Find $int_{gamma} e^zz^n dz$ where $gamma$ is the unit circle, using Cauchy's Integral Formula












2












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I'm been banging my head against the wall trying to solve the following question which ask to solve the following integral using the Cauchy integral formula, and hence evaluating the corresponding real integrals.



$int_{gamma} e^zz^n dz$ where $gamma$ is the unit circle {$e^{itheta}: -pi leq theta leq pi$} and $nin mathbf{Z}$.



To solve the question, I'm attempting to use the generalised form of the Cauchy integral formula. Although to use it, the $z^n$ is normally in the denominator not the numerator.



Any help will be much appreciated, thanks!!






integration complex-analysis contour-integration complex-integration cauchy-integral-formula







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  • $begingroup$
    Hey. Is this from math2621?
    $endgroup$
    – Bell
    Oct 7 '18 at 6:02










  • $begingroup$
    @Bell yep it is!
    $endgroup$
    – Mr Bro
    Oct 7 '18 at 9:52










  • $begingroup$
    I'm guessing you do math2221 as well. Are your group A or B? haha
    $endgroup$
    – Bell
    Oct 7 '18 at 23:24
















2












$begingroup$


I'm been banging my head against the wall trying to solve the following question which ask to solve the following integral using the Cauchy integral formula, and hence evaluating the corresponding real integrals.



$int_{gamma} e^zz^n dz$ where $gamma$ is the unit circle {$e^{itheta}: -pi leq theta leq pi$} and $nin mathbf{Z}$.



To solve the question, I'm attempting to use the generalised form of the Cauchy integral formula. Although to use it, the $z^n$ is normally in the denominator not the numerator.



Any help will be much appreciated, thanks!!






integration complex-analysis contour-integration complex-integration cauchy-integral-formula







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hey. Is this from math2621?
    $endgroup$
    – Bell
    Oct 7 '18 at 6:02










  • $begingroup$
    @Bell yep it is!
    $endgroup$
    – Mr Bro
    Oct 7 '18 at 9:52










  • $begingroup$
    I'm guessing you do math2221 as well. Are your group A or B? haha
    $endgroup$
    – Bell
    Oct 7 '18 at 23:24














2












2








2


0



$begingroup$


I'm been banging my head against the wall trying to solve the following question which ask to solve the following integral using the Cauchy integral formula, and hence evaluating the corresponding real integrals.



$int_{gamma} e^zz^n dz$ where $gamma$ is the unit circle {$e^{itheta}: -pi leq theta leq pi$} and $nin mathbf{Z}$.



To solve the question, I'm attempting to use the generalised form of the Cauchy integral formula. Although to use it, the $z^n$ is normally in the denominator not the numerator.



Any help will be much appreciated, thanks!!






integration complex-analysis contour-integration complex-integration cauchy-integral-formula







share|cite|improve this question











$endgroup$




I'm been banging my head against the wall trying to solve the following question which ask to solve the following integral using the Cauchy integral formula, and hence evaluating the corresponding real integrals.



$int_{gamma} e^zz^n dz$ where $gamma$ is the unit circle {$e^{itheta}: -pi leq theta leq pi$} and $nin mathbf{Z}$.



To solve the question, I'm attempting to use the generalised form of the Cauchy integral formula. Although to use it, the $z^n$ is normally in the denominator not the numerator.



Any help will be much appreciated, thanks!!






integration complex-analysis contour-integration complex-integration cauchy-integral-formula




integration complex-analysis contour-integration complex-integration cauchy-integral-formula






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 11:25









José Carlos Santos

161k22127232




161k22127232










asked Sep 27 '18 at 8:38









Mr BroMr Bro

175




175












  • $begingroup$
    Hey. Is this from math2621?
    $endgroup$
    – Bell
    Oct 7 '18 at 6:02










  • $begingroup$
    @Bell yep it is!
    $endgroup$
    – Mr Bro
    Oct 7 '18 at 9:52










  • $begingroup$
    I'm guessing you do math2221 as well. Are your group A or B? haha
    $endgroup$
    – Bell
    Oct 7 '18 at 23:24


















  • $begingroup$
    Hey. Is this from math2621?
    $endgroup$
    – Bell
    Oct 7 '18 at 6:02










  • $begingroup$
    @Bell yep it is!
    $endgroup$
    – Mr Bro
    Oct 7 '18 at 9:52










  • $begingroup$
    I'm guessing you do math2221 as well. Are your group A or B? haha
    $endgroup$
    – Bell
    Oct 7 '18 at 23:24
















$begingroup$
Hey. Is this from math2621?
$endgroup$
– Bell
Oct 7 '18 at 6:02




$begingroup$
Hey. Is this from math2621?
$endgroup$
– Bell
Oct 7 '18 at 6:02












$begingroup$
@Bell yep it is!
$endgroup$
– Mr Bro
Oct 7 '18 at 9:52




$begingroup$
@Bell yep it is!
$endgroup$
– Mr Bro
Oct 7 '18 at 9:52












$begingroup$
I'm guessing you do math2221 as well. Are your group A or B? haha
$endgroup$
– Bell
Oct 7 '18 at 23:24




