Solving $\sin z = i$ I know that
$$\sin z = \frac{e^{iz}-e^{-iz}}{2i}$$
so:
$$\frac{e^{iz}-e^{-iz}}{2i} = i\implies e^{iz}-e^{-iz} = -2$$
but I can't take anything useful from here. How do I solve such equations?
What about $\tan z = 1$? Is there any solutions?

$\sin z = \cos z $ comes down to ..
$$ e^{iz}-e^{-iz}  = i( e^{iz} + e^{-iz} )  $$
$$  (1-i)e^{iz}= (1+i)e^{-iz} $$
$$  e^{2iz}= \frac{ 1+i}{1-i} =i=e^{i( \frac\pi2+2n\pi )}$$
So $$z=\frac\pi4 +n\pi$$
All solutions are real.
