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In order to evaluate $\int_0^{2 \pi} \frac {\cos 2 \theta}{1 -2a \cos \theta +a^2}$ we can define $$ f(z) := \frac 1 z \cdot \frac { (z^2+z^{-2})/2}{1-2a( \frac {z+z^{-1}} 2) +a^2} $$ I have $0 < a <1$ which gives singular points in $0$ and $a$ which lie in the unit circle. I want to calculate the residues in those points to use the Residue-formula. How can I evaluate the residues here ?

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    You can write $f(z)$ as

    $$f(z)=\frac{1}{z}\frac{(z^{2}+z^{-2})/2}{1-2a\left( \frac{z+z^{-1}}{2} \right) +a^{2}}=-\frac{1}{2a}\frac{z^{4}+1}{z^{2}\left( z-a\right) (z-1/a)}.$$

     – Américo Tavares Jun 09 '13 at 16:05
  • Do I have to use power series or limit laws ? –  Jun 09 '13 at 16:06
  • No, just algebra.

    $$\frac{1}{z}\frac{(z^{2}+z^{-2})/2}{1-2a\left( \frac{z+z^{-1}}{2}\right) +a^{2}}=-\frac{1}{2}\frac{z^{4}+1}{z^{2}\left( az^{2}-\left( a^{2}+1\right) z+a\right) }=-\frac{1}{2a}\frac{z^{4}+1}{z^{2}\left( z-a\right) (z-1/a)},$$

    because $az^{2}-\left( a^{2}+1\right) z+a=a\left( z-a\right) (z-1/a)$, as can be seen by solving $az^{2}-\left( a^{2}+1\right) z+a=0$.

     – Américo Tavares Jun 09 '13 at 16:14
  • Thanks, but how do you know what the residues are ? –  Jun 09 '13 at 16:14
  • You have a double pole at $z=0$ and two simple poles at $z=a$ and $z=1/a$. – Américo Tavares Jun 09 '13 at 16:17

1 Answers1

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With

$$f(z)=-\frac1{2a}\frac{z^4+1}{z^2(z-a)(z-a^{-1})}$$

with a single pole (within the unit circle) at $\,z=a\;$ and a double one at $\,z=0$ :

$$\text{Res}_{z=a}(f)=\lim_{z\to a}(z-a)f(z)=-\frac1{2a}\frac{a^4+1}{a-a^{-1}}=-\frac12\frac{a^4+1}{a^2-1}$$

$$\text{Res}_{z=0}(f)=\lim_{z\to 0}\frac d{dz}\left(z^2f(z)\right)=-\frac{a^2+1}{2a^2}$$

DonAntonio
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