Complex integration problem

Question

Im having a bit trouble getting started in complex integrals.

Im integrating along a circle the integral $\int \frac{e^z}{z^2 + 1}$ where

a) center O radius 1/2 b) center i radius 1 c) center i radius 3

I have set $\gamma(t) = \frac{1}{2}e^{it}$ and $\gamma' = \frac{i}{2}e^{it}$ and $\gamma = z$ for part a) but how do I get around to doing the integral?

Where are you stuck? Also, do you know about the residue theorem? — AlexR, Sep 19 '13 at 11:04
Im stuck on integrating with respect to t. I have heard of the residue theorem but I dont know how to apply it — dingari, Sep 19 '13 at 11:07
@user2750354 Please write the integrals you face into the question so we can see where the problem is. — AlexR, Sep 19 '13 at 11:33

Ron Gordon · Answer 1 · 2013-09-19T14:20:50.320

1

If you know the residue theorem, then these integrals may be done easily. The integrand has poles at $z=i$ and $z=-i$. In part a), neither pole is inside the contour, so the integral is zero. For part b), only the pole at $z=i$ is inside, so the integral is $i 2 \pi$ times the residue at $z=i$, or

$$i 2 \pi \frac{e^i}{2 i} = \pi (\cos{1} + i \sin{1})$$

In part c), both poles are inside the contour, so you will need to sum the residues at both poles. I get that the integral is then $i 2 \pi \sin{1}$.

edited Sep 19 '13 at 14:20

answered Sep 19 '13 at 11:09

Ron Gordon

138,521

Thank you for your reply. How would this be done using Cauchys integral theorem? – dingari Sep 19 '13 at 13:20
@user2750354: Part a) is Cauchy's Theorem. So technically are parts b) and c) - the residue theorem is merely an extension of Cauchy's theorem. – Ron Gordon Sep 19 '13 at 13:23
when I sum them together I get zero. What is the residue at z=-i – dingari Sep 19 '13 at 14:20
@user2750354: the residue there is $(e^{-i})/(-2 i)$. – Ron Gordon Sep 19 '13 at 14:21

Complex integration problem

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