evaluate the integral by using Cauchy's residue theorem












2












$begingroup$


evalulate the integral by using Cauchy's residue theorem



$$int_{|z|=1} frac{1}{e^z -1-2z}dz$$



MY attempt : $ f(z) =frac{1}{e^z -1-2z}$, now put $z= 1$ we get $f(z)=-1$



so $$int_{|z|=1} frac{1}{e^z -1-2z}dz= -2pi i$$



Is it correct?



Any hints/solution



thanks u






complex-analysis







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Residue is not as simple as "the value at $z=1$". For instance, $frac1z^2$ has zero residue.
    $endgroup$
    – Arthur
    21 hours ago








  • 1




    $begingroup$
    You're supposed to calculate the residue at $z=0$, that's where the function has a pole and $z=0$ lies inside $mid zmid=1$.
    $endgroup$
    – Paras Khosla
    21 hours ago
















2












$begingroup$


evalulate the integral by using Cauchy's residue theorem



$$int_{|z|=1} frac{1}{e^z -1-2z}dz$$



MY attempt : $ f(z) =frac{1}{e^z -1-2z}$, now put $z= 1$ we get $f(z)=-1$



so $$int_{|z|=1} frac{1}{e^z -1-2z}dz= -2pi i$$



Is it correct?



Any hints/solution



thanks u






complex-analysis







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Residue is not as simple as "the value at $z=1$". For instance, $frac1z^2$ has zero residue.
    $endgroup$
    – Arthur
    21 hours ago








  • 1




    $begingroup$
    You're supposed to calculate the residue at $z=0$, that's where the function has a pole and $z=0$ lies inside $mid zmid=1$.
    $endgroup$
    – Paras Khosla
    21 hours ago














2












2








2





$begingroup$


evalulate the integral by using Cauchy's residue theorem



$$int_{|z|=1} frac{1}{e^z -1-2z}dz$$



MY attempt : $ f(z) =frac{1}{e^z -1-2z}$, now put $z= 1$ we get $f(z)=-1$



so $$int_{|z|=1} frac{1}{e^z -1-2z}dz= -2pi i$$



Is it correct?



Any hints/solution



thanks u






complex-analysis







share|cite|improve this question











$endgroup$




evalulate the integral by using Cauchy's residue theorem



$$int_{|z|=1} frac{1}{e^z -1-2z}dz$$



MY attempt : $ f(z) =frac{1}{e^z -1-2z}$, now put $z= 1$ we get $f(z)=-1$



so $$int_{|z|=1} frac{1}{e^z -1-2z}dz= -2pi i$$



Is it correct?



Any hints/solution



thanks u






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 20 hours ago









Bernard

121k740116




121k740116










asked 21 hours ago









jasminejasmine

1,781417




1,781417








  • 2




    $begingroup$
    Residue is not as simple as "the value at $z=1$". For instance, $frac1z^2$ has zero residue.
    $endgroup$
    – Arthur
    21 hours ago








  • 1




    $begingroup$
    You're supposed to calculate the residue at $z=0$, that's where the function has a pole and $z=0$ lies inside $mid zmid=1$.
    $endgroup$
    – Paras Khosla
    21 hours ago














  • 2




    $begingroup$
    Residue is not as simple as "the value at $z=1$". For instance, $frac1z^2$ has zero residue.
    $endgroup$
    – Arthur
    21 hours ago








  • 1




    $begingroup$
    You're supposed to calculate the residue at $z=0$, that's where the function has a pole and $z=0$ lies inside $mid zmid=1$.
    $endgroup$
    – Paras Khosla
    21 hours ago








2




2




$begingroup$
Residue is not as simple as "the value at $z=1$". For instance, $frac1z^2$ has zero residue.
$endgroup$
– Arthur
21 hours ago






$begingroup$
Residue is not as simple as "the value at $z=1$". For instance, $frac1z^2$ has zero residue.
$endgroup$
– Arthur
21 hours ago






1




1




$begingroup$
You're supposed to calculate the residue at $z=0$, that's where the function has a pole and $z=0$ lies inside $mid zmid=1$.
$endgroup$
– Paras Khosla
21 hours ago




$begingroup$
You're supposed to calculate the residue at $z=0$, that's where the function has a pole and $z=0$ lies inside $mid zmid=1$.
$endgroup$
– Paras Khosla
21 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

