I'm trying to find all $z \in \mathbb{C}$ such that $\sin(z) = 2$. I also know the following; $\sin(z): \mathbb{C} \to \mathbb{C}$ and $\sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$.
After setting $\sin(z) = \frac{e^{iz}-e^{-iz}}{2i} = 2$, I proceeded with some further manipulation:
$e^{iz}-e^{-iz} = 4i$
$e^{i(a+bi)}-e^{-i(a+bi)} = 4i$
$e^{-b+ai}-e^{b-ai} = 4i$
$e^{-b}\cdot (\cos(a)+i\sin(a))-e^{b}\cdot (\cos(-a)+i\sin(-a)) = 4i$
$\underbrace{e^{-b}\cdot \cos(a) - e^{b}\cdot \cos(-a)}_{\text{0}} + e^{-b}\cdot i\sin(a) -e^{b}\cdot i\sin(-a) = 4i$
$e^{-b}\cdot \cos(a) - e^{b}\cdot \cos(a) = 0$ can further be simplified to due to cosine being an even function.
Thus $\cos(a) \cdot (e^{-b}-e^{b}) = 0$ and
$a= \frac{\pi}{2}+k\pi; k \in \mathbb{Z}$ or $b = 0$.
The imaginary part can be simplified to
$e^{-b}\cdot i\sin(a) - e^{b}\cdot i\sin(-a) = 4i$
$e^{-b}\cdot \sin(a) + e^{b}\cdot \sin(a) = 4$
$\sin(a) \cdot (e^{-b} + e^{b}) = 4$
Now if $b = 0$ then $\\sin(a) \cdot (2) = 4$ and there is no $a \in \mathbb{R}$ that would produce the required identity.
Likewise if $b \neq 0$, then $a= \frac{\pi}{2}+k\pi; k \in \mathbb{Z}$. In this case we get $0 \cdot (e^{-b} + e^{b}) = 4$ and again there is no $b \in \mathbb{R}$ that would suffice.
But apparently there is a solution so I must be making a mistake somewhere? I'd be grateful for any hint.
