Finding roots of $\tan z=2i$ I'm trying to find the roots of $\tan(z) = 2i.$
At the moment I have
$$
\tan(z)=\frac{\sin 2x +i \sinh 2y }{\cos 2x+\cosh2y}=2i
$$
At a loss as to where I can go from here. Any advice would be appreciated.

Recall that $\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$, $\cos(z) = \frac{e^{iz} + e^{-iz}}{2}$, and $\tan(z) = \frac{\sin(z)}{\cos(z)} = \frac{e^{iz} - e^{-iz}}{i(e^{iz} + e^{-iz})}$.  So your equation becomes:
$$\frac{e^{iz} - e^{-iz}}{i(e^{iz} + e^{-iz})} = 2i$$
$$\frac{e^{iz} - e^{-iz}}{e^{iz} + e^{-iz}} = -2$$
Make the substitution $w = e^{iz}$.  Then:
$$\frac{w - \frac{1}{w}}{w + \frac{1}{w}} = -2$$
$$\frac{w^2 - 1}{w^2 + 1} = -2$$
$$w^2 - 1 = -2(w^2 + 1)$$
$$3w^2 + 1 = 0$$
This is a simple quadratic with the roots $w = \pm \frac{i\sqrt{3}}{3}$.  Or in polar form, $w = \frac{\sqrt{3}}{3} \operatorname{cis}(\pm\frac{\pi}{2}) = \frac{\sqrt{3}}{3} e^{\pm i\pi/2}$.  So,
$$e^{iz} = \frac{\sqrt{3}}{3} e^{\pm i\pi/2}$$
$$iz = \ln(\frac{\sqrt{3}}{3}) + i(2\pi k \pm \frac{\pi}{2}), k \in \mathbb{Z}$$
$$iz = \frac{-\ln 3}{2} + i(2\pi k \pm \frac{\pi}{2})$$
$$z = -i\frac{-\ln 3}{2} + 2\pi k \pm \frac{\pi}{2}$$
$$z = 2\pi k \pm \frac{\pi}{2} + i\frac{\ln 3}{2}$$
The real part can be simplified a little by noting that it's just the set of numbers $(n.5) \pi$ for any integer.
$$z = \pi(n + \frac{1}{2}) + i\frac{\ln 3}{2}, n \in \mathbb{Z}$$
