On the residues of $f(z)=(z-z_0)^{-n}$

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Define $f_{n}(z)=(z-z_0)^{-n}$ where $n$ is a positive integer. Notice that $f_n$ has a pole of order $n$ at $z=z_0$, for every positive integer $n$.



My question: Does $f$ has residue $0$ at $z=z_0$ for every positive integer $n$?



Unless i'm mistaken, this seems to be the case, since



$frac{1}{(n-1)!}frac{d^{n} f}{dz^{n}} (z-z_0)^n (z-z_0)^{-n}=0$



for every positive integer $n.$






complex-analysis residue-calculus







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  • There will be a non-zero residue for $n = 1$, and a zero residue for all other integers $n$.
    – Omnomnomnom
    2 days ago






  • 1




    The formula to compute the residue at an $n$th order pole is $$ operatorname{Res}(f,z_0) = frac{1}{(n-1)!} lim_{z to z_0} frac{d^{n-1}}{dz^{n-1}} left( (z-z_0)^n f(z) right) $$ note that for $n=1$, we're taking a "zeroeth derivative"
    – Omnomnomnom
    2 days ago


















0














Define $f_{n}(z)=(z-z_0)^{-n}$ where $n$ is a positive integer. Notice that $f_n$ has a pole of order $n$ at $z=z_0$, for every positive integer $n$.



My question: Does $f$ has residue $0$ at $z=z_0$ for every positive integer $n$?



Unless i'm mistaken, this seems to be the case, since



$frac{1}{(n-1)!}frac{d^{n} f}{dz^{n}} (z-z_0)^n (z-z_0)^{-n}=0$



for every positive integer $n.$






complex-analysis residue-calculus







share|cite|improve this question









New contributor




user632013 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • There will be a non-zero residue for $n = 1$, and a zero residue for all other integers $n$.
    – Omnomnomnom
    2 days ago






  • 1




    The formula to compute the residue at an $n$th order pole is $$ operatorname{Res}(f,z_0) = frac{1}{(n-1)!} lim_{z to z_0} frac{d^{n-1}}{dz^{n-1}} left( (z-z_0)^n f(z) right) $$ note that for $n=1$, we're taking a "zeroeth derivative"
    – Omnomnomnom
    2 days ago
















0












0








0







Define $f_{n}(z)=(z-z_0)^{-n}$ where $n$ is a positive integer. Notice that $f_n$ has a pole of order $n$ at $z=z_0$, for every positive integer $n$.



My question: Does $f$ has residue $0$ at $z=z_0$ for every positive integer $n$?



Unless i'm mistaken, this seems to be the case, since



$frac{1}{(n-1)!}frac{d^{n} f}{dz^{n}} (z-z_0)^n (z-z_0)^{-n}=0$



for every positive integer $n.$






complex-analysis residue-calculus







share|cite|improve this question









New contributor




user632013 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Define $f_{n}(z)=(z-z_0)^{-n}$ where $n$ is a positive integer. Notice that $f_n$ has a pole of order $n$ at $z=z_0$, for every positive integer $n$.



My question: Does $f$ has residue $0$ at $z=z_0$ for every positive integer $n$?



Unless i'm mistaken, this seems to be the case, since



$frac{1}{(n-1)!}frac{d^{n} f}{dz^{n}} (z-z_0)^n (z-z_0)^{-n}=0$



for every positive integer $n.$






complex-analysis residue-calculus




complex-analysis residue-calculus






share|cite|improve this question









New contributor




user632013 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Check out our Code of Conduct.









