Find Solution of trigonometric complex equation Find the solutions of $\sin z = 3$
There are 2 ways to solve this, I know how to do this with: $\sin z = \frac{1}{2i}(e^{iz}-e^{-iz}) = 3$
Now, I am now doing in the way: 
$\sin z = \sin x \cosh y+i \cos x \sinh y$ = 3 
By comparing the terms:
$\sin x \cosh y = 3$
$\cos x \sinh y = 0$
After this part, I have got no idea what to do.
Could anyone help me please?
Thank you.
 A: you have $$\sin x \cosh y = 3, \cos x \sinh y = 0 $$ take the second equation. you have $$\sinh y = 0 \to y = 0 \\ \cos x = 0, x = \pm \pi/2 +2k\pi $$
putting $y = 0,$ in the first equation gives $\sin x = 3$ which has no real solution. we are now left with $$x = \pm \pi/2 +2k\pi \to \cosh y = \pm 3.$$ since $\cosh y \ge 1,$ we only need to solve $$\cosh y = 3 \to e^{2y} - 6e^y+1=0\to e^y = 3\pm2\sqrt2 \to y = \ln(3\pm2\sqrt2)$$
the solutions are $$\sin^{-1}(3)=\pm \pi/2 +2k\pi+i \ln(3\pm2\sqrt2).$$
A: Let you want to solve the equation $sin(z)=\omega$. By the definition of the $sin(z)$ we should have: 
$$\frac{\mathbb e^{iz}-\mathbb e^{-iz}}{2i}=\omega$$
By multiplying to side of above equation in the $\mathbb e^{iz}$ we get:
$$\mathbb e^{2iz}-2i\omega\mathbb e^{iz}-1=0$$
Let $\mathbb e^{iz}=y$, then:
$$y^2-2i\omega y-1=0$$
The last equation is a quadratic equation and you can solve it.
A: $$sin(z)=3$$
Take the inverse sine of both sides, we get:
$$z=2\pi n+\pi-sin^{-1}(3)$$
Or:
$$z=2\pi n+sin^{-1}(3)$$
(with n is the element of Z -> the set of integers)
And if you know maybe from the university, we can write sin^-1(3) in a different way, so we get the following solutions:
$$z=2\pi n+\pi-\left(\frac{1}{2}\pi *i*ln\left(3+2\sqrt{2}\right)\right)$$
Or:
$$z=2\pi n+\left(\frac{1}{2}\pi *i*ln\left(3+2\sqrt{2}\right)\right)$$
A: $$\cos z=\pm\sqrt{1-3^2}=\pm2\sqrt2i$$
$$e^{iz}=\cos z+i\sin z=\pm2\sqrt2i+3i$$
$$\implies iz=Log[i(3\pm2\sqrt2)=Log(3\pm2\sqrt2)+Log(i)$$
As $i=\cos\dfrac\pi2+i\sin\dfrac\pi2=e^{i\dfrac\pi2},Log(i)=2n\pi i+i\dfrac\pi2$ where $n$ is any integer
