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 Complex Analysis Complex Analysis Math Forum

December 4th, 2016, 03:06 PM   #1
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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?
Attached Images Screen Shot 2016-12-05 at 00.04.30.jpg (19.1 KB, 25 views) December 16th, 2016, 05:08 PM #2 Member   Joined: Dec 2016 From: - Posts: 62 Thanks: 10 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, 06:11 PM #3 Member   Joined: Dec 2016 From: - Posts: 62 Thanks: 10 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} Tags analysis, complex, contour, integral, integratiion Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post kennalj Calculus 1 March 29th, 2015 06:01 AM mathematician1234 Complex Analysis 1 November 23rd, 2013 02:16 AM evol_w10lv Real Analysis 2 September 15th, 2013 01:31 AM qwertyuiop89 Complex Analysis 1 November 20th, 2012 07:32 AM evol_w10lv Calculus 0 December 31st, 1969 04:00 PM

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