Why is the derevative of meromorphic function is meromorphic?












0












$begingroup$


I know that if $f$ is meromorphic then $exists Asubset Omega$ s.t $f$ is holomorphic on $Omega setminus A$ and $A$ is discrete, and $A$ are the poles of $f$. I want to show that $f'$ is meromorphic, but this seems trivial for $forall xin Omegasetminus A$ we know that $f$ is analytic on a region of $x$, so $f$ is analytic there and thus $f'$ is holomorphic in $x$. this implies $f'$ holomorphic on $Omegasetminus A$,hence meromorphic. Am I missing something? This seems too obvious.






complex-analysis meromorphic-functions







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You have to show that $f'$ has a pole at each $x in A$.
    $endgroup$
    – Paul Frost
    Dec 8 '18 at 10:10










  • $begingroup$
    Why? The definition just requires that the set of poles is discrete. I showed that the set of poles has to be a subset of $A$, thus has to be discrete.
    $endgroup$
    – Simon Green
    Dec 8 '18 at 11:58










  • $begingroup$
    A meromorphic function $f$ on $Omega$ is a a holomorphic function on $Omega setminus A$, where $A$ is discrete, such that $f$ has a pole at each $x in A$. It is not allowed to have a removable or an essential singularity at $x$. You correctly state that $f'$ is holomorphic on $Omega setminus A$, but you do not analyze the singularities at points of $A$. You have to show that if $f$ has a pole at $x$, then so has $f'$.
    $endgroup$
    – Paul Frost
    Dec 8 '18 at 13:23












  • $begingroup$
    Very satisfaying answer, thank you very much.
    $endgroup$
    – Simon Green
    Dec 8 '18 at 14:02
















0












$begingroup$


I know that if $f$ is meromorphic then $exists Asubset Omega$ s.t $f$ is holomorphic on $Omega setminus A$ and $A$ is discrete, and $A$ are the poles of $f$. I want to show that $f'$ is meromorphic, but this seems trivial for $forall xin Omegasetminus A$ we know that $f$ is analytic on a region of $x$, so $f$ is analytic there and thus $f'$ is holomorphic in $x$. this implies $f'$ holomorphic on $Omegasetminus A$,hence meromorphic. Am I missing something? This seems too obvious.






complex-analysis meromorphic-functions







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You have to show that $f'$ has a pole at each $x in A$.
    $endgroup$
    – Paul Frost
    Dec 8 '18 at 10:10










  • $begingroup$
    Why? The definition just requires that the set of poles is discrete. I showed that the set of poles has to be a subset of $A$, thus has to be discrete.
    $endgroup$
    – Simon Green
    Dec 8 '18 at 11:58










  • $begingroup$
    A meromorphic function $f$ on $Omega$ is a a holomorphic function on $Omega setminus A$, where $A$ is discrete, such that $f$ has a pole at each $x in A$. It is not allowed to have a removable or an essential singularity at $x$. You correctly state that $f'$ is holomorphic on $Omega setminus A$, but you do not analyze the singularities at points of $A$. You have to show that if $f$ has a pole at $x$, then so has $f'$.
    $endgroup$
    – Paul Frost
    Dec 8 '18 at 13:23












  • $begingroup$
    Very satisfaying answer, thank you very much.
    $endgroup$
    – Simon Green
    Dec 8 '18 at 14:02














0












0








0





$begingroup$


I know that if $f$ is meromorphic then $exists Asubset Omega$ s.t $f$ is holomorphic on $Omega setminus A$ and $A$ is discrete, and $A$ are the poles of $f$. I want to show that $f'$ is meromorphic, but this seems trivial for $forall xin Omegasetminus A$ we know that $f$ is analytic on a region of $x$, so $f$ is analytic there and thus $f'$ is holomorphic in $x$. this implies $f'$ holomorphic on $Omegasetminus A$,hence meromorphic. Am I missing something? This seems too obvious.






complex-analysis meromorphic-functions







share|cite|improve this question









$endgroup$




I know that if $f$ is meromorphic then $exists Asubset Omega$ s.t $f$ is holomorphic on $Omega setminus A$ and $A$ is discrete, and $A$ are the poles of $f$. I want to show that $f'$ is meromorphic, but this seems trivial for $forall xin Omegasetminus A$ we know that $f$ is analytic on a region of $x$, so $f$ is analytic there and thus $f'$ is holomorphic in $x$. this implies $f'$ holomorphic on $Omegasetminus A$,hence meromorphic. Am I missing something? This seems too obvious.






complex-analysis meromorphic-functions




complex-analysis meromorphic-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 '18 at 9:26









Simon GreenSimon Green

825




825








  • 1




    $begingroup$
    You have to show that $f'$ has a pole at each $x in A$.
    $endgroup$
    – Paul Frost
    Dec 8 '18 at 10:10










