Rational function with absolute value $1$ on unit circle

If a rational function $R$ satisfies $\lvert R(z)\rvert = 1$ for $\lvert z\rvert = 1$, then the rational function

$$M(z) = R(z)\cdot \overline{R(1/\overline{z})}$$

satisfies $M(z) = 1$ for $\lvert z\rvert = 1$, therefore it is constant (a nonconstant rational function attains every value only finitely often), and $R$ satisfies

$$R(1/\overline{z}) = 1/\overline{R(z)}.$$

Hence the poles and zeros of $R$ are related by reflection in the unit circle; if $\zeta$ is a zero of order $k$, then $1/\overline{\zeta}$ is a pole of order $k$ and vice versa.

Thus, if $(a_n)_{0\leqslant n \leqslant N}$ are the distinct zeros and poles of $R$ in the unit disk, with orders $m_n$ ($m_n > 0$ for zeros, and $m_n < 0$ for poles), and $a_0 = 0$ [$m_0 = 0$ is allowed], the product

$$B(z) = z^{m_0}\cdot \prod_{n=1}^N \left(\frac{z - a_n}{1-\overline{a_n}z}\right)^{m_n}$$

is a rational function having exactly the same zeros and poles as $R$, and also $\lvert B(z)\rvert = 1$ for $\lvert z\rvert = 1$. So the quotient $R(z)/B(z)$ is a rational function without zeros or poles, hence constant, and therefore

$$R(z) = \lambda\cdot B(z)$$

for some $\lambda$ with $\lvert\lambda\rvert = 1$.

You can show that

$$R(z)= \frac{1}{\overline{R\left(\frac{1}{\overline{z}}\right)}}.$$

If $w$ is a zero for $R$, then $\frac{1}{\overline{w}}$ is a pole for $R$. Similarly, the existence of a pole implies the existence of a zero.