Meromorphic Function with a Simple Pole at the Origin

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Let $z_0 in mathbbC$ and $f$ be a meromorphic function with a simple pole at the origin. Show that $dfrac-f^'(z)f(z)-z_0 = dfrac1z + sumlimits_n=0^infty p_n(z_0).z^n$ where each $p_n$ is a polynomial of degree
$n+1$. I am getting nowhere with this problem. I tried to work this out using the Taylor series expansion about $0$ and since $z_0$ is arbitrary, this is making the problem even more difficult for me. Thanks for any help.




complex-analysis




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  • What is $w$ here?. Is it same as $z_0$?
    – Kavi Rama Murthy
    Jul 16 at 8:46











  • Sorry!That's a typo.I have edited.
    – Ester
    Jul 16 at 8:47










  • Something more is wrong, because the new function does not have a pole at 0.
    – AlgebraicsAnonymous
    Jul 16 at 8:52










  • 1/z is present in the new function
    – Ester
    Jul 16 at 8:55










  • Not on the left hand side...
    – AlgebraicsAnonymous
    Jul 16 at 8:59














up vote
0
down vote

favorite












Let $z_0 in mathbbC$ and $f$ be a meromorphic function with a simple pole at the origin. Show that $dfrac-f^'(z)f(z)-z_0 = dfrac1z + sumlimits_n=0^infty p_n(z_0).z^n$ where each $p_n$ is a polynomial of degree
$n+1$. I am getting nowhere with this problem. I tried to work this out using the Taylor series expansion about $0$ and since $z_0$ is arbitrary, this is making the problem even more difficult for me. Thanks for any help.




complex-analysis




share|cite|improve this question





















  • What is $w$ here?. Is it same as $z_0$?
    – Kavi Rama Murthy
    Jul 16 at 8:46











  • Sorry!That's a typo.I have edited.
    – Ester
    Jul 16 at 8:47










  • Something more is wrong, because the new function does not have a pole at 0.
    – AlgebraicsAnonymous
    Jul 16 at 8:52










  • 1/z is present in the new function
    – Ester
    Jul 16 at 8:55










  • Not on the left hand side...
    – AlgebraicsAnonymous
    Jul 16 at 8:59












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $z_0 in mathbbC$ and $f$ be a meromorphic function with a simple pole at the origin. Show that $dfrac-f^'(z)f(z)-z_0 = dfrac1z + sumlimits_n=0^infty p_n(z_0).z^n$ where each $p_n$ is a polynomial of degree
$n+1$. I am getting nowhere with this problem. I tried to work this out using the Taylor series expansion about $0$ and since $z_0$ is arbitrary, this is making the problem even more difficult for me. Thanks for any help.




complex-analysis




share|cite|improve this question













Let $z_0 in mathbbC$ and $f$ be a meromorphic function with a simple pole at the origin. Show that $dfrac-f^'(z)f(z)-z_0 = dfrac1z + sumlimits_n=0^infty p_n(z_0).z^n$ where each $p_n$ is a polynomial of degree
$n+1$. I am getting nowhere with this problem. I tried to work this out using the Taylor series expansion about $0$ and since $z_0$ is arbitrary, this is making the problem even more difficult for me. Thanks for any help.





complex-analysis





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share|cite|improve this question




share|cite|improve this question








edited Jul 16 at 8:46
























asked Jul 16 at 8:38









Ester

8791925




8791925











  • What is $w$ here?. Is it same as $z_0$?
    – Kavi Rama Murthy
    Jul 16 at 8:46











  • Sorry!That's a typo.I have edited.
    – Ester
    Jul 16 at 8:47










  • Something more is wrong, because the new function does not have a pole at 0.
    – AlgebraicsAnonymous
    Jul 16 at 8:52










  • 1/z is present in the new function
    – Ester
    Jul 16 at 8:55










  • Not on the left hand side...
    – AlgebraicsAnonymous
    Jul 16 at 8:59
















  • What is $w$ here?. Is it same as $z_0$?
    – Kavi Rama Murthy
    Jul 16 at 8:46











