"I have encountered a question like this: Suppose a complex function f is analytic in |z|<1, c

Guy Rollins

Guy Rollins

Answered question

2022-02-15

"I have encountered a question like this:
Suppose a complex function f is analytic in |z|<1, continuous in |z|1, and |f|=1 on |z|=1. Show that f can be extended as a rational function.
According to the isolation property of zero points, f has only finite zero points in |z|1. But I don’t know how to show that it’s a rational function. Hope someone could help.

Answer & Explanation

publilandiaik8

publilandiaik8

Beginner2022-02-16Added 5 answers

There are finitely many zeros of f in the unit disc. There is a finite Blaschke product p(z) having the same zeros as f in the unit disc. This is designed so that |p|=1 on the unit circle. Then f=pg where g is continuous and non-zero on the closed unit disc, holomorphic on the open unit disc and has |g|=1 on the unit circle. By the maximum modulus principle, g is constant. So f is a constant multiple of a finite Blaschke product, which is a rational function on C.

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