Suppose R(z) be a rational function such that |R(z)|=1 for |z|=1. Show that \alpha is a

Guinzio3rv

Guinzio3rv

Answered question

2022-02-16

Suppose R(z) be a rational function such that |R(z)|=1 for |z|=1. Show that α is a zero or a pole or order m, if and only if 1α is a pole or zero or order mP respectively.
I am wondering that if I can first show that M(z)=R(z) the complex conjugate of R(1z) is a rational function such that M(z)=1P on |z|=1. Can I assume z=1 and then substitute all z with 1? Thanks.

Answer & Explanation

Chance Mill

Chance Mill

Beginner2022-02-17Added 8 answers

Hint. Note that F(z)=R(z)R1z is a the rational function such that for |z|=1:
F(z)=R(z)R1z=R(z)R(z)=|R(z)|2=1.
Since a non-constant rational function attains a value only finitely often, it follows that F is identically equal to 1. Hence
R1z=1R(z)

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