$begingroup$
I'm guessing you do math2221 as well. Are your group A or B? haha
$endgroup$
– Bell
Oct 7 '18 at 23:24










2 Answers
2






active

oldest

votes


















3












$begingroup$

That integral is equal to $0$ if $ngeqslant0$. In fact,begin{align}int_gamma e^zz^n,mathrm dz&=int_gammafrac{e^zz^{n+1}}z,mathrm dz\&=2pi ie^00^{n+1}\&=0.end{align}On the other hand, if $n<0$, thenbegin{align}int_gamma e^zz^n,mathrm dz&=int_gammafrac{e^z}{z^{-n}},mathrm dz\&=frac{2pi i}{(-n-1)!}exp^{(-n-1)}(0)\&=frac{2pi i}{(-n-1)!}.end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The question is for $n in mathbb Z$ not $n in mathbb N$.
    $endgroup$
    – Kavi Rama Murthy
    Sep 27 '18 at 8:46










  • $begingroup$
    @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
    $endgroup$
    – Mr Bro
    Sep 27 '18 at 8:50










  • $begingroup$
    @KaviRamaMurthy I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Sep 27 '18 at 8:52










  • $begingroup$
    Should there be a (−n−1)! in the denominator of the answer?
    $endgroup$
    – Mr Bro
    Sep 27 '18 at 9:24












  • $begingroup$
    @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Sep 27 '18 at 9:39



















3












$begingroup$

Let $f(z)=e^{z}$. Then $f^{(k)}(0)=frac {k!} {2pi i} int_{gamma} frac {f(z)} {z^{k+1}}dz$. Use this when $n<0$ in your integral.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    That integral is equal to $0$ if $ngeqslant0$. In fact,begin{align}int_gamma e^zz^n,mathrm dz&=int_gammafrac{e^zz^{n+1}}z,mathrm dz\&=2pi ie^00^{n+1}\&=0.end{align}On the other hand, if $n<0$, thenbegin{align}int_gamma e^zz^n,mathrm dz&=int_gammafrac{e^z}{z^{-n}},mathrm dz\&=frac{2pi i}{(-n-1)!}exp^{(-n-1)}(0)\&=frac{2pi i}{(-n-1)!}.end{align}






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The question is for $n in mathbb Z$ not $n in mathbb N$.
      $endgroup$
      – Kavi Rama Murthy
      Sep 27 '18 at 8:46










    • $begingroup$
      @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
      $endgroup$
      – Mr Bro
      Sep 27 '18 at 8:50










    • $begingroup$
      @KaviRamaMurthy I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Sep 27 '18 at 8:52










    • $begingroup$
      Should there be a (−n−1)! in the denominator of the answer?
      $endgroup$
      – Mr Bro
      Sep 27 '18 at 9:24












    • $begingroup$
      @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Sep 27 '18 at 9:39
















    3












    $begingroup$

    That integral is equal to $0$ if $ngeqslant0$. In fact,begin{align}int_gamma e^zz^n,mathrm dz&=int_gammafrac{e^zz^{n+1}}z,mathrm dz\&=2pi ie^00^{n+1}\&=0.end{align}On the other hand, if $n<0$, thenbegin{align}int_gamma e^zz^n,mathrm dz&=int_gammafrac{e^z}{z^{-n}},mathrm dz\&=frac{2pi i}{(-n-1)!}exp^{(-n-1)}(0)\&=frac{2pi i}{(-n-1)!}.end{align}






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The question is for $n in mathbb Z$ not $n in mathbb N$.
      $endgroup$
      – Kavi Rama Murthy
      Sep 27 '18 at 8:46










    • $begingroup$
      @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
      $endgroup$
      – Mr Bro
      Sep 27 '18 at 8:50










    • $begingroup$
      @KaviRamaMurthy I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Sep 27 '18 at 8:52










    • $begingroup$
      Should there be a (−n−1)! in the denominator of the answer?
      $endgroup$
      – Mr Bro
      Sep 27 '18 at 9:24












    • $begingroup$
      @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Sep 27 '18 at 9:39














