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brendijadq

brendijadq

Answered question

2022-05-23

Let
f ( z ) = z a z b , z b a
be a complex valued rational function.
How can I show that, if | a | , | b | < 1 ,, then there is a complex number z 0 satisfying | z 0 | = 1 and f ( z 0 ) R ?
I have tried in many ways, but on success. Basically I tried to show that there is a unimodular complex number such that
a b z b = a ¯ b ¯ z ¯ b ¯ .
I could make a quadratic equation by using the fact that z ¯ = 1 z , z D . Unfortunately I could not solve this question using that. So, I would like to see different (and somewhat general) approach.
Also, I would like to know that what happen if (both or atleast one) a , b D .Any comment or hint will be welcome. Thank you.

Answer & Explanation

aniizl

aniizl

Beginner2022-05-24Added 12 answers

a = p + q i
b = u + v i
z = x + y i
We can always find z so that a , b and z are on a same line (actually whether | z | = 1 or not, or | a | , | b | < 1 or not).
Then,
y q x p = y v x u = m
f ( z ) = x p + ( y q ) i x u + ( y v ) i = x p x u 1 + y q x p i 1 + y v x u i = x p x u 1 + m i 1 + m i = x p x u R

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