Missing $i$ while evaluating $int_{-infty}^{infty}frac{e^{iz}}{(z^2+2z+2)^2}$ using residue theorem












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$begingroup$


Okay, first I'm a bit ashamed to ask because I already asked a question yesterday about a similar question (it's from far not the same integral though), but I'm missing an $i$ somewhere in the process, that's why I'm asking haha.



I'm asked to evaluate the following integral : $$int_{-infty}^{infty}frac{cos(z)}{(z^2+2z+2)^2}dz$$ which is the real part of



$$int_{-infty}^{infty}frac{e^{iz}}{(z^2+2z+2)^2}dz$$ which is much simpler to evaluate obviously.



The poles are at $z=-1pm i$. I evaluate the integral in the upper half circle, so only $-1+i$ is in our domain. When I calculate the derivative (pole of order 2), I get $$frac{1}{1!}lim_{zto-1+i}frac{d}{dz}(z-(-1+i))^2frac{e^{iz}}{(z^2+2z+2)^2} =lim_{zto-1+i}(z-(-1+i))^2 frac{ie^{iz}(z^2+2z+2)-e^{iz}2(2z+2)}{(z^2+2z+2)^3} = lim_{zto-1+i}(z-(-1+i))^2 frac{-e^{iz}2(2z+2)}{(z^2+2z+2)^3}=frac{-4ie^{i(-1+i)}}{8i^3}=frac{e^{-1-i}}{2}$$ But an $i$ seems to be missing given that when I multiply by $2pi i$, I get $pi ie^{-1-i}$ while WolframAlpha gets $pi e^{-1-i}$. A real (and not imaginary) result would also make more sense given that we then have to take the real part of it to get the original integral with $cos(x)$.



As yesterday, I checked my work several times and couldn't find the mistake, that's why I'm asking haha.



Thanks for your help !



Edit : I forgot to include $(z-(-1+i))^2$ in the process. This is actually a typo, not a mistake.






integration complex-analysis improper-integrals residue-calculus







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    0












    $begingroup$


    Okay, first I'm a bit ashamed to ask because I already asked a question yesterday about a similar question (it's from far not the same integral though), but I'm missing an $i$ somewhere in the process, that's why I'm asking haha.



    I'm asked to evaluate the following integral : $$int_{-infty}^{infty}frac{cos(z)}{(z^2+2z+2)^2}dz$$ which is the real part of



    $$int_{-infty}^{infty}frac{e^{iz}}{(z^2+2z+2)^2}dz$$ which is much simpler to evaluate obviously.



    The poles are at $z=-1pm i$. I evaluate the integral in the upper half circle, so only $-1+i$ is in our domain. When I calculate the derivative (pole of order 2), I get $$frac{1}{1!}lim_{zto-1+i}frac{d}{dz}(z-(-1+i))^2frac{e^{iz}}{(z^2+2z+2)^2} =lim_{zto-1+i}(z-(-1+i))^2 frac{ie^{iz}(z^2+2z+2)-e^{iz}2(2z+2)}{(z^2+2z+2)^3} = lim_{zto-1+i}(z-(-1+i))^2 frac{-e^{iz}2(2z+2)}{(z^2+2z+2)^3}=frac{-4ie^{i(-1+i)}}{8i^3}=frac{e^{-1-i}}{2}$$ But an $i$ seems to be missing given that when I multiply by $2pi i$, I get $pi ie^{-1-i}$ while WolframAlpha gets $pi e^{-1-i}$. A real (and not imaginary) result would also make more sense given that we then have to take the real part of it to get the original integral with $cos(x)$.



    As yesterday, I checked my work several times and couldn't find the mistake, that's why I'm asking haha.



    Thanks for your help !



    Edit : I forgot to include $(z-(-1+i))^2$ in the process. This is actually a typo, not a mistake.






    integration complex-analysis improper-integrals residue-calculus







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Okay, first I'm a bit ashamed to ask because I already asked a question yesterday about a similar question (it's from far not the same integral though), but I'm missing an $i$ somewhere in the process, that's why I'm asking haha.



