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?