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.

  • $\begingroup$ 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 $\sqrt{1}=+1$. $\endgroup$ – user10354138 May 23 at 12:04
  • $\begingroup$ Why though? what we plug into the square root are the exact same complex numbers both times, right? Both times it's $\{z\in\mathbb C:\ |z|=1,\ \text{Im}(z)\leq 0\}$. $\endgroup$ – 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 $$\sqrt{e^{it}}=e^{\frac{i}{2}\ln|e^{it}|}e^{\frac{i}{2}\left(\theta+2\pi\right)}=e^{\frac{i}{2}\left(t+2\pi\right)}=-e^{\frac{it}{2}}.$$


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