# Math Help - Simple complex equality driving me nuts

1. ## Simple complex equality driving me nuts

Find all solutions to sinz = 2i.

I've gone down the road of sin (z-z^(-1))/2i and its polar counterpart, and also laurent series, but the solutions I get are very complicated. This is a problem from a competency exam so I'm guessing it should have a clever but simple path and answer. Any guidance?

2. Originally Posted by igopogo
Find all solutions to sinz = 2i.
$\sin(z)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)$
So in this solve this system:
$\sin(x)\cosh(y)=0~\&~\cos(x)\sinh(y)=2$