  • Sorry!That's a typo.I have edited.
    – Ester
    Jul 16 at 8:47










  • Something more is wrong, because the new function does not have a pole at 0.
    – AlgebraicsAnonymous
    Jul 16 at 8:52










  • 1/z is present in the new function
    – Ester
    Jul 16 at 8:55










  • Not on the left hand side...
    – AlgebraicsAnonymous
    Jul 16 at 8:59















What is $w$ here?. Is it same as $z_0$?
– Kavi Rama Murthy
Jul 16 at 8:46





What is $w$ here?. Is it same as $z_0$?
– Kavi Rama Murthy
Jul 16 at 8:46













Sorry!That's a typo.I have edited.
– Ester
Jul 16 at 8:47




Sorry!That's a typo.I have edited.
– Ester
Jul 16 at 8:47












Something more is wrong, because the new function does not have a pole at 0.
– AlgebraicsAnonymous
Jul 16 at 8:52




Something more is wrong, because the new function does not have a pole at 0.
– AlgebraicsAnonymous
Jul 16 at 8:52












1/z is present in the new function
– Ester
Jul 16 at 8:55




1/z is present in the new function
– Ester
Jul 16 at 8:55












Not on the left hand side...
– AlgebraicsAnonymous
Jul 16 at 8:59




Not on the left hand side...
– AlgebraicsAnonymous
Jul 16 at 8:59










1 Answer
1






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2
down vote



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Note that the $m$-th derivative of $p_n$ can be expressed as $$p_n^(m)(z_0)=fracm!2 pi mathrmiint_gammafrac-f’(z)z^n+1(f(z)-z_0)^m+1mathrmdz$$ for some small loop around $z=0$. Now if $mgeq n+2$ then the integrand is holomorphic at $z=0$ and therefore $p_n^(m)(z_0)=0$.






share|cite|improve this answer





















  • What is p_n and how are you getting this expression?Please elaborate a bit.
    – Ester
    Jul 16 at 11:27











  • @Ester These are the Laurent coefficients as in your question (but without the assumption that they are polynomial to start with). See here for the integral expression of Laurent coefficients.
    – WimC
    Jul 16 at 11:29











  • You mean Laurent series expansion of f?
    – Ester
    Jul 16 at 11:31










  • No, not of $f$ but of $-f’(z)/(f(z)-z_0)$.
    – WimC
    Jul 16 at 11:33










  • Ok,let me check this out.If I face further problems,I shall ask you.
    – Ester
    Jul 16 at 11:34










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Note that the $m$-th derivative of $p_n$ can be expressed as $$p_n^(m)(z_0)=fracm!2 pi mathrmiint_gammafrac-f’(z)z^n+1(f(z)-z_0)^m+1mathrmdz$$ for some small loop around $z=0$. Now if $mgeq n+2$ then the integrand is holomorphic at $z=0$ and therefore $p_n^(m)(z_0)=0$.






share|cite|improve this answer





















  • What is p_n and how are you getting this expression?Please elaborate a bit.
    – Ester
    Jul 16 at 11:27











  • @Ester These are the Laurent coefficients as in your question (but without the assumption that they are polynomial to start with). See here for the integral expression of Laurent coefficients.
    – WimC
    Jul 16 at 11:29











  • You mean Laurent series expansion of f?
    – Ester
    Jul 16 at 11:31










  • No, not of $f$ but of $-f’(z)/(f(z)-z_0)$.
    – WimC
    Jul 16 at 11:33










  • Ok,let me check this out.If I face further problems,I shall ask you.
    – Ester
    Jul 16 at 11:34














up vote
2
down vote



accepted










Note that the $m$-th derivative of $p_n$ can be expressed as $$p_n^(m)(z_0)=fracm!2 pi mathrmiint_gammafrac-f’(z)z^n+1(f(z)-z_0)^m+1mathrmdz$$ for some small loop around $z=0$. Now if $mgeq n+2$ then the integrand is holomorphic at $z=0$ and therefore $p_n^(m)(z_0)=0$.






share|cite|improve this answer





















  • What is p_n and how are you getting this expression?Please elaborate a bit.
    – Ester
    Jul 16 at 11:27