  • $begingroup$
    Why? The definition just requires that the set of poles is discrete. I showed that the set of poles has to be a subset of $A$, thus has to be discrete.
    $endgroup$
    – Simon Green
    Dec 8 '18 at 11:58










  • $begingroup$
    A meromorphic function $f$ on $Omega$ is a a holomorphic function on $Omega setminus A$, where $A$ is discrete, such that $f$ has a pole at each $x in A$. It is not allowed to have a removable or an essential singularity at $x$. You correctly state that $f'$ is holomorphic on $Omega setminus A$, but you do not analyze the singularities at points of $A$. You have to show that if $f$ has a pole at $x$, then so has $f'$.
    $endgroup$
    – Paul Frost
    Dec 8 '18 at 13:23












  • $begingroup$
    Very satisfaying answer, thank you very much.
    $endgroup$
    – Simon Green
    Dec 8 '18 at 14:02














  • 1




    $begingroup$
    You have to show that $f'$ has a pole at each $x in A$.
    $endgroup$
    – Paul Frost
    Dec 8 '18 at 10:10










  • $begingroup$
    Why? The definition just requires that the set of poles is discrete. I showed that the set of poles has to be a subset of $A$, thus has to be discrete.
    $endgroup$
    – Simon Green
    Dec 8 '18 at 11:58










  • $begingroup$
    A meromorphic function $f$ on $Omega$ is a a holomorphic function on $Omega setminus A$, where $A$ is discrete, such that $f$ has a pole at each $x in A$. It is not allowed to have a removable or an essential singularity at $x$. You correctly state that $f'$ is holomorphic on $Omega setminus A$, but you do not analyze the singularities at points of $A$. You have to show that if $f$ has a pole at $x$, then so has $f'$.
    $endgroup$
    – Paul Frost
    Dec 8 '18 at 13:23












  • $begingroup$
    Very satisfaying answer, thank you very much.
    $endgroup$
    – Simon Green
    Dec 8 '18 at 14:02








1




1




$begingroup$
You have to show that $f'$ has a pole at each $x in A$.
$endgroup$
– Paul Frost
Dec 8 '18 at 10:10




$begingroup$
You have to show that $f'$ has a pole at each $x in A$.
$endgroup$
– Paul Frost
Dec 8 '18 at 10:10












$begingroup$
Why? The definition just requires that the set of poles is discrete. I showed that the set of poles has to be a subset of $A$, thus has to be discrete.
$endgroup$
– Simon Green
Dec 8 '18 at 11:58




$begingroup$
Why? The definition just requires that the set of poles is discrete. I showed that the set of poles has to be a subset of $A$, thus has to be discrete.
$endgroup$
– Simon Green
Dec 8 '18 at 11:58












$begingroup$
A meromorphic function $f$ on $Omega$ is a a holomorphic function on $Omega setminus A$, where $A$ is discrete, such that $f$ has a pole at each $x in A$. It is not allowed to have a removable or an essential singularity at $x$. You correctly state that $f'$ is holomorphic on $Omega setminus A$, but you do not analyze the singularities at points of $A$. You have to show that if $f$ has a pole at $x$, then so has $f'$.
$endgroup$
– Paul Frost
Dec 8 '18 at 13:23






$begingroup$
A meromorphic function $f$ on $Omega$ is a a holomorphic function on $Omega setminus A$, where $A$ is discrete, such that $f$ has a pole at each $x in A$. It is not allowed to have a removable or an essential singularity at $x$. You correctly state that $f'$ is holomorphic on $Omega setminus A$, but you do not analyze the singularities at points of $A$. You have to show that if $f$ has a pole at $x$, then so has $f'$.
$endgroup$
– Paul Frost
Dec 8 '18 at 13:23














$begingroup$
Very satisfaying answer, thank you very much.
$endgroup$
– Simon Green
Dec 8 '18 at 14:02




$begingroup$
Very satisfaying answer, thank you very much.
$endgroup$
– Simon Green
Dec 8 '18 at 14:02










2 Answers
2






active

oldest

votes


















0












$begingroup$

Hint

Expand the function as Laurent series at the poles. You know the rest part is holomorphic, and expanding it as Laurent series gives the meromorphicity of the poles in the original function.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The word should be meromorphicity.
    $endgroup$
    – Szeto
    Dec 8 '18 at 10:43










  • $begingroup$
    Thanks for your help!
    $endgroup$
    – Simon Green
    Dec 8 '18 at 12:15



















0












$begingroup$

Your argument is incomplete. You have to show that $f'$ doesn't have an essential singularity at points of $A$. Let $c in A$. We can write $(z-c)^{n}f(z)=g(z)$ for some non-negative integer $n$ and some holomorphic function $g$ in a neighborhood of $c$. We then have $(z-c)^{n}f'(z)+n(z-c)^{n-1}f(z)=g'(z)$ from which it is easy to see that $c$ is indeed a pole of $f'$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – Simon Green
    Dec 8 '18 at 12:14











