Let f ( z ) be a rational function in the complex plane such that f does not ha

cambrassk3 2022-07-08 Answered
Let f ( z ) be a rational function in the complex plane such that f does not have any poles in { z : z 0 }.
Prove that sup { | f ( z ) | : z 0 } = sup { | f ( z ) | : z = 0 }.
Let Γ r be a half circle counter such that Γ r = Γ r 1 + Γ r 2 when Γ r 1 = { z : z = 0 , | z | = r }, Γ r 2 = { z : z > 0 , | z | = r }. Using the Maximum modulus principle on the insides of Γ r , | f | Gets is maximum value on Γ r . As r gets bigger if |f| got it's maximum on Γ r 2 than it's still smaller the the value on Γ r + 1 which does not contain Γ r 2 . I would like a hint on how to proceed.
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Answers (1)

potamanixv
Answered 2022-07-09 Author has 15 answers
To develop the comment of Daniel Fischer. Let T ( z ) = i z 1 i z . Then T is a conformal mapping of D(0,1) onto the upper half plane { z : z > 0 }. f T is a rational function which is analytic in D(0,1) and continuous on D ( 0 , 1 ) ¯ , according to the assumption. So the maximum of | f T | on D ( 0 , 1 ) ¯ is attained on the boundary of the disk that is
sup D ( 0 , 1 ) | f T | = sup C ( 0 , 1 ) | f T |
or equivalently
sup { z : z > 0 } | f ( z ) | = sup { z : z = 0 } | f ( z ) | .
Because T ( D ( 0 , 1 ) ) = { z : z > 0 }, and T ( C ( 0 , 1 ) { i } ) = { z : z = 0 }.

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