      I'm asked to evaluate the following integral : $$int_{-infty}^{infty}frac{cos(z)}{(z^2+2z+2)^2}dz$$ which is the real part of



      $$int_{-infty}^{infty}frac{e^{iz}}{(z^2+2z+2)^2}dz$$ which is much simpler to evaluate obviously.



      The poles are at $z=-1pm i$. I evaluate the integral in the upper half circle, so only $-1+i$ is in our domain. When I calculate the derivative (pole of order 2), I get $$frac{1}{1!}lim_{zto-1+i}frac{d}{dz}(z-(-1+i))^2frac{e^{iz}}{(z^2+2z+2)^2} =lim_{zto-1+i}(z-(-1+i))^2 frac{ie^{iz}(z^2+2z+2)-e^{iz}2(2z+2)}{(z^2+2z+2)^3} = lim_{zto-1+i}(z-(-1+i))^2 frac{-e^{iz}2(2z+2)}{(z^2+2z+2)^3}=frac{-4ie^{i(-1+i)}}{8i^3}=frac{e^{-1-i}}{2}$$ But an $i$ seems to be missing given that when I multiply by $2pi i$, I get $pi ie^{-1-i}$ while WolframAlpha gets $pi e^{-1-i}$. A real (and not imaginary) result would also make more sense given that we then have to take the real part of it to get the original integral with $cos(x)$.



      As yesterday, I checked my work several times and couldn't find the mistake, that's why I'm asking haha.



      Thanks for your help !



      Edit : I forgot to include $(z-(-1+i))^2$ in the process. This is actually a typo, not a mistake.






      integration complex-analysis improper-integrals residue-calculus







      share|cite|improve this question











      $endgroup$




      Okay, first I'm a bit ashamed to ask because I already asked a question yesterday about a similar question (it's from far not the same integral though), but I'm missing an $i$ somewhere in the process, that's why I'm asking haha.



      I'm asked to evaluate the following integral : $$int_{-infty}^{infty}frac{cos(z)}{(z^2+2z+2)^2}dz$$ which is the real part of



      $$int_{-infty}^{infty}frac{e^{iz}}{(z^2+2z+2)^2}dz$$ which is much simpler to evaluate obviously.



      The poles are at $z=-1pm i$. I evaluate the integral in the upper half circle, so only $-1+i$ is in our domain. When I calculate the derivative (pole of order 2), I get $$frac{1}{1!}lim_{zto-1+i}frac{d}{dz}(z-(-1+i))^2frac{e^{iz}}{(z^2+2z+2)^2} =lim_{zto-1+i}(z-(-1+i))^2 frac{ie^{iz}(z^2+2z+2)-e^{iz}2(2z+2)}{(z^2+2z+2)^3} = lim_{zto-1+i}(z-(-1+i))^2 frac{-e^{iz}2(2z+2)}{(z^2+2z+2)^3}=frac{-4ie^{i(-1+i)}}{8i^3}=frac{e^{-1-i}}{2}$$ But an $i$ seems to be missing given that when I multiply by $2pi i$, I get $pi ie^{-1-i}$ while WolframAlpha gets $pi e^{-1-i}$. A real (and not imaginary) result would also make more sense given that we then have to take the real part of it to get the original integral with $cos(x)$.



      As yesterday, I checked my work several times and couldn't find the mistake, that's why I'm asking haha.



      Thanks for your help !



