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Suppose I wish to compute the integral $$ I = \oint_C dz f(z,{\bar z}) $$ where the contour $C$ is given. Obviously, the answer will depend on the contour chosen. I know the standard way to do this (standard from where I learnt it anyway). One simply takes $z = re^{i \theta}$ and ${\bar z} = r e^{- i \theta}$. The contour is then given by $r = r(\theta)$. The integral then becomes $$ I = \int_0^{2\pi} d \left( r(\theta)e^{i \theta} \right) f \left( r(\theta) e^{i \theta} , r(\theta) e^{- i \theta} \right) $$

QUESTION:

I am wondering if the following method also works. In the complex coordinates, the contour may be given by ${\bar z} = g(z)$. (Let us for simplicity, consider only contours that can be written in the form above) We can then write the integral as $$ I = \oint_C dz f \left(z, g(z) \right) $$ Now we have a contour integral over a holomorphic function. We can now simply use Cauchy residue theorem to evaluate the contour integral. Will this give the correct answer?

Prahar
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  • If $g(z) = \bar{z}$, then $g$ cannot be a holomorphic function. So it is impossible to reduce down to the holomorphic case in general. – Christopher A. Wong Jul 23 '13 at 22:28
  • I am not saying that. Let me explain what I mean using an example. Let us look at the integral $f(z,{\bar z}) = \frac{1}{z} + {\bar z}$. Let us evaluate the integral for the contour given by $z {\bar z} = 1 \implies {\bar z} = \frac{1}{z}$. We then find that $I = \oint_C dz \frac{2}{z} = 2$. – Prahar Jul 23 '13 at 23:12
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    In that case, yes, your method works. If you can find a holomorphic function identically equal to the integrand along the contour, then you can always exchange them. – Christopher A. Wong Jul 23 '13 at 23:16

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