Rational function with absolute value $1$ on unit circle

complex-analysis

What is the general form of a rational function which has absolute value $1$ on the circle $|z|=1$? In particular, how are the zeros and poles related to each other?

So, write $R(z)=\dfrac{P(z)}{Q(z)}$, where $P,Q$ are polynomials in $z$. The condition specifies that $|R(z)|=1$ for all $z$ such that $|z|=1$. In other words, $|P(z)|=|Q(z)|$ for all $z$ such that $|z|=1$. What can we say about $P$ and $Q$?

Solution 1:

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 $\lvert\lambda\rvert = 1$.