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):
$$-i\ln (\sqrt2 +1)+\left(2k+\frac12\right)\pi, \quad -i\ln(\sqrt 2 - 1) +\left(2k-\frac12\right)\pi, \qquad k\in \Bbb Z.$$

\begin{align}
& e^{iz} + e^{-iz} = 2i \\
& e^{2iz} + 1 = 2ie^{iz} \qquad \text{(Both sides were multiplied by $e^{iz}$.)} \\
& w^2 + 1 = 2iw \\
& w^2 -2iw = -1 \\
& w^2 - 2iw - 1 = -2 \qquad\text{(completing the square)} \\
& (w- i)^2 = -2 \\
& w-i = \pm i\sqrt 2 \\
& w = i \pm i\sqrt 2 \\
& e^{iz} = i(1\pm\sqrt 2) = e^{i\pi/2} e^{\log(1\pm\sqrt 2)}
\end{align}
etc.
(And notice that $\log(1-\sqrt 2) = -\log(1+\sqrt 2)$.)