      Edit : I forgot to include $(z-(-1+i))^2$ in the process. This is actually a typo, not a mistake.






      integration complex-analysis improper-integrals residue-calculus




      integration complex-analysis improper-integrals residue-calculus






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 12 '18 at 18:17









      José Carlos Santos

      163k22131234




      163k22131234










      asked Dec 12 '18 at 17:34









      PoujhPoujh

      616516




      616516






















          2 Answers
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          1












          $begingroup$

          I don't understand your method of computing that residue. Since$$frac{e^{iz}}{(z^2+2z+2)^2}=frac{e^{iz}}{(z+1+i)^2(z+1-i)^2}=frac{frac{e^{iz}}{(z+1+i)^2}}{(z+1-i)^2}$$and since the Taylor series of $dfrac{e^{iz}}{(z+1+i)^2}$ about $-1+i$ is$$-frac{e^{-1-i}}4-frac{e^{-1-i}i}2(z+1-i)+cdots,$$the residue that you're after is $-dfrac{e^{-1-i}i}2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Well, I used the residue formula for pole of order n, like here en.wikipedia.org/wiki/… $frac{1}{(n-1)!}lim_{z to z_0}frac{d^{n-1}}{dz}((z-z_0)^n f(z))$
            $endgroup$
            – Poujh
            Dec 12 '18 at 17:49








          • 1




            $begingroup$
            I see. But then you should have obtained $g'(-1+i)$, with $g(z)=dfrac{e^{iz}}{(z+1+i)^2}$. Since $g'(z)=dfrac{ie^{iz}(z+1+3i)}{(z+1+i)^3}$, you should have got $g'(-1+i)=-dfrac{ie^{-1-i}}2$, like me.
            $endgroup$
            – José Carlos Santos
            Dec 12 '18 at 18:01










          • $begingroup$
            Okay, I think I first took the derivative and then multiplied by $(z-z_0)^2$ Which is obviously wrong. I don't see how I didn't noticed it in the two hours I was stuck. Thanks a lot !
            $endgroup$
            – Poujh
            Dec 12 '18 at 18:14



















          1












          $begingroup$

          $$begin{align}mathrm{Res}left(frac{e^{iz}}{(z^2+2z+1)^2},-1+iright) &=frac 1{1!}lim_{zto -1+i}frac d{dz}left(color{red}{(z-(-1+i))^2}frac{e^{iz}} {(z^2+2z+1)^2}right)\ &=lim_{zto -1+i}frac d{dz}left(frac{e^{iz}}{(z+1+i)^2}right)\ &=lim_{zto -1+i}frac{ie^{iz}(z+1+i)^2-2e^{iz}(z+1+i)}{(z+1+i)^4}\&=lim_{zto -1+i}frac{ie^{iz}(z+1+3i)}{(z+1+i)^3} \&=frac{-ie^{-1-i}}2end{align}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Uh sorry this is actually a typo not a mistake I made but thanks for pointing it out. When we evaluate it we also get $8i^3$ in the denominator which doesn't get back that extra $i$.
            $endgroup$
            – Poujh
            Dec 12 '18 at 17:54








          • 1




            $begingroup$
            I've modified my comment.
            $endgroup$
            – Philippe Malot
            Dec 12 '18 at 18:09










          • $begingroup$
            Thank you, somehow I did the same thing wrong (deriving then multiplying by $(z-z_0)^2$) for two hours withour noticing it. It seems obvious now. Thanks !
            $endgroup$
            – Poujh
            Dec 12 '18 at 18:16













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          2 Answers
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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          I don't understand your method of computing that residue. Since$$frac{e^{iz}}{(z^2+2z+2)^2}=frac{e^{iz}}{(z+1+i)^2(z+1-i)^2}=frac{frac{e^{iz}}{(z+1+i)^2}}{(z+1-i)^2}$$and since the Taylor series of $dfrac{e^{iz}}{(z+1+i)^2}$ about $-1+i$ is$$-frac{e^{-1-i}}4-frac{e^{-1-i}i}2(z+1-i)+cdots,$$the residue that you're after is $-dfrac{e^{-1-i}i}2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Well, I used the residue formula for pole of order n, like here en.wikipedia.org/wiki/… $frac{1}{(n-1)!}lim_{z to z_0}frac{d^{n-1}}{dz}((z-z_0)^n f(z))$
            $endgroup$
            – Poujh
            Dec 12 '18 at 17:49