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Hint

Expand the function as Laurent series at the poles. You know the rest part is holomorphic, and expanding it as Laurent series gives the meromorphicity of the poles in the original function.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The word should be meromorphicity.
    $endgroup$
    – Szeto
    Dec 8 '18 at 10:43










  • $begingroup$
    Thanks for your help!
    $endgroup$
    – Simon Green
    Dec 8 '18 at 12:15
















0












$begingroup$

Hint

Expand the function as Laurent series at the poles. You know the rest part is holomorphic, and expanding it as Laurent series gives the meromorphicity of the poles in the original function.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The word should be meromorphicity.
    $endgroup$
    – Szeto
    Dec 8 '18 at 10:43










  • $begingroup$
    Thanks for your help!
    $endgroup$
    – Simon Green
    Dec 8 '18 at 12:15














0












0








0





$begingroup$

Hint

Expand the function as Laurent series at the poles. You know the rest part is holomorphic, and expanding it as Laurent series gives the meromorphicity of the poles in the original function.






share|cite|improve this answer











$endgroup$



Hint

Expand the function as Laurent series at the poles. You know the rest part is holomorphic, and expanding it as Laurent series gives the meromorphicity of the poles in the original function.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 10:44

























answered Dec 8 '18 at 9:39









Kemono ChenKemono Chen

3,0721743




3,0721743












  • $begingroup$
    The word should be meromorphicity.
    $endgroup$
    – Szeto
    Dec 8 '18 at 10:43










  • $begingroup$
    Thanks for your help!
    $endgroup$
    – Simon Green
    Dec 8 '18 at 12:15


















  • $begingroup$
    The word should be meromorphicity.
    $endgroup$
    – Szeto
    Dec 8 '18 at 10:43










  • $begingroup$
    Thanks for your help!
    $endgroup$
    – Simon Green
    Dec 8 '18 at 12:15
















$begingroup$
The word should be meromorphicity.
$endgroup$
– Szeto
Dec 8 '18 at 10:43




$begingroup$
The word should be meromorphicity.
$endgroup$
– Szeto
Dec 8 '18 at 10:43












$begingroup$
Thanks for your help!
$endgroup$
– Simon Green
Dec 8 '18 at 12:15




$begingroup$
Thanks for your help!
$endgroup$
– Simon Green
Dec 8 '18 at 12:15











0












$begingroup$

Your argument is incomplete. You have to show that $f'$ doesn't have an essential singularity at points of $A$. Let $c in A$. We can write $(z-c)^{n}f(z)=g(z)$ for some non-negative integer $n$ and some holomorphic function $g$ in a neighborhood of $c$. We then have $(z-c)^{n}f'(z)+n(z-c)^{n-1}f(z)=g'(z)$ from which it is easy to see that $c$ is indeed a pole of $f'$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – Simon Green
    Dec 8 '18 at 12:14
















0












$begingroup$

Your argument is incomplete. You have to show that $f'$ doesn't have an essential singularity at points of $A$. Let $c in A$. We can write $(z-c)^{n}f(z)=g(z)$ for some non-negative integer $n$ and some holomorphic function $g$ in a neighborhood of $c$. We then have $(z-c)^{n}f'(z)+n(z-c)^{n-1}f(z)=g'(z)$ from which it is easy to see that $c$ is indeed a pole of $f'$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – Simon Green
    Dec 8 '18 at 12:14














0












0








0





$begingroup$

Your argument is incomplete. You have to show that $f'$ doesn't have an essential singularity at points of $A$. Let $c in A$. We can write $(z-c)^{n}f(z)=g(z)$ for some non-negative integer $n$ and some holomorphic function $g$ in a neighborhood of $c$. We then have $(z-c)^{n}f'(z)+n(z-c)^{n-1}f(z)=g'(z)$ from which it is easy to see that $c$ is indeed a pole of $f'$.






share|cite|improve this answer









$endgroup$



Your argument is incomplete. You have to show that $f'$ doesn't have an essential singularity at points of $A$. Let $c in A$. We can write $(z-c)^{n}f(z)=g(z)$ for some non-negative integer $n$ and some holomorphic function $g$ in a neighborhood of $c$. We then have $(z-c)^{n}f'(z)+n(z-c)^{n-1}f(z)=g'(z)$ from which it is easy to see that $c$ is indeed a pole of $f'$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '18 at 12:10









Kavi Rama MurthyKavi Rama Murthy

59.8k42161




59.8k42161












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Simon Green
    Dec 8 '18 at 12:14


















  • $begingroup$
    Thank you very much!
    $endgroup$
    – Simon Green
    Dec 8 '18 at 12:14
















$begingroup$
Thank you very much!
$endgroup$
– Simon Green
Dec 8 '18 at 12:14




$begingroup$
Thank you very much!
$endgroup$
– Simon Green
Dec 8 '18 at 12:14


















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