The function $f(z) =frac{1}{e^z -1-2z}$ has a pole of order $1$ in $z=0.$ Hence



$Res(f;0)= lim_{z to 0}z f(z)=-1.$ This gives



$int_{|z|=1} frac{1}{e^z -1-2z}dz= 2 pi i Res(f;0)=-2 pi i.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you know it has no other poles?
    $endgroup$
    – José Carlos Santos
    20 hours ago










  • $begingroup$
    @ Jose: do you agree that $e^z-1-2z ne 0$ if $|z|=1$ ? If yes, then there is $ delta >0$ such that $e^z-1-2z ne 0$ if $0< |z|<1+ delta$. Now aply the residue theorem in the region ${z: |z|<1+ delta}.$
    $endgroup$
    – Fred
    20 hours ago






  • 1




    $begingroup$
    I agree that $lvert zrvert=1implies e^z-1-2zneq0$, yes, but I don't see how is it that you deduce from that that $e^z-1-2zneq0$ in a region $D(0,1+delta)setminus{0}$, for some $delta>0$.
    $endgroup$
    – José Carlos Santos
    20 hours ago



















2












$begingroup$

Hint:



$$int_{mid zmid=1}dfrac{mathrm dz}{e^z-1-2z}=2pi icdot text{Res}_{z=0}$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    The function $f(z) =frac{1}{e^z -1-2z}$ has a pole of order $1$ in $z=0.$ Hence



    $Res(f;0)= lim_{z to 0}z f(z)=-1.$ This gives



    $int_{|z|=1} frac{1}{e^z -1-2z}dz= 2 pi i Res(f;0)=-2 pi i.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How do you know it has no other poles?
      $endgroup$
      – José Carlos Santos
      20 hours ago










    • $begingroup$
      @ Jose: do you agree that $e^z-1-2z ne 0$ if $|z|=1$ ? If yes, then there is $ delta >0$ such that $e^z-1-2z ne 0$ if $0< |z|<1+ delta$. Now aply the residue theorem in the region ${z: |z|<1+ delta}.$
      $endgroup$
      – Fred
      20 hours ago






    • 1




      $begingroup$
      I agree that $lvert zrvert=1implies e^z-1-2zneq0$, yes, but I don't see how is it that you deduce from that that $e^z-1-2zneq0$ in a region $D(0,1+delta)setminus{0}$, for some $delta>0$.
      $endgroup$
      – José Carlos Santos
      20 hours ago
















    4












    $begingroup$

    The function $f(z) =frac{1}{e^z -1-2z}$ has a pole of order $1$ in $z=0.$ Hence



    $Res(f;0)= lim_{z to 0}z f(z)=-1.$ This gives



    $int_{|z|=1} frac{1}{e^z -1-2z}dz= 2 pi i Res(f;0)=-2 pi i.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How do you know it has no other poles?
      $endgroup$
      – José Carlos Santos
      20 hours ago










    • $begingroup$
      @ Jose: do you agree that $e^z-1-2z ne 0$ if $|z|=1$ ? If yes, then there is $ delta >0$ such that $e^z-1-2z ne 0$ if $0< |z|<1+ delta$. Now aply the residue theorem in the region ${z: |z|<1+ delta}.$
      $endgroup$
      – Fred
      20 hours ago






    • 1




      $begingroup$
      I agree that $lvert zrvert=1implies e^z-1-2zneq0$, yes, but I don't see how is it that you deduce from that that $e^z-1-2zneq0$ in a region $D(0,1+delta)setminus{0}$, for some $delta>0$.
      $endgroup$
      – José Carlos Santos
      20 hours ago














    4












    4








    4





    $begingroup$

    The function $f(z) =frac{1}{e^z -1-2z}$ has a pole of order $1$ in $z=0.$ Hence



    $Res(f;0)= lim_{z to 0}z f(z)=-1.$ This gives



    $int_{|z|=1} frac{1}{e^z -1-2z}dz= 2 pi i Res(f;0)=-2 pi i.$






    share|cite|improve this answer









    $endgroup$



    The function $f(z) =frac{1}{e^z -1-2z}$ has a pole of order $1$ in $z=0.$ Hence



    $Res(f;0)= lim_{z to 0}z f(z)=-1.$ This gives



    $int_{|z|=1} frac{1}{e^z -1-2z}dz= 2 pi i Res(f;0)=-2 pi i.$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 21 hours ago









