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.
 A: The other comments and answers all have good alternative suggestions on how to solve the problem. That said, your method can also reach the solution. Continuing from where you left off, take real and imaginary parts of both sides:
$$
\frac{\sin(2x)}{\cos(2x)+\cosh(2y)} = 0\;\;,\;\; \frac{\sinh(2y)}{\cos(2x)+\cosh(2y)}= 2
$$
From the first equation, we have $\sin(2x) = 0$. That is, $x = n\pi/2$. Then the second equation is $\sinh(2y)/[\cosh(2y)+(-1)^n] = 2$. Now note that if $n$ is even, the denominator is always greater than the numerator. Therefore $n$ must be odd (and equal to $2m+1$ for some $m$), and we have
$$
\frac{\sinh(2y)}{\cosh(2y) - 1} = \coth y = 2\Longrightarrow y = \tanh^{-1}\frac{1}{2} = \frac{1}{2}\ln 3.
$$
Putting the results for $x$ and $y$ together gives
$$
z = \frac{\pi}{2}+\frac{i}{2}\ln 3 + m\pi.
$$
Having a solutions spaced apart by $\pi$ makes sense, as $\tan$ has period $\pi$.
A: 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.
$$z = \pi(n + \frac{1}{2}) + i\frac{\ln 3}{2}, n \in \mathbb{Z}$$
A: Note that
$$2i=\tan x=\tan(n\pi+ x)=-i\tanh(i(n\pi+x))
$$
Thus
$$i(n\pi+x)=-\tanh^{-1}2=-\frac12\ln(-3)
=-\frac{i\pi}2-\frac12\ln3
$$
which leads to the roots
$$x=-n\pi-\frac\pi2+\frac i2\ln3$$
