Solving $\cos z = i$ for $z$ 
Solve $\cos z = i$ for $z$.

What I have tried:
$$\cos z = i$$
$$\frac{e^{-zi}+e^{zi}}{2}=i$$
$$e^{-zi}+e^{zi}=2i=2e^{\frac\pi 2 + 2\pi k},\quad k\in \Bbb Z$$
I would take logs, but then I would have two terms in the left log, what now?

I have the solution for the problem already (this hasn't helped me since they have shown no work):
$$-i\ln (\sqrt2 +1)+\left(2k+\frac12\right)\pi, \quad -i\ln(\sqrt 2 - 1) +\left(2k-\frac12\right)\pi, \qquad k\in \Bbb Z.$$

You got to $e^{iz}+e^{-iz}=2i$. Let $u=e^{iz}$, this becomes:
$$u+\frac{1}{u}=2i$$
$$u^2+1=2iu$$
$$u^2-2iu+1=0$$
Thus 
$$u=\frac{2i\pm\sqrt{-4-4}}{2}$$
$$ u=\frac{2i\pm i\sqrt{8}}{2}$$
$$ u = i(1\pm\sqrt{2})$$
Then $$e^{iz}=i(1\pm\sqrt{2})$$
