Compute the integral $$\int\limits_{\Gamma} \frac{1}{(z-2\sqrt{2}i)(z-\sqrt{2}i)^3}dz$$ with $\Gamma=2e^{-it}+2\left(i+\frac{1}{2}e^{it}\right)+3\left(-i+\frac{1}{2}e^{it}\right)$. Let $G=\mathbb{C}\setminus\{i,-i\}$

I'm not sure if one can use the general Cauchy integral theorem, because the cycle $n(\Gamma, z)\neq 0$? In a former task I show, that for this paths the coefficients ($\Gamma=k\alpha+l\beta_1+m\beta_2)$ can be choose by $$(k+l)-(k+m)=0$$ so that the $n(\Gamma, z)=0$. Any hints? Thank you!


The singularities of your function are $\sqrt{2}i$ and $2\sqrt{2}i$.

Now simplify your curve using $e^{it}=\cos(t)+\sin(t)i$ and $2\cos(t)=e^{it}+e^{-it}$ \begin{align} \Gamma&=2e^{-it}+2\left(i+\frac12e^{it}\right)+3\left(-i+\frac12e^{it}\right)\\ &=-i+\frac52e^{it}+2e^{-it}\\ &=-i+\frac12e^{it}+2(e^{it}+e^{-it})\\ &=-i+\frac12e^{it}+4\cos(t)\\ &=-i+\frac12\left(\cos(t)+\sin(t)i\right)+4\cos(t)\\ &=\frac92\cos(t)+\left(-1+\frac12\sin(t)\right)i. \end{align} This implies $\Im(\Gamma)=-1+\frac12\sin(t)\in\left[-\frac{3}2,-\frac12\right]$. Since $\Im(\sqrt{2}i)=\sqrt{2}>-\frac12$ and $\Im(2\sqrt{2}i)=2\sqrt{2}>-\frac12$ the curve $\Gamma$ doesn't run around the singularities and the Cauchy integral theorem yields that your integral has to be $0$.

  • $\begingroup$ Thank you, Mundron. How you get $\Im(\Gamma)=-1+\frac{1}{2}\sin(t)$ and $-i+\frac{1}{2}e^{it}+4\cos(t)$? $\endgroup$ – jacmeird Jul 14 '17 at 7:53
  • 1
    $\begingroup$ Using $e^{it}=\cos(t)+\sin(t)i$ and $2\cos(t)=e^{it}+e^{-it}$ yields $\Gamma=-i+\frac12\cos(t)+\frac12\sin(t)i+4\cos(t)=\frac92\cos(t)+\left(-1+\frac12\sin(t)\right)i$ $\endgroup$ – Mundron Schmidt Jul 14 '17 at 8:01

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