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I tried solving the integral over the circle with radius 2. $$ \int_{|z|=2}{\bar{z}^{\frac{1}{2}}}dz $$ By using the branch $(-\pi,\pi)$ I obtain $8\sqrt{2}i$

Namely

$$ \int_{|z|=2}{\bar{z}^{\frac{1}{2}}}dz=\int_{-\pi}^{\pi}{e^{\frac{1}{2}ln(2)-\frac{\phi}{2}i}}dz=\int_{-\pi}^{\pi}{e^{\frac{1}{2}ln(2)-\frac{\phi}{2}i}}\ 2ie^{\phi i}d\phi=8\sqrt2i $$

However by using the branch $(0,2\pi)$ I obtain $-8\sqrt2$

$$ \int_{|z|=2}{\bar{z}^{\frac{1}{2}}}dz=\int_{0}^{2\pi}{e^{\frac{1}{2}ln(2)-\frac{\phi}{2}i}}dz=\int_{0}^{2\pi}{e^{\frac{1}{2}ln(2)-\frac{\phi}{2}i}}\ 2ie^{\phi i}d\phi=-8\sqrt2 $$

How is it that these two integrals yield different results?

Bosnan
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    You have integrated two distinct functions and you got two distinct answers. What's remarkable about it? – José Carlos Santos Sep 29 '22 at 19:22
  • As @JoséCarlosSantos implied, for $|z|=2$ the function $\sqrt{\bar z}$, $\arg(z)\in (-\pi,\pi]$ is not the same as the function $\sqrt{\bar z}$, $\arg(z)\in [0,2\pi)$. This is due to the fact the the complex logarithm is multivalued on the plane. By choosing a specific branch cut, the complex logarithm becomes a distince single-valued function on each corresponding branch. – Mark Viola Sep 29 '22 at 19:28
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    Dear @MarkViola and thank you for the clarification, I always assumed functions that use the complex logarithm, with exception of the branch of discontinuity, were the same regardless of the branch chosen. This is due to the fact that if I were to integrate over $\frac{1}{z}$ I seem to get the same answer regardless of the branch I chose. A wrong assumption of mine as it seems.

    Again thank you kindly for the clarification

     – Bosnan Sep 29 '22 at 19:36
  • You're welcome. My pleasure. And when you integrate $1/z$ over a contour with winding number equal to $1$, then you always get $2\pi i$ because the natural logarithm has a discontinuity of $2\pi i$ as we cross the branch cut. – Mark Viola Sep 29 '22 at 20:53

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