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edited 2 days ago







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user632013 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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user632013 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • There will be a non-zero residue for $n = 1$, and a zero residue for all other integers $n$.
    – Omnomnomnom
    2 days ago






  • 1




    The formula to compute the residue at an $n$th order pole is $$ operatorname{Res}(f,z_0) = frac{1}{(n-1)!} lim_{z to z_0} frac{d^{n-1}}{dz^{n-1}} left( (z-z_0)^n f(z) right) $$ note that for $n=1$, we're taking a "zeroeth derivative"
    – Omnomnomnom
    2 days ago




















  • There will be a non-zero residue for $n = 1$, and a zero residue for all other integers $n$.
    – Omnomnomnom
    2 days ago






  • 1




    The formula to compute the residue at an $n$th order pole is $$ operatorname{Res}(f,z_0) = frac{1}{(n-1)!} lim_{z to z_0} frac{d^{n-1}}{dz^{n-1}} left( (z-z_0)^n f(z) right) $$ note that for $n=1$, we're taking a "zeroeth derivative"
    – Omnomnomnom
    2 days ago


















There will be a non-zero residue for $n = 1$, and a zero residue for all other integers $n$.
– Omnomnomnom
2 days ago




There will be a non-zero residue for $n = 1$, and a zero residue for all other integers $n$.
– Omnomnomnom
2 days ago




1




1




The formula to compute the residue at an $n$th order pole is $$ operatorname{Res}(f,z_0) = frac{1}{(n-1)!} lim_{z to z_0} frac{d^{n-1}}{dz^{n-1}} left( (z-z_0)^n f(z) right) $$ note that for $n=1$, we're taking a "zeroeth derivative"
– Omnomnomnom
2 days ago






The formula to compute the residue at an $n$th order pole is $$ operatorname{Res}(f,z_0) = frac{1}{(n-1)!} lim_{z to z_0} frac{d^{n-1}}{dz^{n-1}} left( (z-z_0)^n f(z) right) $$ note that for $n=1$, we're taking a "zeroeth derivative"
– Omnomnomnom
2 days ago












2 Answers
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No. The residue is $1$ if $n=1$ (and $0$ on all other cases). Your formula is wrong. It should have $dfrac{mathrm d^{n-1}f}{mathrm dx^{n-1}}$ instead of $dfrac{mathrm d^nf}{mathrm dx^n}$.






share|cite|improve this answer





















  • Oh, that was a typo, anyway thanks.
    – user632013
    2 days ago












  • @user632013 If my answer was useful perhaps that you could mark it as the accepted one.
    – José Carlos Santos
    2 days ago



















0














The residue at a point $z_0$ of some function it is just the coefficient $c_{-1}$ of the Laurent expansion of the function around $z_0$. A Laurent expansion in an annulus around $z=z_0$ for some function $g$ have the form



$$g(z):=sum_{kinBbb Z} c_k (z-z_0)^k$$



In your case $f_n$ is already in the form of a Laurent series around $z_0$ such that $c_k=0$ for all $kneq -n$ and $c_{-n}=1$. Then $f_n$ have residue zero at $z_0$ for each $nneq 1$, and for $f_1$ we have that $c_{-1}=1$, so the residue of $f_1$ at $z_0$ is $1$.






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    2 Answers
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    2 Answers
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    active

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    active

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    No. The residue is $1$ if $n=1$ (and $0$ on all other cases). Your formula is wrong. It should have $dfrac{mathrm d^{n-1}f}{mathrm dx^{n-1}}$ instead of $dfrac{mathrm d^nf}{mathrm dx^n}$.






    share|cite|improve this answer





















    • Oh, that was a typo, anyway thanks.
      – user632013
      2 days ago












    • @user632013 If my answer was useful perhaps that you could mark it as the accepted one.
      – José Carlos Santos
      2 days ago
















    2














    No. The residue is $1$ if $n=1$ (and $0$ on all other cases). Your formula is wrong. It should have $dfrac{mathrm d^{n-1}f}{mathrm dx^{n-1}}$ instead of $dfrac{mathrm d^nf}{mathrm dx^n}$.






    share|cite|improve this answer





















    • Oh, that was a typo, anyway thanks.
      – user632013
      2 days ago












    • @user632013 If my answer was useful perhaps that you could mark it as the accepted one.
      – José Carlos Santos
      2 days ago














    2












    2








    2






    No. The residue is $1$ if $n=1$ (and $0$ on all other cases). Your formula is wrong. It should have $dfrac{mathrm d^{n-1}f}{mathrm dx^{n-1}}$ instead of $dfrac{mathrm d^nf}{mathrm dx^n}$.






