Is $int frace^z^2z^3 dz=pi i$? I got $2 pi i$.

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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka



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The answer key says $pi i$. By Cauchy's Integral Formula 5.1 for $f''(w)$, I think the answer is $2 pi i$:



$$int_square frace^z^2z^3 dz = pi i fracd^2dz^2 e^z^2|_z=0 $$ = $ pi i (2) = 2 pi i$.



I also tried on Wolfram Alpha 1 2 for circles which I believe are homotopic to the square. I think I get the same answer with Residue Thm 9.10 and Prop 9.11:



$$int_square frace^z^2z^3 dz = 2 pi i frac12! lim_z to 0 fracd^2dz^2 (z-0)^3 frace^z^2z^3$$



$$ = pi i (2)$$



What am I doing wrong?




complex-analysis proof-verification contour-integration residue-calculus complex-integration




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  • 3




    $2pi i$ is the correct answer.
    – Kavi Rama Murthy
    Jul 23 at 8:46










  • @KaviRamaMurthy Thanks! ^-^ Post as answer?
    – BCLC
    Jul 23 at 8:48














up vote
0
down vote

favorite
1












A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka



enter image description here



enter image description here



The answer key says $pi i$. By Cauchy's Integral Formula 5.1 for $f''(w)$, I think the answer is $2 pi i$:



$$int_square frace^z^2z^3 dz = pi i fracd^2dz^2 e^z^2|_z=0 $$ = $ pi i (2) = 2 pi i$.



I also tried on Wolfram Alpha 1 2 for circles which I believe are homotopic to the square. I think I get the same answer with Residue Thm 9.10 and Prop 9.11:



$$int_square frace^z^2z^3 dz = 2 pi i frac12! lim_z to 0 fracd^2dz^2 (z-0)^3 frace^z^2z^3$$



$$ = pi i (2)$$



What am I doing wrong?




complex-analysis proof-verification contour-integration residue-calculus complex-integration




share|cite|improve this question

















  • 3




    $2pi i$ is the correct answer.
    – Kavi Rama Murthy
    Jul 23 at 8:46










  • @KaviRamaMurthy Thanks! ^-^ Post as answer?
    – BCLC
    Jul 23 at 8:48












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka



enter image description here



enter image description here



The answer key says $pi i$. By Cauchy's Integral Formula 5.1 for $f''(w)$, I think the answer is $2 pi i$:



$$int_square frace^z^2z^3 dz = pi i fracd^2dz^2 e^z^2|_z=0 $$ = $ pi i (2) = 2 pi i$.



I also tried on Wolfram Alpha 1 2 for circles which I believe are homotopic to the square. I think I get the same answer with Residue Thm 9.10 and Prop 9.11:



$$int_square frace^z^2z^3 dz = 2 pi i frac12! lim_z to 0 fracd^2dz^2 (z-0)^3 frace^z^2z^3$$



$$ = pi i (2)$$



What am I doing wrong?




complex-analysis proof-verification contour-integration residue-calculus complex-integration




share|cite|improve this question













A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka



enter image description here



enter image description here



The answer key says $pi i$. By Cauchy's Integral Formula 5.1 for $f''(w)$, I think the answer is $2 pi i$:



$$int_square frace^z^2z^3 dz = pi i fracd^2dz^2 e^z^2|_z=0 $$ = $ pi i (2) = 2 pi i$.



I also tried on Wolfram Alpha 1 2 for circles which I believe are homotopic to the square. I think I get the same answer with Residue Thm 9.10 and Prop 9.11:



$$int_square frace^z^2z^3 dz = 2 pi i frac12! lim_z to 0 fracd^2dz^2 (z-0)^3 frace^z^2z^3$$



$$ = pi i (2)$$



What am I doing wrong?





complex-analysis proof-verification contour-integration residue-calculus complex-integration





share|cite|improve this question












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asked Jul 23 at 8:41









BCLC

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  • 3




    $2pi i$ is the correct answer.
    – Kavi Rama Murthy
    Jul 23 at 8:46










  • @KaviRamaMurthy Thanks! ^-^ Post as answer?
    – BCLC
    Jul 23 at 8:48












  • 3




    $2pi i$ is the correct answer.
    – Kavi Rama Murthy
    Jul 23 at 8:46










  • @KaviRamaMurthy Thanks! ^-^ Post as answer?
    – BCLC
    Jul 23 at 8:48







3




3




$2pi i$ is the correct answer.
– Kavi Rama Murthy
Jul 23 at 8:46




$2pi i$ is the correct answer.
– Kavi Rama Murthy
Jul 23 at 8:46












@KaviRamaMurthy Thanks! ^-^ Post as answer?
– BCLC
Jul 23 at 8:48




@KaviRamaMurthy Thanks! ^-^ Post as answer?
– BCLC
Jul 23 at 8:48










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The correct answer is $2pi i$. I will suggest another method: expand the exponential in a power series and look at the coefficient of $frac 1 z$. This shows that the residue at $0$ is $1$ so the integral is $2pi i$.






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    up vote
    2
    down vote



    accepted










    The correct answer is $2pi i$. I will suggest another method: expand the exponential in a power series and look at the coefficient of $frac 1 z$. This shows that the residue at $0$ is $1$ so the integral is $2pi i$.






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      The correct answer is $2pi i$. I will suggest another method: expand the exponential in a power series and look at the coefficient of $frac 1 z$. This shows that the residue at $0$ is $1$ so the integral is $2pi i$.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        The correct answer is $2pi i$. I will suggest another method: expand the exponential in a power series and look at the coefficient of $frac 1 z$. This shows that the residue at $0$ is $1$ so the integral is $2pi i$.






        share|cite|improve this answer













        The correct answer is $2pi i$. I will suggest another method: expand the exponential in a power series and look at the coefficient of $frac 1 z$. This shows that the residue at $0$ is $1$ so the integral is $2pi i$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 23 at 8:51









        Kavi Rama Murthy

        20.4k2830




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