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Let $0 < r < 1$. Compute $$\frac{1}{2 \pi} \int_0^{2 \pi} \frac{1 - r^2}{1 - 2r \cos(\theta) +r^2}d\theta$$ The hint is rewrite this integral as a complex line, but I still don't know how to to it

Alexander
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  • Maybe $\cos t = \frac{e^{it}+e^{-it}}{2}$ can help. – Siminore Jul 22 '15 at 08:22
  • hint :$$z=\frac{(-r-cos\theta)+i sin \theta}{(r-cos\theta) +isin \theta} \Re(z)=?\Re(z)=Re\frac{(-r-cos\theta)+i sin \theta}{(r-cos\theta) +isin \theta}*\frac{(r-cos\theta)-i sin \theta}{(r-cos\theta) -isin \theta}=\\frac{((-r-cos\theta)(r-cos\theta)-i^2sin^2 \theta}{((r-cos\theta)^2-i^2sin^2 \theta}=\frac{-r^2+cos^2 \theta +sin^2 \theta}{r^2-2rcos \theta +cos^2 \theta +sin^2 \theta}=\frac{-r^2+1}{r^2+1-2rcos \theta}$$ – Khosrotash Jul 22 '15 at 08:48
  • @daryakhosrotash Is this a hint? – Did Jul 22 '15 at 08:59
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    This question was discussed many times on this site, please have a close look. – tired Jul 22 '15 at 09:38
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    @tired: I just found this question, which is closely related and my answer there answers this question using contour integration rather than the mean value property. – robjohn Jul 22 '15 at 10:20
  • @Math-fun: one could use the indefinite integral, but since this is a definite integral, there are often other (possibly simpler) methods of integration. – robjohn Jul 22 '15 at 14:33
  • @robjohn this is true, I agree. I was not sure if this is an exact duplicate, though in the link I thought relevant, there is a solution to this question with the same bounds. Many thanks. – Math-fun Jul 22 '15 at 15:54

2 Answers2

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Using the Mean Value Property, we have $$ \begin{align} \frac1{2\pi}\int_0^{2\pi}\frac{1-r^2}{1-2r\cos(\theta)+r^2} \,\mathrm{d}\theta &=\frac1{2\pi}\int_0^{2\pi}\frac{1-r^2}{(1-re^{i\theta})(1-re^{-i\theta})} \,\mathrm{d}\theta\\ &=\frac1{2\pi}\int_0^{2\pi}u\left(re^{i\theta}\right)\mathrm{d}\theta\\[6pt] &=u(0)\\[12pt] &=1 \end{align} $$ where $u(z)=\frac{1-|z|^2}{|1-z|^2}=\mathrm{Re}\left(\frac{1+z}{1-z}\right)$ is a Harmonic Function.

robjohn
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Hint: The denominator can be expressed as $(1-e^{i\theta}r)(1-e^{-i\theta}r)$.

Alijah Ahmed
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