Solving $\sinh z = 2i$ This is my attempt at the question (I stopped early because it starts to not work out...)
$$\sinh z = 2i \\ e^{iz} - e^{-iz} = 4i \\ = e^{2iz} - 4ie^{iz} - 1 = 0 $$
solving the quadratic gives
$$e^{iz} = i(2\pm \sqrt{3}).$$
I stop here to check:
$$\sinh z = \frac{e^{iz} - e^{-iz}}{2}= \frac{i(2\pm\sqrt{3}) + i(2\pm \sqrt{3})}{2}= i(2\pm \sqrt{3}) \neq 2i...$$  

I'm not sure if I'm doing something really wrong... because I've redone this a few times and I can't see it.
Use $\sinh^{-1} w = \ln(w+\sqrt{1+w^2})$

$$z = \sinh^{-1} 2i=\ln(2i+\sqrt{1-4})=\ln(2i\pm i\sqrt3)=\ln i + \ln(2\pm\sqrt3)$$ 

Since $\ln i = \ln e^{i\frac\pi2 + i2n\pi} = i\frac\pi2 + i2n\pi$, the solutions are

$$z = i\frac\pi2 + i2n\pi + \ln(2\pm\sqrt3)$$
Actually, correct version of the last expression is 

$$ \sinh(z)=\frac{e^{iz}-e^{-iz}}{2}=\frac{i(2\pm\sqrt{3})+i(2\mp\sqrt{3})}{2}=2i $$

Use $z=(2+\sqrt{3})i$ or $z=(2-\sqrt{3})i $ uniformly in the all part of expression