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I have to calculate the integral

$$I=\int_{0}^{2\pi}\frac{\cos^2(3\theta)}{5-4\cos(2\theta)}d\theta$$

using residues, but I'm having trouble calculating the residue in $z=0$. If you rewrite the integral by stating $z=exp(i\theta)$, and letting gamma be the unitary circle, then:

$$I=\int_{\gamma} \frac {(\frac {z^9+1}{2z^3})^2}{5-2(z^2+z^{-2})}\frac{1}{zi}dz$$

Well, if you do a little algebra you have that

$$\frac {(\frac {z^9+1}{2z^3})^2}{5-2(z^2+z^{-2})}\frac{1}{zi}=\frac{1}{4i}\frac{(z^9+1)^2}{z^5(-2z^4+5z^2-2)}$$

I'm struggling with the residue of this function in $z=0$. Any ideas?

This is an exercise taken from the book Churchill's about Complex Analysis, in case that is useful.

PD: I have made a mistake when replacing the cosines, and the function is a bit different. Iǘe already edited it.

Lessa121
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2 Answers2

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After normalising a little, you need the coefficient of $z^4$ in the Taylor series of

$$\frac{1+2z^9+z^{18}}{1 - \frac{5}{2}z^2 + z^4},$$

if you haven't made any mistakes in your computation thus far (which I haven't checked). Then expand the denominator in a geometric series. Since $9 > 4$, only the $1$ of the numerator is relevant for the coefficient of $z^4$, and

$$\frac{1}{1 - \frac{5}{2}z^2 + z^4} = 1 + \left(\frac{5}{2}z^2-z^4\right) + \left(\frac{5}{2}z^2-z^4\right)^2 + O(z^6)$$

makes it easy to see the coefficient of $z^4$ is $\frac{25}{4} - 1$.

Daniel Fischer
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  • I'd indeed made a mistake. I've corrected it, but i'm guessing that I could use the same principle you explained here to solve it. – Lessa121 Oct 18 '14 at 13:00
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    Yes, that principle works generally. It may not always be the simplest method to obtain the relevant coefficient, but for coefficients of small exponents, it's never too much worse than other methods. – Daniel Fischer Oct 18 '14 at 13:03
  • Required only minimal changes ;) – Daniel Fischer Oct 18 '14 at 13:06
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May be, this could help

If you use partial fraction decomposition, you get $$\frac{1+2z^3+z^9}{1 - \frac{5}{2}z^2 + z^4}=z^5+\frac{5 z^3}{2}+\frac{2 (20 z+1)}{3 \left(z^2-2\right)}+\frac{-17 z-16}{12 \left(2 z^2-1\right)}+\frac{21 z}{4}$$

Claude Leibovici
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