          • 1




            $begingroup$
            I see. But then you should have obtained $g'(-1+i)$, with $g(z)=dfrac{e^{iz}}{(z+1+i)^2}$. Since $g'(z)=dfrac{ie^{iz}(z+1+3i)}{(z+1+i)^3}$, you should have got $g'(-1+i)=-dfrac{ie^{-1-i}}2$, like me.
            $endgroup$
            – José Carlos Santos
            Dec 12 '18 at 18:01










          • $begingroup$
            Okay, I think I first took the derivative and then multiplied by $(z-z_0)^2$ Which is obviously wrong. I don't see how I didn't noticed it in the two hours I was stuck. Thanks a lot !
            $endgroup$
            – Poujh
            Dec 12 '18 at 18:14
















          1












          $begingroup$

          I don't understand your method of computing that residue. Since$$frac{e^{iz}}{(z^2+2z+2)^2}=frac{e^{iz}}{(z+1+i)^2(z+1-i)^2}=frac{frac{e^{iz}}{(z+1+i)^2}}{(z+1-i)^2}$$and since the Taylor series of $dfrac{e^{iz}}{(z+1+i)^2}$ about $-1+i$ is$$-frac{e^{-1-i}}4-frac{e^{-1-i}i}2(z+1-i)+cdots,$$the residue that you're after is $-dfrac{e^{-1-i}i}2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Well, I used the residue formula for pole of order n, like here en.wikipedia.org/wiki/… $frac{1}{(n-1)!}lim_{z to z_0}frac{d^{n-1}}{dz}((z-z_0)^n f(z))$
            $endgroup$
            – Poujh
            Dec 12 '18 at 17:49








          • 1




            $begingroup$
            I see. But then you should have obtained $g'(-1+i)$, with $g(z)=dfrac{e^{iz}}{(z+1+i)^2}$. Since $g'(z)=dfrac{ie^{iz}(z+1+3i)}{(z+1+i)^3}$, you should have got $g'(-1+i)=-dfrac{ie^{-1-i}}2$, like me.
            $endgroup$
            – José Carlos Santos
            Dec 12 '18 at 18:01










          • $begingroup$
            Okay, I think I first took the derivative and then multiplied by $(z-z_0)^2$ Which is obviously wrong. I don't see how I didn't noticed it in the two hours I was stuck. Thanks a lot !
            $endgroup$
            – Poujh
            Dec 12 '18 at 18:14














          1












          1








          1





          $begingroup$

          I don't understand your method of computing that residue. Since$$frac{e^{iz}}{(z^2+2z+2)^2}=frac{e^{iz}}{(z+1+i)^2(z+1-i)^2}=frac{frac{e^{iz}}{(z+1+i)^2}}{(z+1-i)^2}$$and since the Taylor series of $dfrac{e^{iz}}{(z+1+i)^2}$ about $-1+i$ is$$-frac{e^{-1-i}}4-frac{e^{-1-i}i}2(z+1-i)+cdots,$$the residue that you're after is $-dfrac{e^{-1-i}i}2$.






          share|cite|improve this answer









          $endgroup$



          I don't understand your method of computing that residue. Since$$frac{e^{iz}}{(z^2+2z+2)^2}=frac{e^{iz}}{(z+1+i)^2(z+1-i)^2}=frac{frac{e^{iz}}{(z+1+i)^2}}{(z+1-i)^2}$$and since the Taylor series of $dfrac{e^{iz}}{(z+1+i)^2}$ about $-1+i$ is$$-frac{e^{-1-i}}4-frac{e^{-1-i}i}2(z+1-i)+cdots,$$the residue that you're after is $-dfrac{e^{-1-i}i}2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 17:46









