Principal branch of square root

I thought I understood what happened here when I asked this question yesterday. However now I'm not so sure anymore. My explanation was: The first parametrization uses $$[-\pi,0]$$ which happens to be included in the domain of the principal branch of the logarithm and thus the corresponding integral yields something different than the one using $$[\pi,2\pi]$$ which is included in the domain of a side branch (if you call it like that in english). But actually the square root must be taken of $$\gamma$$ which in both cases (once you plugged t into the exponential function) is the lower arc of the unit circle, so we shouldn't have two different domains yielding two different branches.

Long story short: Where exactly in those two integrals is the point the different branch cuts step in? I just can't see it.

• when you write $e^{it/2}$ for $\sqrt{e^{it}}$ and the choice of $t\in[\pi,2\pi]$, you are explicitly choosing the branch that gives $\sqrt{1}=-1$. Similarly, for $t\in[-\pi,0]$ you are choosing $\sqrt{1}=+1$. – user10354138 May 23 at 12:04
• Why though? what we plug into the square root are the exact same complex numbers both times, right? Both times it's $\{z\in\mathbb C:\ |z|=1,\ \text{Im}(z)\leq 0\}$. – RedLantern May 23 at 12:09

The principle branch of the logarithm has the form $$\log z=\ln |z|+i \theta,$$ with $$\theta\in (-\pi,\pi]$$ being the argument of $$z$$. The choice of branch boils down, essentially, to choosing the right $$k$$ in the multi-valued form $$\log z=\ln|z|+i(\theta+2k\pi).$$ If you want $$z\in (\pi,2\pi],$$ then we want to take $$k=1$$. Then, $$\sqrt{z}=e^{i\log (z^{1/2})}=e^{\frac{i}{2}\log z}=e^{\frac{i}{2}\left(\ln|z|+i(\theta+2\pi)\right)}=e^{\frac{i}{2}\ln|z|}e^{\frac{i}{2}\left(\theta+2\pi\right)}.$$ In your post, you wanted to take the square root of $$e^{it}.$$ Using the above, we get $$\sqrt{e^{it}}=e^{\frac{i}{2}\ln|e^{it}|}e^{\frac{i}{2}\left(\theta+2\pi\right)}=e^{\frac{i}{2}\left(t+2\pi\right)}=-e^{\frac{it}{2}}.$$