December 4th, 2016, 04:06 PM  #1 
Newbie Joined: Dec 2016 From: England Posts: 4 Thanks: 0  Contour Integral
I am really stuck on this question and can't find appropriate help anywhere please can someone help! I think the first part is a circle with centre origin but not sure on the radius? What cauchy theorem do i use for part 2? 
December 16th, 2016, 06:08 PM  #2 
Member Joined: Dec 2016 From:  Posts: 36 Thanks: 7 
The contour $\gamma(t)$ defines an ellipse in the complex plane, where $t$ is the angle formed with the $x$ axis when you move along the ellipse. For $t=\frac{\pi}{2}\to\gamma(\frac{\pi}{2})=ib$, being that one semiaxis of the ellipse (vertical). To complete the contour integral, you need to apply Cauchy's theorem of residues, being $\gamma$ a closed path, this states that: \begin{eqnarray} \int_{\gamma}dz f(z)=2\pi i\sum \text{Res}f(z) \end{eqnarray} where $\gamma$ needs tto be a closed path. To perform the intergral, realize that $z=re^{i\theta}$, where $r(t)$ since it is the distance of the point $z$ respect the origin (z lies on the curve now). Hope this helps! 
December 16th, 2016, 07:11 PM  #3 
Member Joined: Dec 2016 From:  Posts: 36 Thanks: 7 
Ok, couldnt resist to solve it at the end, here is your solution: Applying Cauchy theorem of residues, it is straightforward that: \begin{eqnarray} \int_{\gamma}\frac{dz}{z}=2\pi i \end{eqnarray} since the only pole inside the region enclosed by $\gamma$ is the point $z=0$. This solves part b). To solve part c): Notice that on $\gamma$, $z=a\cos(t)+ib\sin(t)$, which implies $dz=dt(a\sin(t)+ib\cos(t))$. Performing the integral \begin{eqnarray} I=\int_{0}^{2\pi}\frac{(a\sin(t)+ib\cos(t))(a\cos(t)ib\sin(t))}{r(t)^{2}}dt \end{eqnarray} where we define $r(t)^{2}=a^{2}\cos^{2}(t)+b^{2}\sin^{2}(t)$ Taking that into account we have: \begin{eqnarray} I=\int_{0}^{2\pi}\frac{(b^{2}a^{2})\sin(t)\cos(t)}{r(t)^{2}}dt+iba\int_{0}^{2\p i}\frac{\cos^{2}(t)+\sin^{2}(t)}{r(t)^{2}}dt \end{eqnarray} The integral on the left is equal to $0$, and using the relation $\sin^{2}(t)+\cos^{2}(t)=1$, the right hand side is: \begin{eqnarray} I=iba\int_{0}^{2\pi}\frac{dt}{r(t)^{2}}=2\pi i \end{eqnarray} according to the result obtained in b). Therefore, we conclude: \begin{eqnarray} \int_{0}^{2\pi}\frac{dt}{a^{2}\cos^{2}(t)+b^{2}\si n^{2}(t)}=\frac{2\pi}{ab} \end{eqnarray} 

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analysis, complex, contour, integral, integratiion 
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