          José Carlos SantosJosé Carlos Santos

          163k22131234




          163k22131234












          • $begingroup$
            Well, I used the residue formula for pole of order n, like here en.wikipedia.org/wiki/… $frac{1}{(n-1)!}lim_{z to z_0}frac{d^{n-1}}{dz}((z-z_0)^n f(z))$
            $endgroup$
            – Poujh
            Dec 12 '18 at 17:49








          • 1




            $begingroup$
            I see. But then you should have obtained $g'(-1+i)$, with $g(z)=dfrac{e^{iz}}{(z+1+i)^2}$. Since $g'(z)=dfrac{ie^{iz}(z+1+3i)}{(z+1+i)^3}$, you should have got $g'(-1+i)=-dfrac{ie^{-1-i}}2$, like me.
            $endgroup$
            – José Carlos Santos
            Dec 12 '18 at 18:01










          • $begingroup$
            Okay, I think I first took the derivative and then multiplied by $(z-z_0)^2$ Which is obviously wrong. I don't see how I didn't noticed it in the two hours I was stuck. Thanks a lot !
            $endgroup$
            – Poujh
            Dec 12 '18 at 18:14


















          • $begingroup$
            Well, I used the residue formula for pole of order n, like here en.wikipedia.org/wiki/… $frac{1}{(n-1)!}lim_{z to z_0}frac{d^{n-1}}{dz}((z-z_0)^n f(z))$
            $endgroup$
            – Poujh
            Dec 12 '18 at 17:49








          • 1




            $begingroup$
            I see. But then you should have obtained $g'(-1+i)$, with $g(z)=dfrac{e^{iz}}{(z+1+i)^2}$. Since $g'(z)=dfrac{ie^{iz}(z+1+3i)}{(z+1+i)^3}$, you should have got $g'(-1+i)=-dfrac{ie^{-1-i}}2$, like me.
            $endgroup$
            – José Carlos Santos
            Dec 12 '18 at 18:01










          • $begingroup$
            Okay, I think I first took the derivative and then multiplied by $(z-z_0)^2$ Which is obviously wrong. I don't see how I didn't noticed it in the two hours I was stuck. Thanks a lot !
            $endgroup$
            – Poujh
            Dec 12 '18 at 18:14
















          $begingroup$
          Well, I used the residue formula for pole of order n, like here en.wikipedia.org/wiki/… $frac{1}{(n-1)!}lim_{z to z_0}frac{d^{n-1}}{dz}((z-z_0)^n f(z))$
          $endgroup$
          – Poujh
          Dec 12 '18 at 17:49






          $begingroup$
          Well, I used the residue formula for pole of order n, like here en.wikipedia.org/wiki/… $frac{1}{(n-1)!}lim_{z to z_0}frac{d^{n-1}}{dz}((z-z_0)^n f(z))$
          $endgroup$
          – Poujh
          Dec 12 '18 at 17:49






          1




          1




          $begingroup$
          I see. But then you should have obtained $g'(-1+i)$, with $g(z)=dfrac{e^{iz}}{(z+1+i)^2}$. Since $g'(z)=dfrac{ie^{iz}(z+1+3i)}{(z+1+i)^3}$, you should have got $g'(-1+i)=-dfrac{ie^{-1-i}}2$, like me.
          $endgroup$
          – José Carlos Santos
          Dec 12 '18 at 18:01




          $begingroup$
          I see. But then you should have obtained $g'(-1+i)$, with $g(z)=dfrac{e^{iz}}{(z+1+i)^2}$. Since $g'(z)=dfrac{ie^{iz}(z+1+3i)}{(z+1+i)^3}$, you should have got $g'(-1+i)=-dfrac{ie^{-1-i}}2$, like me.
          $endgroup$
          – José Carlos Santos
          Dec 12 '18 at 18:01












