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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 $Q$?

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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 $\lvert\lambda\rvert = 1$.

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  • $\begingroup$ Thank you. For the definition of $(a_n)_{0\leq n\leq N}$, should it be just the distinct zeros of $R$ (not poles)? Because if $a_n$ is a zero, then $1/\overline{a_n}$ is automatically a pole, and you've taken care of it in the denominator of the expression for $B(z)$. $\endgroup$ – PJ Miller Aug 24 '13 at 21:36
  • $\begingroup$ Also, I get it that $M(z)=1$ for $|z|=1$, but why does that imply $M(z)$ is constant? $\endgroup$ – PJ Miller Aug 24 '13 at 21:39
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    $\begingroup$ @Earthliŋ Ah, I have the third edition, but I don't think that makes much of a difference. Ahlfors starts by giving you lots of geometric stuff (and assumes one knows a lot of geometry) in the first chapter. He's probably after a more geometric argument. I'll read and see whether I can find what argument he may have had in mind. $\endgroup$ – Daniel Fischer Apr 21 '16 at 18:42
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    $\begingroup$ @Earthliŋ I'm not sure, but I think he was aiming for roughly that argument, though using the (stronger) identity theorem for rational functions [a nonconstant rational function has only finitely many zeros, hence if two rational functions coincide on an infinite set, the two functions are identical]. That gives you $\overline{R(1/\overline{z})} = 1/R(z)$, and hence the symmetry of zeros and poles. From that, one gets the general form more or less elegantly. If one recalls that $\frac{a-b}{1-\overline{a}b}$ was dealt with in exercises in Chapter 1, it helps. $\endgroup$ – Daniel Fischer Apr 21 '16 at 19:31
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    $\begingroup$ @MathewJames That follows from the relation $\overline{R(1/\overline{z})} = 1/R(z)$ which is a consequence of $\lvert z\rvert = 1 \implies \lvert R(z)\rvert = 1$. If $\lvert z\rvert > 1$, then $\lvert 1/\overline{z}\rvert < 1$, so we have a direct way to write the behaviour outside the unit disk in terms of the behaviour inside the unit disk. $\endgroup$ – Daniel Fischer Sep 26 '17 at 9:40
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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.

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  • $\begingroup$ Thank you. I get it that your equation holds for $|z|=1$, but why must it be true for all $z$? $\endgroup$ – PJ Miller Aug 24 '13 at 21:40
  • $\begingroup$ @PJMiller You can just verify it directly (write $P$ and $Q$ as arbitrary polynomials and do the substations), but see the other answer for a slicker approach. $\endgroup$ – Potato Aug 24 '13 at 21:43
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Answer your question about why $M$ is constant: it's simply because $M$ is a quotient of two polynomials. If the quotient is $1$ on the unit circle, it means these two polynomials are equal at all the points of the circle. This implies that these two polynomials are the same. So $M$ is identically $1$.

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Define $M(z)$ the same as the high-voted answer. And we know (if you want to prove it, it's easy), if $f(z)$ is analytic and so is $\ \overline{f(\bar z)}\ $analytic. We choose $f(z) = R(1/z)$, so $\overline{R(1/\bar z )}$ analytic and thus $M(z)$ is analytic.

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For your doubt regarding the polynomial turning out to be constant, observe that a rational function has finitely many zeros. So any value must be attained finitely many times(equal to the order of said rational function)

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