Finding $z\in \mathbb C$ with $\sin z = 2$? I'm trying to find all $z\in \mathbb C$ with $\sin z = 2$ but I'm stuck.
I tried writing $\sin z = {e^{iz} - e^{-iz}\over 2i} = 2$ and then writing $z = x + iy$ but the expression I got was $e^{ix}e^{-y} - e^{-ix}e^{y}=4i$ and I don't see how it's possible to solve this for $x$ and $y$.

Please could someone tell me how to calculate the inverse values of
  the complex sine function?

I had also tried letting $w=e^{iz}$. Then I got the quadratic 
$$ w^2 - 4i w + 1 = 0$$
and using the formula for quadratics I got
$$ w_{1,2} = {4i \pm i \sqrt{20}\over 2}$$
Taking the logarithm I got
$$ i z_{1,2} = \ln{{4i \pm i \sqrt{20}\over 2}}$$
whereas I should be getting 
$$ z = ({\pi \over 2}  + 2\pi k ) - i \ln(2 \pm \sqrt{3})$$
for $k \in \mathbb Z$.
 A: You have 
$$
e^{iz}-e^{-iz}=4i
$$
For the sake of sanity, define $e^{iz}=x$, then multiply by $x$ to get a quadratic you can solve
$$
x-x^{-1}=4i\Rightarrow x^2-4ix-1=0
$$
Solve for $x$, then plug back in.
A: If $z = x + i y$, then $\sin(z) = \sin(x) \cos(iy) + \cos(x) \sin(iy) = \sin(x) \cosh(y) + i \cos(x) \sinh(y)$.  You want $\sin(x) \cosh(y) = 2$ and $\cos(x) \sinh(y) = 0$, where $x$ and $y$ are real. $\sinh(y)=0$ would imply $y=0$, and then you would have $|\sin(z)| \le 1$.  So you want $\cos(x) = 0$, meaning $x = (n + 1/2) \pi$ for some integer $n$, and $\sin(x) = (-1)^n$ (but you want this to be positive).  So $n$ should be even, making $\sin(x) = 1$, and then you want $\cosh(y) = 2$.  If $t = e^y$, that says $(t+1/t)/2 = 2$, 
which has positive solutions $t = 2 \pm \sqrt{3}$.  Thus 
$$z = \left(2k+\frac{1}{2}\right) \pi + i \ln(2 \pm \sqrt{3})$$
for integers $k$.
A: There is also a slight different approach to do this problem.
Using $\sin z=2$ and the Pythagorean theorem $\sin^2z+\cos^2z=1$, we find $\cos z=  \pm i\sqrt{3}$. Now using $e^{iz}=\cos z+i\sin z$ we can put in our results: $e^{iz}=\pm i\sqrt{3}+2i = (2\pm \sqrt{3})i $. Taking natural log on both sides solves it for $iz$. If you know how to take the log of a complex number, you are good to go.
