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I have a homework problem to evaluate the integral $$ \oint_{\gamma}\frac{\cos z}{(z+i)^3}dz $$ along the curve $\gamma(t)=-i+e^{it}, t\in[0,2\pi]$. I proceeded to plug the given information into the definition of a contour integral and got to the expression $$ \oint_{\gamma}=i\int_{0}^{2\pi}\cos(-i+e^{it})e^{-2it}dt $$ which seems hardly helpful. I don't know how to evaluate this or manipulate it any further and I suppose there is some trick earlier on to make the integration more manageable. I just can't figure it out so I'd be grateful for any help.

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    $\begingroup$ This is a typical exercise about Cauchy's integral formula. $\endgroup$ – José Carlos Santos Sep 28 '18 at 14:12
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With rediue theorem $$\oint_{\gamma}\frac{\cos z}{(z+i)^3}dz=\dfrac{2\pi i}{2!}\lim_{z\to-i}\dfrac{d^2}{dz^2}\cos z=-\dfrac{2\pi i}{2!}\cos i=\color{blue}{-\dfrac{\pi i}{2}\left(e+\dfrac1e\right)}$$

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Hint: As Jose hints in the comments, we want to use the fact that $$f^{(k)}(z_0)=\frac{k!}{2\pi i}\oint_{\partial B_r(z_0)}\frac{f(z)}{(z-z_0)^{k+1}}dz$$ with the correct holomorphic function $f(z)$

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