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December 4th, 2016, 04:06 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?
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December 16th, 2016, 06:08 PM   #2
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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!
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December 16th, 2016, 07:11 PM   #3
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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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