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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

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    $\begingroup$ May be you need to have $e^{i(5\pi /4+2n\pi)}$. $\endgroup$ – velut luna Sep 25 '17 at 10:45
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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.

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    $\begingroup$ Or, simplified, $z=\frac{(-1 + i)}{8}\bigg((5+8k) i π-2\ln(2)\bigg)$ $\endgroup$ – Jam Sep 25 '17 at 11:12

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