          $begingroup$
          Okay, I think I first took the derivative and then multiplied by $(z-z_0)^2$ Which is obviously wrong. I don't see how I didn't noticed it in the two hours I was stuck. Thanks a lot !
          $endgroup$
          – Poujh
          Dec 12 '18 at 18:14




          $begingroup$
          Okay, I think I first took the derivative and then multiplied by $(z-z_0)^2$ Which is obviously wrong. I don't see how I didn't noticed it in the two hours I was stuck. Thanks a lot !
          $endgroup$
          – Poujh
          Dec 12 '18 at 18:14











          1












          $begingroup$

          $$begin{align}mathrm{Res}left(frac{e^{iz}}{(z^2+2z+1)^2},-1+iright) &=frac 1{1!}lim_{zto -1+i}frac d{dz}left(color{red}{(z-(-1+i))^2}frac{e^{iz}} {(z^2+2z+1)^2}right)\ &=lim_{zto -1+i}frac d{dz}left(frac{e^{iz}}{(z+1+i)^2}right)\ &=lim_{zto -1+i}frac{ie^{iz}(z+1+i)^2-2e^{iz}(z+1+i)}{(z+1+i)^4}\&=lim_{zto -1+i}frac{ie^{iz}(z+1+3i)}{(z+1+i)^3} \&=frac{-ie^{-1-i}}2end{align}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Uh sorry this is actually a typo not a mistake I made but thanks for pointing it out. When we evaluate it we also get $8i^3$ in the denominator which doesn't get back that extra $i$.
            $endgroup$
            – Poujh
            Dec 12 '18 at 17:54








          • 1




            $begingroup$
            I've modified my comment.
            $endgroup$
            – Philippe Malot
            Dec 12 '18 at 18:09










          • $begingroup$
            Thank you, somehow I did the same thing wrong (deriving then multiplying by $(z-z_0)^2$) for two hours withour noticing it. It seems obvious now. Thanks !
            $endgroup$
            – Poujh
            Dec 12 '18 at 18:16


















          1












          $begingroup$

          $$begin{align}mathrm{Res}left(frac{e^{iz}}{(z^2+2z+1)^2},-1+iright) &=frac 1{1!}lim_{zto -1+i}frac d{dz}left(color{red}{(z-(-1+i))^2}frac{e^{iz}} {(z^2+2z+1)^2}right)\ &=lim_{zto -1+i}frac d{dz}left(frac{e^{iz}}{(z+1+i)^2}right)\ &=lim_{zto -1+i}frac{ie^{iz}(z+1+i)^2-2e^{iz}(z+1+i)}{(z+1+i)^4}\&=lim_{zto -1+i}frac{ie^{iz}(z+1+3i)}{(z+1+i)^3} \&=frac{-ie^{-1-i}}2end{align}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Uh sorry this is actually a typo not a mistake I made but thanks for pointing it out. When we evaluate it we also get $8i^3$ in the denominator which doesn't get back that extra $i$.
            $endgroup$
            – Poujh
            Dec 12 '18 at 17:54








          • 1




            $begingroup$
            I've modified my comment.
            $endgroup$
            – Philippe Malot
            Dec 12 '18 at 18:09










          • $begingroup$
            Thank you, somehow I did the same thing wrong (deriving then multiplying by $(z-z_0)^2$) for two hours withour noticing it. It seems obvious now. Thanks !
            $endgroup$
            – Poujh
            Dec 12 '18 at 18:16
















