Find the residue of the function $f(z) = e^{(z^2)/(z^n)}$ at all its poles.












0














I'm currently working on this problem, and I think there is a trick to it because of the $z^2$ in the exponent of e. I fail to see it though.



Find the residue of the function $f(z) = e^{(z^2)/(z^n)}$ at all its poles.










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  • What do we know about $n$? Is $n$ an integer?
    – Omnomnomnom
    Jan 4 at 15:51






  • 2




    What poles? This has no poles. $0$ is an essential singularity (if $n$ is an integer $>2$), or a branch point if $n$ is not an integer.
    – Robert Israel
    Jan 4 at 15:56


















0














I'm currently working on this problem, and I think there is a trick to it because of the $z^2$ in the exponent of e. I fail to see it though.



Find the residue of the function $f(z) = e^{(z^2)/(z^n)}$ at all its poles.










share|cite|improve this question






















  • What do we know about $n$? Is $n$ an integer?
    – Omnomnomnom
    Jan 4 at 15:51






  • 2




    What poles? This has no poles. $0$ is an essential singularity (if $n$ is an integer $>2$), or a branch point if $n$ is not an integer.
    – Robert Israel
    Jan 4 at 15:56
















0












0








0







I'm currently working on this problem, and I think there is a trick to it because of the $z^2$ in the exponent of e. I fail to see it though.



Find the residue of the function $f(z) = e^{(z^2)/(z^n)}$ at all its poles.










share|cite|improve this question













I'm currently working on this problem, and I think there is a trick to it because of the $z^2$ in the exponent of e. I fail to see it though.



Find the residue of the function $f(z) = e^{(z^2)/(z^n)}$ at all its poles.







real-analysis complex-analysis analysis






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asked Jan 4 at 15:48









user628226user628226

525




525












  • What do we know about $n$? Is $n$ an integer?
    – Omnomnomnom
    Jan 4 at 15:51






  • 2




    What poles? This has no poles. $0$ is an essential singularity (if $n$ is an integer $>2$), or a branch point if $n$ is not an integer.
    – Robert Israel
    Jan 4 at 15:56




















  • What do we know about $n$? Is $n$ an integer?
    – Omnomnomnom
    Jan 4 at 15:51






  • 2




    What poles? This has no poles. $0$ is an essential singularity (if $n$ is an integer $>2$), or a branch point if $n$ is not an integer.
    – Robert Israel
    Jan 4 at 15:56


















What do we know about $n$? Is $n$ an integer?
– Omnomnomnom
Jan 4 at 15:51




What do we know about $n$? Is $n$ an integer?
– Omnomnomnom
Jan 4 at 15:51




2




2




What poles? This has no poles. $0$ is an essential singularity (if $n$ is an integer $>2$), or a branch point if $n$ is not an integer.
– Robert Israel
Jan 4 at 15:56






What poles? This has no poles. $0$ is an essential singularity (if $n$ is an integer $>2$), or a branch point if $n$ is not an integer.
– Robert Israel
Jan 4 at 15:56












1 Answer
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I am assuming that $n$ is supposed to be an integer.



Hint: $f$ only fails to be complex-differentiable at $z = 0$ (which is a singularity, but not a pole). Using the series $e^x = sum_k frac 1{k!}x^k$, we can express $f(z)$ as a Laurent series centered at $z = 0$. Using the Laurent series, we can find the desired residue.






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    I am assuming that $n$ is supposed to be an integer.



    Hint: $f$ only fails to be complex-differentiable at $z = 0$ (which is a singularity, but not a pole). Using the series $e^x = sum_k frac 1{k!}x^k$, we can express $f(z)$ as a Laurent series centered at $z = 0$. Using the Laurent series, we can find the desired residue.






    share|cite|improve this answer


























      2














      I am assuming that $n$ is supposed to be an integer.



      Hint: $f$ only fails to be complex-differentiable at $z = 0$ (which is a singularity, but not a pole). Using the series $e^x = sum_k frac 1{k!}x^k$, we can express $f(z)$ as a Laurent series centered at $z = 0$. Using the Laurent series, we can find the desired residue.






      share|cite|improve this answer
























        2












        2








        2






        I am assuming that $n$ is supposed to be an integer.



        Hint: $f$ only fails to be complex-differentiable at $z = 0$ (which is a singularity, but not a pole). Using the series $e^x = sum_k frac 1{k!}x^k$, we can express $f(z)$ as a Laurent series centered at $z = 0$. Using the Laurent series, we can find the desired residue.






        share|cite|improve this answer












        I am assuming that $n$ is supposed to be an integer.



        Hint: $f$ only fails to be complex-differentiable at $z = 0$ (which is a singularity, but not a pole). Using the series $e^x = sum_k frac 1{k!}x^k$, we can express $f(z)$ as a Laurent series centered at $z = 0$. Using the Laurent series, we can find the desired residue.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 15:55









        OmnomnomnomOmnomnomnom

        127k788176




        127k788176






























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