    FredFred

    46.9k1848




    46.9k1848












    • $begingroup$
      How do you know it has no other poles?
      $endgroup$
      – José Carlos Santos
      20 hours ago










    • $begingroup$
      @ Jose: do you agree that $e^z-1-2z ne 0$ if $|z|=1$ ? If yes, then there is $ delta >0$ such that $e^z-1-2z ne 0$ if $0< |z|<1+ delta$. Now aply the residue theorem in the region ${z: |z|<1+ delta}.$
      $endgroup$
      – Fred
      20 hours ago






    • 1




      $begingroup$
      I agree that $lvert zrvert=1implies e^z-1-2zneq0$, yes, but I don't see how is it that you deduce from that that $e^z-1-2zneq0$ in a region $D(0,1+delta)setminus{0}$, for some $delta>0$.
      $endgroup$
      – José Carlos Santos
      20 hours ago


















    • $begingroup$
      How do you know it has no other poles?
      $endgroup$
      – José Carlos Santos
      20 hours ago










    • $begingroup$
      @ Jose: do you agree that $e^z-1-2z ne 0$ if $|z|=1$ ? If yes, then there is $ delta >0$ such that $e^z-1-2z ne 0$ if $0< |z|<1+ delta$. Now aply the residue theorem in the region ${z: |z|<1+ delta}.$
      $endgroup$
      – Fred
      20 hours ago






    • 1




      $begingroup$
      I agree that $lvert zrvert=1implies e^z-1-2zneq0$, yes, but I don't see how is it that you deduce from that that $e^z-1-2zneq0$ in a region $D(0,1+delta)setminus{0}$, for some $delta>0$.
      $endgroup$
      – José Carlos Santos
      20 hours ago
















    $begingroup$
    How do you know it has no other poles?
    $endgroup$
    – José Carlos Santos
    20 hours ago




    $begingroup$
    How do you know it has no other poles?
    $endgroup$
    – José Carlos Santos
    20 hours ago












    $begingroup$
    @ Jose: do you agree that $e^z-1-2z ne 0$ if $|z|=1$ ? If yes, then there is $ delta >0$ such that $e^z-1-2z ne 0$ if $0< |z|<1+ delta$. Now aply the residue theorem in the region ${z: |z|<1+ delta}.$
    $endgroup$
    – Fred
    20 hours ago




    $begingroup$
    @ Jose: do you agree that $e^z-1-2z ne 0$ if $|z|=1$ ? If yes, then there is $ delta >0$ such that $e^z-1-2z ne 0$ if $0< |z|<1+ delta$. Now aply the residue theorem in the region ${z: |z|<1+ delta}.$
    $endgroup$
    – Fred
    20 hours ago




    1




    1




    $begingroup$
    I agree that $lvert zrvert=1implies e^z-1-2zneq0$, yes, but I don't see how is it that you deduce from that that $e^z-1-2zneq0$ in a region $D(0,1+delta)setminus{0}$, for some $delta>0$.
    $endgroup$
    – José Carlos Santos
    20 hours ago




    $begingroup$
    I agree that $lvert zrvert=1implies e^z-1-2zneq0$, yes, but I don't see how is it that you deduce from that that $e^z-1-2zneq0$ in a region $D(0,1+delta)setminus{0}$, for some $delta>0$.
    $endgroup$
    – José Carlos Santos
    20 hours ago











    2












    $begingroup$

    Hint:



    $$int_{mid zmid=1}dfrac{mathrm dz}{e^z-1-2z}=2pi icdot text{Res}_{z=0}$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint:



      $$int_{mid zmid=1}dfrac{mathrm dz}{e^z-1-2z}=2pi icdot text{Res}_{z=0}$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint:



        $$int_{mid zmid=1}dfrac{mathrm dz}{e^z-1-2z}=2pi icdot text{Res}_{z=0}$$






        share|cite|improve this answer









        $endgroup$



        Hint:



        $$int_{mid zmid=1}dfrac{mathrm dz}{e^z-1-2z}=2pi icdot text{Res}_{z=0}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 21 hours ago









        Paras KhoslaParas Khosla

        679213




        679213






























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