    share|cite|improve this answer












    No. The residue is $1$ if $n=1$ (and $0$ on all other cases). Your formula is wrong. It should have $dfrac{mathrm d^{n-1}f}{mathrm dx^{n-1}}$ instead of $dfrac{mathrm d^nf}{mathrm dx^n}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 days ago









    José Carlos SantosJosé Carlos Santos

    152k22123226




    152k22123226












    • Oh, that was a typo, anyway thanks.
      – user632013
      2 days ago












    • @user632013 If my answer was useful perhaps that you could mark it as the accepted one.
      – José Carlos Santos
      2 days ago


















    • Oh, that was a typo, anyway thanks.
      – user632013
      2 days ago












    • @user632013 If my answer was useful perhaps that you could mark it as the accepted one.
      – José Carlos Santos
      2 days ago
















    Oh, that was a typo, anyway thanks.
    – user632013
    2 days ago






    Oh, that was a typo, anyway thanks.
    – user632013
    2 days ago














    @user632013 If my answer was useful perhaps that you could mark it as the accepted one.
    – José Carlos Santos
    2 days ago




    @user632013 If my answer was useful perhaps that you could mark it as the accepted one.
    – José Carlos Santos
    2 days ago











    0














    The residue at a point $z_0$ of some function it is just the coefficient $c_{-1}$ of the Laurent expansion of the function around $z_0$. A Laurent expansion in an annulus around $z=z_0$ for some function $g$ have the form



    $$g(z):=sum_{kinBbb Z} c_k (z-z_0)^k$$



    In your case $f_n$ is already in the form of a Laurent series around $z_0$ such that $c_k=0$ for all $kneq -n$ and $c_{-n}=1$. Then $f_n$ have residue zero at $z_0$ for each $nneq 1$, and for $f_1$ we have that $c_{-1}=1$, so the residue of $f_1$ at $z_0$ is $1$.






    share|cite|improve this answer


























      0














      The residue at a point $z_0$ of some function it is just the coefficient $c_{-1}$ of the Laurent expansion of the function around $z_0$. A Laurent expansion in an annulus around $z=z_0$ for some function $g$ have the form



      $$g(z):=sum_{kinBbb Z} c_k (z-z_0)^k$$



      In your case $f_n$ is already in the form of a Laurent series around $z_0$ such that $c_k=0$ for all $kneq -n$ and $c_{-n}=1$. Then $f_n$ have residue zero at $z_0$ for each $nneq 1$, and for $f_1$ we have that $c_{-1}=1$, so the residue of $f_1$ at $z_0$ is $1$.






      share|cite|improve this answer
























        0












        0








        0






        The residue at a point $z_0$ of some function it is just the coefficient $c_{-1}$ of the Laurent expansion of the function around $z_0$. A Laurent expansion in an annulus around $z=z_0$ for some function $g$ have the form



        $$g(z):=sum_{kinBbb Z} c_k (z-z_0)^k$$



        In your case $f_n$ is already in the form of a Laurent series around $z_0$ such that $c_k=0$ for all $kneq -n$ and $c_{-n}=1$. Then $f_n$ have residue zero at $z_0$ for each $nneq 1$, and for $f_1$ we have that $c_{-1}=1$, so the residue of $f_1$ at $z_0$ is $1$.






        share|cite|improve this answer












        The residue at a point $z_0$ of some function it is just the coefficient $c_{-1}$ of the Laurent expansion of the function around $z_0$. A Laurent expansion in an annulus around $z=z_0$ for some function $g$ have the form



        $$g(z):=sum_{kinBbb Z} c_k (z-z_0)^k$$



        In your case $f_n$ is already in the form of a Laurent series around $z_0$ such that $c_k=0$ for all $kneq -n$ and $c_{-n}=1$. Then $f_n$ have residue zero at $z_0$ for each $nneq 1$, and for $f_1$ we have that $c_{-1}=1$, so the residue of $f_1$ at $z_0$ is $1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        MasacrosoMasacroso

        13k41746




        13k41746






















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