Are we to presume that "Log", without a base shown, is base "e"?
Then \(\displaystyle Log(-e^{-i\tau}z)= Log(-1)+ Log(e^{-i\tau})+ Log(z)= i\pi- i\tau+ Log(z)\)
Adding \(\displaystyle i(\tau+ \pi)= i\tau+ i\pi\) gives
\(\displaystyle Log(-e^{-i\tau}z)= 2i\pi+ Log(z)\)
But "Log(z)", like most complex functions is "multivalued". Specifically, if \(\displaystyle z= re^{i\theta}\), \(\displaystyle \theta\) is "modulo 2\pi" so we can write \(\displaystyle z= re^{i(\theta+ 2n\pi)}\) for any integer, n. Then \(\displaystyle log(z)= log(r)+ i(\theta+ 2n\pi)\) so that we can add any multiple of \(\displaystyle 2i\pi\) to the logarithm.