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Suppose I have a closed-loop counter-clockwise contour integral over a function $f(z,t)$:

$$F(t)=\oint_C dz\enspace f(z,t)$$

Then suppose I want to know the complex conjugate of $F(t)$. What happens to the contour itself? does the contour flip orientation (i.e. become clockwise)?

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Write out the integral as $\int_0^1 f(z(u),t)z'(u)\,du$, where $z(u)$ is a parametrization of the contour, then it's easy to take the complex conjugate. – Kirill May 31 '13 at 0:28
This is true if $f(z,t)$ is holomorphic with respect to $z$, but otherwise I am not sure. Also I think you assume the circle is centered at origin. – Bombyx mori May 31 '13 at 1:10
That is the definition of a (complex) path integral. It doesn't matter if $f$ is holomorphic in $z$, and it doesn't matter what the path is so long as there is a differentiable map $[0,1]\to\mathbb{C}$ given by $z(t)$. – Kirill May 31 '13 at 1:35

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