Find all the complex solutions z of
$$e^{-iz} = \frac{e^z}{i-1}$$
So far I tried the following
$$e^{z(-1-i)}=\frac{1}{i-1}$$ $$e^{z(-1-i)}=\frac{-1-i}{2}$$ Putting the right hand side into polar form gives $$\frac{1}{\sqrt 2}\left(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}\right)$$ Which is equal to $$e^{\ln\left(\frac{1}{\sqrt 2}\right)}e^{i\frac{5\pi}{4}}$$ $$e^{\ln\left(\frac{1}{\sqrt 2}\right)}e^{i\frac{5\pi}{4}} = e^{z(-1-i)}$$ $$z=\left(\ln\left(\frac{1}{\sqrt2}\right)+i\frac{5\pi}{4}\right)/(-1-i)$$
Now, obviously this is incorrect, because the result is wrong. I would very much appreciate it if someone told me where I went wrong and what I might do to correct my mistake. Thank you for your time and help
Your solution is correct, although incomplete. That is, your solution is indeed a solution, but when you got$$e^{\ln\left(\frac1{\sqrt2}\right)}e^{i\frac{5\pi}4}=e^{z(-1-i)},$$what you should have deduced from this was that$$z(-1-i)=\ln\left(\frac1{\sqrt2}\right)+i\frac{5\pi}4+2k\pi i,$$for some $k\in\mathbb Z$. Indeed, there are infinitely many solutions.