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I can not find a way to justify rl as the singularities $z = -1 + i$ and $z = 1 + i$ are inside the semicircle.

Can you help me please?


Consider $~\int\limits_C\frac{z^3e^{iaz}}{z^4+4}\,\mathrm dx=\int\limits_Cf(z)\,\mathrm dz~$where $C$ is the contour consisting of the semi-circle

$C_R$ of the radius R together with the part of the real axis from $-R$ to $R$


To find the poles $\begin{align} \text{Let }z^4+4=0\\ \Rightarrow z&=(-4)^{\frac14}\\ &=\sqrt2(-1)^{\frac14}\\ &=\sqrt 2\big[\cos(2k+1)\pi+i\sin(2k+1)\pi\big]^{\frac14}\\ &=\sqrt 2\left[\cos\frac{(2k+1)}4\pi+i\sin\frac{(2k+1)}4\pi\right];\quad k=0,1,2,3\\ &=\sqrt2\left(\pm\frac1{\sqrt2}\pm\frac1{\sqrt2}i\right)\\ &=(\pm1\pm i)\\ &=1+i,1-i,-1+i,-1-i\\ \end{align}$ are simple poles and $z=1+i,-1+i$ lies in side

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    @Miguel Someone typed the problem based on the images you have uploaded: [1](http://i.stack.imgur.com/U0L5x.png), [2](http://i.stack.imgur.com/C5vRA.png). You should check whether the edited versions is what you wanted. (They [mentioned](http://math.stackexchange.com/revisions/252912/2) that: *I am not sure if I read the exponent in the integral right*.)2012-12-28

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