1$ for all $z$ with $|z|=1$.Prove that $f$ is a rational function. 

Since the closed unit disk is compact, $f$ can have only finitely many zeros and poles in the unit disk. Let $k$ be the order of $f$ in $0$, that is, $f(z) = z^k\cdot g(z)$ where $g$ is holomorphic in a neighbourhood of $0$ with $g(0) \neq 0$. Let $\zeta_1,\dotsc, \zeta_k$ be the zeros of $f$ in the punctured unit disk, with multiplicities $\mu_1,\dotsc,\mu_k$. Let $\pi_1,\dotsc,\pi_r$ be the poles of $f$ in the punctured unit disk with orders $\nu_1,\dotsc,\nu_r$. Consider the finite Blaschke products
$$Z(z) = \prod_{\kappa=1}^k \left(\frac{z - \zeta_\kappa}{1 - \overline{\zeta}_\kappa z}\right)^{\mu_\kappa}$$
and
$$P(z) = \prod_{\rho = 1}^r \left(\frac{z-\pi_\rho}{1-\overline{\pi}_\rho z}\right)^{\nu_\rho}.$$
Evidently,
$$h(z) = z^k\frac{Z(z)}{P(z)}$$
is a rational function, and it has the same zeros and poles in the unit disk as $f$.
Since every factor in $h$ has modulus $1$ on the unit circle, we have $\lvert h(z)\rvert = 1$ for all $z$ with $\lvert z\rvert = 1$, and hence
$$\frac{f(z)}{h(z)}$$
is a zero-free holomorphic function on the unit disk with $\left\lvert \frac{f(z)}{h(z)}\right\rvert = 1$ for $\lvert z\rvert = 1$, thus constant.

An alternative way to obtain the result is by using the reflection principle:
$$F(z) = \begin{cases} f(z) &, \lvert z\rvert \leqslant 1 \\ \dfrac{1}{\overline{f(1/\overline{z})}} &, \lvert z\rvert > 1\end{cases}$$
defines a function that is meromorphic on $\widehat{\mathbb{C}}$ by the reflection principle. A function that is meromorphic on the entire sphere is a rational function, so $F$, and hence $f$ is rational.
