Let f ( z ) be a rational function in the complex plane such that f does not have any poles in { z : ℑ z ≥ 0 }.Prove that sup { | f ( z ) | : ℑ z ≥ 0 } = sup { | f ( z ) | : ℑ 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 : ℑ z = 0 , | z | = r }, Γ r 2 = { z : ℑ z &gt; 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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Let   f  (  z  ) be a rational function in the complex plane such that   f does not have any poles in   {  z  :  ℑ  z  ≥  0  }.Prove that   sup  {      |    f  (  z  )      |    :  ℑ  z  ≥  0  }  =  sup  {      |    f  (  z  )      |    :  ℑ  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  :  ℑ  z  =  0  ,      |    z      |    =  r  },       Γ                  r        2              =  {  z  :  ℑ  z  &amp;gt;  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&#039;s maximum on       Γ                  r        2             than it&#039;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.

potamanixv · Accepted Answer

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  :  ℑ  z  &amp;gt;  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      :      ℑ      z      &amp;gt;      0      }            |    f  (  z  )      |    =      sup          {      z      :      ℑ      z      =      0      }            |    f  (  z  )      |    .Because   T  (  D  (  0  ,  1  )  )  =  {  z  :  ℑ  z  &amp;gt;  0  }, and   T  (  C  (  0  ,  1  )  ∖  {  −  i  }  )  =  {  z  :  ℑ  z  =  0  }.

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

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