    3












    3








    3





    $begingroup$

    That integral is equal to $0$ if $ngeqslant0$. In fact,begin{align}int_gamma e^zz^n,mathrm dz&=int_gammafrac{e^zz^{n+1}}z,mathrm dz\&=2pi ie^00^{n+1}\&=0.end{align}On the other hand, if $n<0$, thenbegin{align}int_gamma e^zz^n,mathrm dz&=int_gammafrac{e^z}{z^{-n}},mathrm dz\&=frac{2pi i}{(-n-1)!}exp^{(-n-1)}(0)\&=frac{2pi i}{(-n-1)!}.end{align}






    share|cite|improve this answer











    $endgroup$



    That integral is equal to $0$ if $ngeqslant0$. In fact,begin{align}int_gamma e^zz^n,mathrm dz&=int_gammafrac{e^zz^{n+1}}z,mathrm dz\&=2pi ie^00^{n+1}\&=0.end{align}On the other hand, if $n<0$, thenbegin{align}int_gamma e^zz^n,mathrm dz&=int_gammafrac{e^z}{z^{-n}},mathrm dz\&=frac{2pi i}{(-n-1)!}exp^{(-n-1)}(0)\&=frac{2pi i}{(-n-1)!}.end{align}







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 27 '18 at 11:16

























    answered Sep 27 '18 at 8:43









    José Carlos SantosJosé Carlos Santos

    161k22127232




    161k22127232












    • $begingroup$
      The question is for $n in mathbb Z$ not $n in mathbb N$.
      $endgroup$
      – Kavi Rama Murthy
      Sep 27 '18 at 8:46










    • $begingroup$
      @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
      $endgroup$
      – Mr Bro
      Sep 27 '18 at 8:50










    • $begingroup$
      @KaviRamaMurthy I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Sep 27 '18 at 8:52










    • $begingroup$
      Should there be a (−n−1)! in the denominator of the answer?
      $endgroup$
      – Mr Bro
      Sep 27 '18 at 9:24












    • $begingroup$
      @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Sep 27 '18 at 9:39


















    • $begingroup$
      The question is for $n in mathbb Z$ not $n in mathbb N$.
      $endgroup$
      – Kavi Rama Murthy
      Sep 27 '18 at 8:46










    • $begingroup$
      @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
      $endgroup$
      – Mr Bro
      Sep 27 '18 at 8:50










    • $begingroup$
      @KaviRamaMurthy I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Sep 27 '18 at 8:52










    • $begingroup$
      Should there be a (−n−1)! in the denominator of the answer?
      $endgroup$
      – Mr Bro
      Sep 27 '18 at 9:24












    • $begingroup$
      @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
      $endgroup$
      – José Carlos Santos
      Sep 27 '18 at 9:39
















    $begingroup$
    The question is for $n in mathbb Z$ not $n in mathbb N$.
    $endgroup$
    – Kavi Rama Murthy
    Sep 27 '18 at 8:46




    $begingroup$
    The question is for $n in mathbb Z$ not $n in mathbb N$.
    $endgroup$
    – Kavi Rama Murthy
    Sep 27 '18 at 8:46












    $begingroup$
    @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
    $endgroup$
    – Mr Bro
    Sep 27 '18 at 8:50




    $begingroup$
    @KaviRamaMurthy is right. since, if n is negative, then the function is not differentiable at z=0 meaning that the integral cannot be evaluated this way
    $endgroup$
    – Mr Bro
    Sep 27 '18 at 8:50












    $begingroup$
    @KaviRamaMurthy I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Sep 27 '18 at 8:52




    $begingroup$
    @KaviRamaMurthy I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Sep 27 '18 at 8:52












    $begingroup$
    Should there be a (−n−1)! in the denominator of the answer?
    $endgroup$
    – Mr Bro
    Sep 27 '18 at 9:24






    $begingroup$
    Should there be a (−n−1)! in the denominator of the answer?
    $endgroup$
    – Mr Bro
    Sep 27 '18 at 9:24














    $begingroup$
    @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Sep 27 '18 at 9:39




    $begingroup$
    @MrBro Indeed. Sorry about that. I've edited my answer. Thank you.
    $endgroup$
    – José Carlos Santos
    Sep 27 '18 at 9:39











    3












    $begingroup$

    Let $f(z)=e^{z}$. Then $f^{(k)}(0)=frac {k!} {2pi i} int_{gamma} frac {f(z)} {z^{k+1}}dz$. Use this when $n<0$ in your integral.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Let $f(z)=e^{z}$. Then $f^{(k)}(0)=frac {k!} {2pi i} int_{gamma} frac {f(z)} {z^{k+1}}dz$. Use this when $n<0$ in your integral.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Let $f(z)=e^{z}$. Then $f^{(k)}(0)=frac {k!} {2pi i} int_{gamma} frac {f(z)} {z^{k+1}}dz$. Use this when $n<0$ in your integral.






        share|cite|improve this answer









        $endgroup$



        Let $f(z)=e^{z}$. Then $f^{(k)}(0)=frac {k!} {2pi i} int_{gamma} frac {f(z)} {z^{k+1}}dz$. Use this when $n<0$ in your integral.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 27 '18 at 8:51









        Kavi Rama MurthyKavi Rama Murthy

        59.8k42161




        59.8k42161






























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