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is the real part of a holomorphic function holomorphic?

real part of a holomorphic function from a PDEHolomorphic function with given real part on unit circleReal and imaginary part of holomorphic functionholomorphic function with real arguments smooth?proving that $f(z)=1/Gamma(z)$ is a entire functionEntire function doesn't cut real axis.Measure of set where holomorphic function is largeExample of entire function on $mathbb C$ such that which does not take only one value in $mathbb C$When is a real-valued continuous function on the unit circle the real part of a holomorphic function on the unit disk?bijective holomorphic entire functions

0

$begingroup$

Say we have some entire function $f:mathbb{C} rightarrow mathbb{C}$. Does this guarantee that the function $Re(f)$ will also be entire?

complex-analysis holomorphic-functions entire-functions

share|cite|improve this question

edited Mar 10 at 22:26

José Carlos Santos

167k22132235

asked Mar 10 at 22:05

D.DogD.Dog

207

$endgroup$

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  It will be entire iff $f$ is constant.
  $endgroup$
  – Severin Schraven
  Mar 10 at 22:07
  

  
  
  
  


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  $begingroup$
  Ah. Does this come from Cauchy Riemann equations or something different?
  $endgroup$
  – D.Dog
  Mar 10 at 22:08
  

  
  
  
  


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  $begingroup$
  Yes, it exactly follows from Cauchy Riemann equations. If $f=u+iv$ is defined in an open connected domain and $v$ is constant then the partial derivatives of $v$ are zero everywhere. By Cauchy Riemann we conclude that the partial derivatives of $u$ are zero as well. So $f$ is constant.
  $endgroup$
  – Mark
  Mar 10 at 22:17
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you explain why we require that v is constant for u to be holomorphic?
  $endgroup$
  – D.Dog
  Mar 10 at 22:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The real part of $f$ has $0$ imaginary part. Apply C-R to $Re(f)$
  $endgroup$
  – saulspatz
  Mar 10 at 22:20
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Say we have some entire function $f:mathbb{C} rightarrow mathbb{C}$. Does this guarantee that the function $Re(f)$ will also be entire?

complex-analysis holomorphic-functions entire-functions

share|cite|improve this question

edited Mar 10 at 22:26

José Carlos Santos

167k22132235

asked Mar 10 at 22:05

D.DogD.Dog

207

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It will be entire iff $f$ is constant.
  $endgroup$
  – Severin Schraven
  Mar 10 at 22:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah. Does this come from Cauchy Riemann equations or something different?
  $endgroup$
  – D.Dog
  Mar 10 at 22:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it exactly follows from Cauchy Riemann equations. If $f=u+iv$ is defined in an open connected domain and $v$ is constant then the partial derivatives of $v$ are zero everywhere. By Cauchy Riemann we conclude that the partial derivatives of $u$ are zero as well. So $f$ is constant.
  $endgroup$
  – Mark
  Mar 10 at 22:17
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you explain why we require that v is constant for u to be holomorphic?
  $endgroup$
  – D.Dog
  Mar 10 at 22:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The real part of $f$ has $0$ imaginary part. Apply C-R to $Re(f)$
  $endgroup$
  – saulspatz
  Mar 10 at 22:20
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Say we have some entire function $f:mathbb{C} rightarrow mathbb{C}$. Does this guarantee that the function $Re(f)$ will also be entire?

complex-analysis holomorphic-functions entire-functions

share|cite|improve this question

edited Mar 10 at 22:26

José Carlos Santos

167k22132235

asked Mar 10 at 22:05

D.DogD.Dog

207

$endgroup$

share|cite|improve this question

edited Mar 10 at 22:26

José Carlos Santos

167k22132235

asked Mar 10 at 22:05

D.DogD.Dog

207

share|cite|improve this question

edited Mar 10 at 22:26

José Carlos Santos

167k22132235

asked Mar 10 at 22:05

D.DogD.Dog

207

edited Mar 10 at 22:26

José Carlos Santos

167k22132235

edited Mar 10 at 22:26

José Carlos Santos

167k22132235

asked Mar 10 at 22:05

D.DogD.Dog

207

asked Mar 10 at 22:05

D.DogD.Dog

207

2 Answers
2

active

oldest

votes

2

$begingroup$

The only entire holomorphic functions whose imaginary parts are constant are the constant functions. So $operatorname{Re}(f)$ is holomorphic if and only if it is constant (which implies that $f$ itself is constant).

share|cite|improve this answer

answered Mar 10 at 22:13

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I follow the first line but I don't understand why Re(f) being holomorphic requires that Im(f) is constant, could you explain?
  $endgroup$
  – D.Dog
  Mar 10 at 22:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $Im(f)$ is not necessary constant. But $Im(Re(f))$ is constant because $Re(f)$ is always a real number. This is why $Re(f)$ being holomorphic implies that it is constant.
  $endgroup$
  – Mark
  Mar 10 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ahhh of course. thanks for the explanation.
  $endgroup$
  – D.Dog
  Mar 10 at 22:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But yes, after we found out that $Re(f)$ is constant then since $f$ is holomorphic with constant real part we conclude that $f$ is constant. So in the end $Im(f)$ is constant as well. But we conclude that only after we proved $Re(f)$ is constant.
  $endgroup$
  – Mark
  Mar 10 at 22:23
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

