cos z = 2 i

where z is a complex number, find z?
and i is (iota)

Hint :

$\cos z=\frac{e^{iz}+e^{-iz}}{2}$

I assume you mean that i is the imaginary unit $\displaystyle \sqrt{-1}$. Anyway

\displaystyle \begin{align*} \cos{(z)} &= 2i \\ \cos{( x + i\, y)} &= 2i \\ \cos{(x)}\cos{(i\,y)} - \sin{(x)}\sin{(i\,y)} &= 2i \\ \cos{(x)}\cosh{(y)} - i\sin{(x)}\sinh{(y)} &= 0 + 2i \\ \cos{(x)}\cosh{(y)} = 0 \textrm{ and } -\sin{(x)}\sinh{(y)} &= 2 \end{align*}

From the first equation, we have $\displaystyle \cos{(x)} = 0 \implies x = \frac{(2n+1)\pi}{2} \textrm{ where } n \in \mathbf{Z}$ or $\displaystyle \cosh{(y)} = 0$ which does not have any real solutions. Substituting x we find

\displaystyle \begin{align*} -\sin{\left[ \frac{(2n+1)\pi}{2} \right] } \sinh{(y)} &= 2 \\ (-1)^{n+1} \sinh{(y)} &= 2 \\ \sinh{(y)} &= \frac{2}{(-1)^{n+1}} \\ y &= \sinh^{-1}{\left[ \frac{2}{(-1)^{n+1}} \right]} \end{align*}

So there you have it, the solutions are $\displaystyle z = \frac{(2n+1)\pi}{2} + i\sinh^{-1}{\left[ \frac{2}{(-1)^{n+1}} \right]} \textrm{ where } n \in \mathbf{Z}$.