  • @Ester These are the Laurent coefficients as in your question (but without the assumption that they are polynomial to start with). See here for the integral expression of Laurent coefficients.
    – WimC
    Jul 16 at 11:29











  • You mean Laurent series expansion of f?
    – Ester
    Jul 16 at 11:31










  • No, not of $f$ but of $-f’(z)/(f(z)-z_0)$.
    – WimC
    Jul 16 at 11:33










  • Ok,let me check this out.If I face further problems,I shall ask you.
    – Ester
    Jul 16 at 11:34












up vote
2
down vote



accepted







up vote
2
down vote



accepted






Note that the $m$-th derivative of $p_n$ can be expressed as $$p_n^(m)(z_0)=fracm!2 pi mathrmiint_gammafrac-f’(z)z^n+1(f(z)-z_0)^m+1mathrmdz$$ for some small loop around $z=0$. Now if $mgeq n+2$ then the integrand is holomorphic at $z=0$ and therefore $p_n^(m)(z_0)=0$.






share|cite|improve this answer













Note that the $m$-th derivative of $p_n$ can be expressed as $$p_n^(m)(z_0)=fracm!2 pi mathrmiint_gammafrac-f’(z)z^n+1(f(z)-z_0)^m+1mathrmdz$$ for some small loop around $z=0$. Now if $mgeq n+2$ then the integrand is holomorphic at $z=0$ and therefore $p_n^(m)(z_0)=0$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 16 at 10:43









WimC

23.7k22860




23.7k22860











  • What is p_n and how are you getting this expression?Please elaborate a bit.
    – Ester
    Jul 16 at 11:27











  • @Ester These are the Laurent coefficients as in your question (but without the assumption that they are polynomial to start with). See here for the integral expression of Laurent coefficients.
    – WimC
    Jul 16 at 11:29











  • You mean Laurent series expansion of f?
    – Ester
    Jul 16 at 11:31










  • No, not of $f$ but of $-f’(z)/(f(z)-z_0)$.
    – WimC
    Jul 16 at 11:33










  • Ok,let me check this out.If I face further problems,I shall ask you.
    – Ester
    Jul 16 at 11:34
















  • What is p_n and how are you getting this expression?Please elaborate a bit.
    – Ester
    Jul 16 at 11:27











  • @Ester These are the Laurent coefficients as in your question (but without the assumption that they are polynomial to start with). See here for the integral expression of Laurent coefficients.
    – WimC
    Jul 16 at 11:29











  • You mean Laurent series expansion of f?
    – Ester
    Jul 16 at 11:31










  • No, not of $f$ but of $-f’(z)/(f(z)-z_0)$.
    – WimC
    Jul 16 at 11:33










  • Ok,let me check this out.If I face further problems,I shall ask you.
    – Ester
    Jul 16 at 11:34















What is p_n and how are you getting this expression?Please elaborate a bit.
– Ester
Jul 16 at 11:27





What is p_n and how are you getting this expression?Please elaborate a bit.
– Ester
Jul 16 at 11:27













@Ester These are the Laurent coefficients as in your question (but without the assumption that they are polynomial to start with). See here for the integral expression of Laurent coefficients.
– WimC
Jul 16 at 11:29





@Ester These are the Laurent coefficients as in your question (but without the assumption that they are polynomial to start with). See here for the integral expression of Laurent coefficients.
– WimC
Jul 16 at 11:29













You mean Laurent series expansion of f?
– Ester
Jul 16 at 11:31




You mean Laurent series expansion of f?
– Ester
Jul 16 at 11:31












No, not of $f$ but of $-f’(z)/(f(z)-z_0)$.
– WimC
Jul 16 at 11:33




No, not of $f$ but of $-f’(z)/(f(z)-z_0)$.
– WimC
Jul 16 at 11:33












Ok,let me check this out.If I face further problems,I shall ask you.
– Ester
Jul 16 at 11:34




Ok,let me check this out.If I face further problems,I shall ask you.
– Ester
Jul 16 at 11:34












 

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