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When we evaluate the integrals of the type $\displaystyle\int_0^{2\pi}f(\cos\theta,\sin\theta) \, d\theta$, we use Cauchy-Residue theorem. My question is what's the significance of evaluating such integrals over the unit circle, i.e. $|z|=1$?

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  • $\begingroup$ If I had to compute such an integral, I would think of a change of variables rather than of the Cauchy theorem. $\endgroup$ – jibounet Sep 7 '17 at 9:26
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It is just because the complex numbers $z$ such that $|z|=1$ are precisely those complex numbers which are of the form $e^{i\theta}$ ($\theta\in[0,2\pi]$). So, if you want to compute that integral and if the domain of $f$ is an open set $U\subset\mathbb{C}^2$, if you define$$V=\left\{z\in\mathbb{C}\setminus\{0\}:\left(\frac{z+z^{-1}}2,\frac{z-z^{-1}}{2i}\right)\in U\right\},$$and if you define $F\colon V\longrightarrow\mathbb C$ by$$F(z)=\frac1zf\left(\frac{z+z^{-1}}2,\frac{z-z^{-1}}{2i}\right),$$then$$\bigl(\forall\theta\in[0,2\pi]\bigr):F(e^{i\theta})=e^{-i\theta}f(\cos\theta,\sin\theta).$$So, if you define $\gamma(\theta)=e^{i\theta}$ ($\theta\in[0,2\pi]$), then$$\int_\gamma F=\int_0^{2\pi}F\bigl(\gamma(\theta)\bigr)\gamma'(\theta)\,\mathrm d\theta=\int_0^{2\pi}f(\cos\theta,\sin\theta)\,\mathrm d\theta$$and you can therefore use the residue theorem to compute the integral.

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  • $\begingroup$ Thank you @José Carlos Santos for the reply, I checked the answer for different examples by taking $|z|=r$ instead of $|z|=1$ answer is same, but I am not clear with the reason why I need to take $|z|=1$ ??, Is it only for simplicity?? $\endgroup$ – Rag Sep 7 '17 at 12:36
  • $\begingroup$ @Rag My answer uses the fact that, if $z=e^{i\theta}$, then $\cos\theta=\frac12(z+z^{-1})$ and $\sin\theta=\frac1{2i}(z-z^{-1})$. This requires that $|z|=1$, since $|z|=|e^{i\theta}|=|\cos\theta+i\sin\theta|=1$. $\endgroup$ – José Carlos Santos Sep 7 '17 at 13:04

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