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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 $Q$?

PJ Miller
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5 Answers5

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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 $\lvert\lambda\rvert = 1$.

Daniel Fischer
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  • 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 $B(z)$. – PJ Miller Aug 24 '13 at 21:36
  • Also, I get it that $M(z)=1$ for $|z|=1$, but why does that imply $M(z)$ is constant? – PJ Miller Aug 24 '13 at 21:39
  • The zeros and poles of $R$ that lie in the unit disk. One could also choose only the zeros, but then one would have to include the ones outside the unit disk. $M$ takes the value $1$ on a non-discrete set, the unit circle. By the identity theorem, $M$ is constant. – Daniel Fischer Aug 24 '13 at 21:43
  • @DanielFischer I don't quite understand the "identity theorem" you cite. We have that $M (z) = R (z) . \overline{R (1/\overline z)}$ satisfies $M (z) = 1$ for $\lvert z \rvert = 1$. But the set of such $z$ is not open, so the usual identity theorem does not apply. Also, at the point of the book, the identity theorem hasn't been mentioned, so I was wondering if there is an easier argument. (The previous exercise asks to construct the Lagrange interpolation polynomial.) – Earthliŋ Apr 21 '16 at 17:55
  • @Earthliŋ The identity theorem needs much weaker premises than a nonempty open set. In dimension $1$, it suffices that the two functions coincide on a set that has an accumulation point in the domain. Thus if two holomorphic functions $f,g\colon U \to \mathbb{C}$ coincide on some nondegenerate curve in $U$ (like the unit circle), they coincide everywhere (assuming $U$ is connected). For me, that's the usual identity theorem. Avoiding the identity theorem surely doesn't give an easier argument. Off the top of my head, I don't see an elementary argument avoiding it. Let me think. – Daniel Fischer Apr 21 '16 at 18:10
  • What tools are available at that point of the book? – Daniel Fischer Apr 21 '16 at 18:10
  • The previous section has as exercise "verify the Cauchy-Riemann equations for $z^2$ and $z^3$". In other words, almost no "tools", only the definition of a holomorphic/rational function. There is a theorem "If all zeros of a polynomial lie in a half plane, then all zeros of the derivative lie in the same half plane", but I don't think it's relevant. The reader also learned how to develop partial fractions (and construct the Lagrange interpolation polynomial by proving $\frac{P (z)}{Q (z)} = \sum_{k=1}^n \frac{P (a_k)}{Q' (a_k)(z-a_k)}$, where $\deg P < n$ and $a_i$'s are the roots of $Q$.) – Earthliŋ Apr 21 '16 at 18:19
  • @Earthliŋ Indeed that doesn't seem relevant. Nor actually does Lagrange interpolation look pertinent. Which book is it? – Daniel Fischer Apr 21 '16 at 18:22
  • The book is Ahlfors Complex analysis (I have the second edition). The Lagrange interpolation says that for any $n$ values, you can find a polynomial P of degree $< n$ such that P takes the specified values. If one could show that the points in the unit circle are "generic enough", one might argue that there couldn't be any other rational function, satisfying the uncountably many "constraints" $M (z) = 1$ for $\lvert z \rvert = 1$... (But I don't know if I'm on the right track...) – Earthliŋ Apr 21 '16 at 18:30
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    @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. – Daniel Fischer Apr 21 '16 at 18:42
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    @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. – Daniel Fischer Apr 21 '16 at 19:31
  • Great, thank you very much. I guess this stronger identity theorem is implicit in his "counting of poles/zeros" of a rational function, which for him are always finite determined by the degrees of the polynomials in the numerator/denominator. $M (z) - 1$ is rational and has infinitely many zeros (at least on the unit circle), so it has to be the zero function. (I totally forgot about $\frac{a−b}{1 - \overline a b}$, thank you! Now there are no more loose ends.) – Earthliŋ Apr 21 '16 at 20:12
  • Could please one clarify, how do we know that we have all the zeros and poles of R in B, since we only take those that lie in the unit circle? I don't get that point. Thanks – perlman Sep 21 '17 at 03:05
  • @MathewJames That's due to the relation between zeros and poles (which is a consequence of $\lvert z\rvert = 1 \implies \lvert R(z)\rvert = 1$). If $z$ is a zero of order $k$ of $R$, then $1/\overline{z}$ is a pole of order $k$ and vice versa. So the zeros and poles of $R$ in the unit disk completely determine all zeros and poles of $R$. Since a Blaschke product has the same relation between zeros and poles, it follows that $B$ has the same zeros and poles as $R$ in all of $\mathbb{C}\cup {\infty}$. – Daniel Fischer Sep 25 '17 at 11:27
  • @DanielFischer I understand everything in your comment but the fact that "the zeros and poles of $R$ in the unit disk completely determines all zeros and poles of $R$." If we know the zeros and poles of $R$ in the unit disk how can we find the zeros and poles of $R$ not in the unit disk? – perlman Sep 25 '17 at 23:48
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    @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. – Daniel Fischer Sep 26 '17 at 09:40
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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.

Potato
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  • Thank you. I get it that your equation holds for $|z|=1$, but why must it be true for all $z$? – PJ Miller Aug 24 '13 at 21:40
  • @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. – Potato Aug 24 '13 at 21:43
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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)

A C
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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 $1$.

Roy
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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.

Hance Wu
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