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Use the residue theorem to calculate $$\int_{0}^{2\pi} \frac {27} {(5+4\sin\theta)^2} d\theta $$

I know $$ \operatorname{Res}_{z_0} f = \frac 1 {2\pi i} \int_\gamma d\theta f(\theta) $$

My question is how do I plug in that function in this formula?

Thanks!

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The integral between $0$ and $2\pi$ should make you think of the unit circle $C(0,1)$. The parametrization is done by $e^{i\theta}$ with $\theta \in [0,2\pi].$ Hence you can write \begin{align*} \int_{0}^{2\pi} \frac {27} {(5+4\sin\theta)^2} d\theta & = \int_{0}^{2\pi} \frac {27} {(5-2i(e^{i\theta}-e^{-i\theta}))^2} d\theta\\ & =\frac{1}{i} \int_{C(0,1)} \frac{27}{z(5-2i(z-1/z))^2} dz \\ & =\frac{1}{i} \int_{C(0,1)} \frac{27z}{(-2iz^2+5z+2i)^2} dz \end{align*}

The two poles of order $2$ are $\frac{-5\pm\sqrt{21}}{-4i}$. The only pole in the unit disk is $\frac{-5+\sqrt{21}}{-4i}$. Applying the usual formula for residue, you find that $$Res\left(\frac{27z}{(-2iz^2+5z+2i)^2},\frac{-5+\sqrt{21}}{-4i}\right) = \lim_{z\to \frac{-5+\sqrt{21}}{-4i}}\left(\frac{27z}{z-\frac{5-\sqrt{21}}{4i} }\right)'=5.$$ Hence $$\int_{0}^{2\pi} \frac {27} {(5+4\sin\theta)^2} d\theta = 10\pi,$$ which is the good answer according to mathematica.

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  • $\begingroup$ Now complete the square in the denominator and solve for z to identify the singularities, then see which ones are inside the unit circle to find the residue of the function at these points. Add the residues, then multiply that by $2\pi i$ and that would be the value of the integral, correct? $\endgroup$ – Josh Apr 26 '16 at 5:54
  • $\begingroup$ Absolutely ! :) $\endgroup$ – C. Dubussy Apr 26 '16 at 5:59
  • $\begingroup$ Awesome. I tried solving for z in the denominator, but only got 2 values $ z = \frac {5}{4i} + \frac {\sqrt{-21}} {2} $ and $ z = \frac {5}{4i} - \frac {\sqrt{-21}} {2} $. Shouldn't I get 4 solutions since the z power in the denominator is 4? $\endgroup$ – Josh Apr 28 '16 at 8:15
  • $\begingroup$ Each of them are poles of order $2$, so you have $4$ roots with multiplicity. $\endgroup$ – C. Dubussy Apr 28 '16 at 8:16
  • $\begingroup$ The solutions I got the first time were wrong. They should be $ z=-2i , z = \frac {-i}{2} $. The only pole inside the circle is $z=\frac{-i}{2}$. I calculated the integral. Is there a way to check my answer? Thank you!! $\endgroup$ – Josh May 2 '16 at 3:23

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