If $mathfrak{R}(f)$ is entire, then $exp(imathfrak{R}(f))$ is also entire. But $|exp(i mathfrak{R}(f))| leq 1$, so $exp(imathfrak{R}(f))$ is constant (Liouville's theorem) and thus so is $mathfrak{R}(f)$. Hence, $f$ has constant real part, and the C-R equations get you that $f$ has constant imaginary part as well.

share|cite|improve this answer

answered Mar 10 at 22:16

rubikscube09rubikscube09

1,556720

$endgroup$

add a comment | 

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2

$begingroup$

The only entire holomorphic functions whose imaginary parts are constant are the constant functions. So $operatorname{Re}(f)$ is holomorphic if and only if it is constant (which implies that $f$ itself is constant).

share|cite|improve this answer

answered Mar 10 at 22:13

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I follow the first line but I don't understand why Re(f) being holomorphic requires that Im(f) is constant, could you explain?
  $endgroup$
  – D.Dog
  Mar 10 at 22:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $Im(f)$ is not necessary constant. But $Im(Re(f))$ is constant because $Re(f)$ is always a real number. This is why $Re(f)$ being holomorphic implies that it is constant.
  $endgroup$
  – Mark
  Mar 10 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ahhh of course. thanks for the explanation.
  $endgroup$
  – D.Dog
  Mar 10 at 22:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But yes, after we found out that $Re(f)$ is constant then since $f$ is holomorphic with constant real part we conclude that $f$ is constant. So in the end $Im(f)$ is constant as well. But we conclude that only after we proved $Re(f)$ is constant.
  $endgroup$
  – Mark
  Mar 10 at 22:23
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

The only entire holomorphic functions whose imaginary parts are constant are the constant functions. So $operatorname{Re}(f)$ is holomorphic if and only if it is constant (which implies that $f$ itself is constant).

share|cite|improve this answer

answered Mar 10 at 22:13

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I follow the first line but I don't understand why Re(f) being holomorphic requires that Im(f) is constant, could you explain?
  $endgroup$
  – D.Dog
  Mar 10 at 22:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $Im(f)$ is not necessary constant. But $Im(Re(f))$ is constant because $Re(f)$ is always a real number. This is why $Re(f)$ being holomorphic implies that it is constant.
  $endgroup$
  – Mark
  Mar 10 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ahhh of course. thanks for the explanation.
  $endgroup$
  – D.Dog
  Mar 10 at 22:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But yes, after we found out that $Re(f)$ is constant then since $f$ is holomorphic with constant real part we conclude that $f$ is constant. So in the end $Im(f)$ is constant as well. But we conclude that only after we proved $Re(f)$ is constant.
  $endgroup$
  – Mark
  Mar 10 at 22:23
  
  
  

  
  
  
  

add a comment | 

$begingroup$

The only entire holomorphic functions whose imaginary parts are constant are the constant functions. So $operatorname{Re}(f)$ is holomorphic if and only if it is constant (which implies that $f$ itself is constant).

share|cite|improve this answer

answered Mar 10 at 22:13

José Carlos SantosJosé Carlos Santos

167k22132235

$endgroup$

share|cite|improve this answer

answered Mar 10 at 22:13

José Carlos SantosJosé Carlos Santos

167k22132235

answered Mar 10 at 22:13

José Carlos SantosJosé Carlos Santos

167k22132235

answered Mar 10 at 22:13

José Carlos SantosJosé Carlos Santos

167k22132235

3

$begingroup$

If $mathfrak{R}(f)$ is entire, then $exp(imathfrak{R}(f))$ is also entire. But $|exp(i mathfrak{R}(f))| leq 1$, so $exp(imathfrak{R}(f))$ is constant (Liouville's theorem) and thus so is $mathfrak{R}(f)$. Hence, $f$ has constant real part, and the C-R equations get you that $f$ has constant imaginary part as well.

share|cite|improve this answer

answered Mar 10 at 22:16

rubikscube09rubikscube09

1,556720

$endgroup$

add a comment | 

3

$begingroup$

If $mathfrak{R}(f)$ is entire, then $exp(imathfrak{R}(f))$ is also entire. But $|exp(i mathfrak{R}(f))| leq 1$, so $exp(imathfrak{R}(f))$ is constant (Liouville's theorem) and thus so is $mathfrak{R}(f)$. Hence, $f$ has constant real part, and the C-R equations get you that $f$ has constant imaginary part as well.

share|cite|improve this answer

answered Mar 10 at 22:16

rubikscube09rubikscube09

1,556720

$endgroup$

add a comment | 

$begingroup$

If $mathfrak{R}(f)$ is entire, then $exp(imathfrak{R}(f))$ is also entire. But $|exp(i mathfrak{R}(f))| leq 1$, so $exp(imathfrak{R}(f))$ is constant (Liouville's theorem) and thus so is $mathfrak{R}(f)$. Hence, $f$ has constant real part, and the C-R equations get you that $f$ has constant imaginary part as well.

share|cite|improve this answer

answered Mar 10 at 22:16

rubikscube09rubikscube09

1,556720

$endgroup$

share|cite|improve this answer

answered Mar 10 at 22:16

rubikscube09rubikscube09

1,556720

answered Mar 10 at 22:16

rubikscube09rubikscube09

1,556720

answered Mar 10 at 22:16

rubikscube09rubikscube09

1,556720