Find the residue of the function $f(z) = frac{z^3}{(z-1)(z^4+2)}$ at $z=0$












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Problem: Calculate the residue of the function $f(z) = frac{z^3}{(z-1)(z^4+2)}$ at $z=0$



I am confused on where to even start on this question since there is no pole at $z=0$. Is $z=0$ even a singularity of this function?










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  • Suppose you were a mathematician in the early 19th century and you had defined the concept of a residue at the poles of meromorphic functions. How would you have generalized your definition in a consistent way such that it also applies to points where the function doesn't have a pole?
    – Count Iblis
    Jul 24 at 22:07






  • 1




    @CountIblis I am guessing your wanting me to notice that if a function $f$ doesn't have a pole at $z_0$, then the coefficient $a_{-1}$ in the Laurent expansion $f(z) = sum_{n=-infty}^{infty} a_n(z - z_0)^n$ is always going to be zero?
    – Fiticous
    Jul 24 at 22:17












  • That's right! :)
    – Count Iblis
    Jul 24 at 23:12
















1














Problem: Calculate the residue of the function $f(z) = frac{z^3}{(z-1)(z^4+2)}$ at $z=0$



I am confused on where to even start on this question since there is no pole at $z=0$. Is $z=0$ even a singularity of this function?










share|cite|improve this question
























  • Suppose you were a mathematician in the early 19th century and you had defined the concept of a residue at the poles of meromorphic functions. How would you have generalized your definition in a consistent way such that it also applies to points where the function doesn't have a pole?
    – Count Iblis
    Jul 24 at 22:07






  • 1




    @CountIblis I am guessing your wanting me to notice that if a function $f$ doesn't have a pole at $z_0$, then the coefficient $a_{-1}$ in the Laurent expansion $f(z) = sum_{n=-infty}^{infty} a_n(z - z_0)^n$ is always going to be zero?
    – Fiticous
    Jul 24 at 22:17












  • That's right! :)
    – Count Iblis
    Jul 24 at 23:12














1












1








1







Problem: Calculate the residue of the function $f(z) = frac{z^3}{(z-1)(z^4+2)}$ at $z=0$



I am confused on where to even start on this question since there is no pole at $z=0$. Is $z=0$ even a singularity of this function?










share|cite|improve this question















Problem: Calculate the residue of the function $f(z) = frac{z^3}{(z-1)(z^4+2)}$ at $z=0$



I am confused on where to even start on this question since there is no pole at $z=0$. Is $z=0$ even a singularity of this function?







complex-analysis residue-calculus






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edited Nov 30 at 20:19

























asked Jul 24 at 22:01









Fiticous

17218




17218












  • Suppose you were a mathematician in the early 19th century and you had defined the concept of a residue at the poles of meromorphic functions. How would you have generalized your definition in a consistent way such that it also applies to points where the function doesn't have a pole?
    – Count Iblis
    Jul 24 at 22:07






  • 1




    @CountIblis I am guessing your wanting me to notice that if a function $f$ doesn't have a pole at $z_0$, then the coefficient $a_{-1}$ in the Laurent expansion $f(z) = sum_{n=-infty}^{infty} a_n(z - z_0)^n$ is always going to be zero?
    – Fiticous
    Jul 24 at 22:17












  • That's right! :)
    – Count Iblis
    Jul 24 at 23:12


















  • Suppose you were a mathematician in the early 19th century and you had defined the concept of a residue at the poles of meromorphic functions. How would you have generalized your definition in a consistent way such that it also applies to points where the function doesn't have a pole?
    – Count Iblis
    Jul 24 at 22:07






  • 1




    @CountIblis I am guessing your wanting me to notice that if a function $f$ doesn't have a pole at $z_0$, then the coefficient $a_{-1}$ in the Laurent expansion $f(z) = sum_{n=-infty}^{infty} a_n(z - z_0)^n$ is always going to be zero?
    – Fiticous
    Jul 24 at 22:17












  • That's right! :)
    – Count Iblis
    Jul 24 at 23:12
















Suppose you were a mathematician in the early 19th century and you had defined the concept of a residue at the poles of meromorphic functions. How would you have generalized your definition in a consistent way such that it also applies to points where the function doesn't have a pole?
– Count Iblis
Jul 24 at 22:07




Suppose you were a mathematician in the early 19th century and you had defined the concept of a residue at the poles of meromorphic functions. How would you have generalized your definition in a consistent way such that it also applies to points where the function doesn't have a pole?
– Count Iblis
Jul 24 at 22:07




1




1




@CountIblis I am guessing your wanting me to notice that if a function $f$ doesn't have a pole at $z_0$, then the coefficient $a_{-1}$ in the Laurent expansion $f(z) = sum_{n=-infty}^{infty} a_n(z - z_0)^n$ is always going to be zero?
– Fiticous
Jul 24 at 22:17






@CountIblis I am guessing your wanting me to notice that if a function $f$ doesn't have a pole at $z_0$, then the coefficient $a_{-1}$ in the Laurent expansion $f(z) = sum_{n=-infty}^{infty} a_n(z - z_0)^n$ is always going to be zero?
– Fiticous
Jul 24 at 22:17














That's right! :)
– Count Iblis
Jul 24 at 23:12




That's right! :)
– Count Iblis
Jul 24 at 23:12










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No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.






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    1 Answer
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    1 Answer
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    No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.






    share|cite|improve this answer


























      1














      No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.






      share|cite|improve this answer
























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        No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.






        share|cite|improve this answer












        No, $0$ is not a singularity of $f$. Therefore, the residue at $0$ is $0$.







        share|cite|improve this answer












        share|cite|improve this answer



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        answered Jul 24 at 22:04









        José Carlos Santos

        149k22119221




        149k22119221






























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