R(z) be a rational function satisfies|R(z)|=1 when |z|=1. Prove that R(z) can be written as R(z)=czm∏n=1kz−an1−an―z, c is a constant and 0&lt;|an|≠1. Hint: try to prove that if z0 is a zero of R(z) then 1z0― is a pole of R(z)

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R(z) be a rational function satisfies|R(z)|=1 when |z|=1. Prove that R(z) can be written as R(z)=czm∏n=1kz−an1−an―z, c is a constant and 0&amp;lt;|an|≠1. Hint: try to prove that if z0 is a zero of R(z) then 1z0― is a pole of R(z)

Arlene Patel · Accepted Answer

1) R(z) has only finitely many zeroes and poles in |z|&amp;lt;1 since |R(z)|=1 when |z|=1.  2) Each z−an1−an―z has a zero at z=an if |an|&amp;lt;1 anf a pole at 1an― if |an|&amp;gt;1. It also has magnitude 1 when |z|=1.

emeriinb4r · Accepted Answer

This follows from a categorization theorem of Blaschke. Essentially, every analytic function that maps the unit disc onto itself, must be a Blaschke product.Preliminary exercises to this include the following.1) Let f:D―→D― is analytic where |f(z)|=1 for |z|=1.Show that if f has no zeros in D then f must be a constant. (Use maximum modulus principle).2) Let f:D―→D― be an analytic function where f(0)=0 and |f(z)|=1 for |z|=1. Then f(z)=αz for some α∈CC. (Schwarz Lemma is a special case of Blaschke&#039;s result)3) For some |w|&amp;lt;1, define a Blaschke factor,f(z)=z−21−w―z.Show that f(z) maps D―→D― and that |f(z)=1 for every |z|=1.4) Suppose f:D―→D― is analytic with |f(z)|=1 and |z|=1. By the uniqueness theorem f has finitely many zeros in D,{a1,...,an}. Show that f is a Blaschke product of it&#039;s zeros,f(z)=c∏k=1n1−ak―zz−ak.h(z)=f(z)∏k=1n1−ak―zz−ak.Certainly, h(z) is analytic on D and has no zeros (since we canceled them). By (3), h:D―→D― and |h(z)|=1 when |z|=1. Thus, we conclude that h(z) must be a constant by (1). Hence, f(z) is indeed a Blaschke product.I am sure a slight modification of this trick will solve your problem.

R(z) be a rational function satisfies|R(z)|=1 when |z|=1. Prove that

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