          1












          1








          1





          $begingroup$

          $$begin{align}mathrm{Res}left(frac{e^{iz}}{(z^2+2z+1)^2},-1+iright) &=frac 1{1!}lim_{zto -1+i}frac d{dz}left(color{red}{(z-(-1+i))^2}frac{e^{iz}} {(z^2+2z+1)^2}right)\ &=lim_{zto -1+i}frac d{dz}left(frac{e^{iz}}{(z+1+i)^2}right)\ &=lim_{zto -1+i}frac{ie^{iz}(z+1+i)^2-2e^{iz}(z+1+i)}{(z+1+i)^4}\&=lim_{zto -1+i}frac{ie^{iz}(z+1+3i)}{(z+1+i)^3} \&=frac{-ie^{-1-i}}2end{align}$$






          share|cite|improve this answer











          $endgroup$



          $$begin{align}mathrm{Res}left(frac{e^{iz}}{(z^2+2z+1)^2},-1+iright) &=frac 1{1!}lim_{zto -1+i}frac d{dz}left(color{red}{(z-(-1+i))^2}frac{e^{iz}} {(z^2+2z+1)^2}right)\ &=lim_{zto -1+i}frac d{dz}left(frac{e^{iz}}{(z+1+i)^2}right)\ &=lim_{zto -1+i}frac{ie^{iz}(z+1+i)^2-2e^{iz}(z+1+i)}{(z+1+i)^4}\&=lim_{zto -1+i}frac{ie^{iz}(z+1+3i)}{(z+1+i)^3} \&=frac{-ie^{-1-i}}2end{align}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 18:12

























          answered Dec 12 '18 at 17:52









          Philippe MalotPhilippe Malot

          2,276824




          2,276824












          • $begingroup$
            Uh sorry this is actually a typo not a mistake I made but thanks for pointing it out. When we evaluate it we also get $8i^3$ in the denominator which doesn't get back that extra $i$.
            $endgroup$
            – Poujh
            Dec 12 '18 at 17:54








          • 1




            $begingroup$
            I've modified my comment.
            $endgroup$
            – Philippe Malot
            Dec 12 '18 at 18:09










          • $begingroup$
            Thank you, somehow I did the same thing wrong (deriving then multiplying by $(z-z_0)^2$) for two hours withour noticing it. It seems obvious now. Thanks !
            $endgroup$
            – Poujh
            Dec 12 '18 at 18:16




















          • $begingroup$
            Uh sorry this is actually a typo not a mistake I made but thanks for pointing it out. When we evaluate it we also get $8i^3$ in the denominator which doesn't get back that extra $i$.
            $endgroup$
            – Poujh
            Dec 12 '18 at 17:54








          • 1




            $begingroup$
            I've modified my comment.
            $endgroup$
            – Philippe Malot
            Dec 12 '18 at 18:09










          • $begingroup$
            Thank you, somehow I did the same thing wrong (deriving then multiplying by $(z-z_0)^2$) for two hours withour noticing it. It seems obvious now. Thanks !
            $endgroup$
            – Poujh
            Dec 12 '18 at 18:16


















          $begingroup$
          Uh sorry this is actually a typo not a mistake I made but thanks for pointing it out. When we evaluate it we also get $8i^3$ in the denominator which doesn't get back that extra $i$.
          $endgroup$
          – Poujh
          Dec 12 '18 at 17:54






          $begingroup$
          Uh sorry this is actually a typo not a mistake I made but thanks for pointing it out. When we evaluate it we also get $8i^3$ in the denominator which doesn't get back that extra $i$.
          $endgroup$
          – Poujh
          Dec 12 '18 at 17:54






          1




          1




          $begingroup$
          I've modified my comment.
          $endgroup$
          – Philippe Malot
          Dec 12 '18 at 18:09




          $begingroup$
          I've modified my comment.
          $endgroup$
          – Philippe Malot
          Dec 12 '18 at 18:09












          $begingroup$
          Thank you, somehow I did the same thing wrong (deriving then multiplying by $(z-z_0)^2$) for two hours withour noticing it. It seems obvious now. Thanks !
          $endgroup$
          – Poujh
          Dec 12 '18 at 18:16






          $begingroup$
          Thank you, somehow I did the same thing wrong (deriving then multiplying by $(z-z_0)^2$) for two hours withour noticing it. It seems obvious now. Thanks !
          $endgroup$
          – Poujh
          Dec 12 '18 at 18:16




















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