Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question
It might help to make more explicit whether $C$ is in the upper half or lower half plane. From context it appears to be in the upper half plane. Notice that $1+i$ and $-1+i$ are the poles in the upper half plane. – Jonas Meyer Dec 7 '12 at 7:26
Also you need to know that $R$ is sufficiently large (larger than $\sqrt2$), but this is probably a part of a problem, where you intend to let $R \to \infty$ eventually, and in that case it won't matter. – mrf Dec 7 '12 at 12:44
@Miguel Someone typed the problem based on the images you have uploaded: 1, 2. You should check whether the edited versions is what you wanted. (They mentioned that: I am not sure if I read the exponent in the integral right.) – Martin Sleziak Dec 28 '12 at 8:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.