Find all complex solutions of $\sin(z)=1$ Find all complex solutions of $\sin(z)=1$.
How would I go about this?
 A: First, let$$sin(z) = {e^{iz} - e^{-iz}\over 2i}.$$ Then we see that $${e^{iz} - e^{-iz}\over 2i} = 1.$$ This gives us $${e^{iz} - e^{-iz} - 2i = 0}.$$ Now we will multiply both sides of this equation by $e^{iz}$  to obtain $${e^{2iz} -2ie^{iz} - 1 = 0}.$$ We can use a substitution to simplify this equation into a quadratic equation. Let $w = e^{iz}$. $$w^2 -2iw -1 = 0.$$ This quadratic equation factors to become $$(w - i)^2 = 0.$$ Thus $e^{iz} = i$ and so $e^{iz} = e^{i\pi\over2}$ which implies that $e^{{i}({{z} - {\pi\over2}})} = 1$ where $z- {\pi\over 2} \in 2k\pi$ for $k\in\mathbb{Z}.$
A: Hint: set $w=e^{iz}$; then, by definition of $\sin$ on the complex plane,
$$
\sin z=\frac{w-w^{-1}}{2i}
$$
so your equation becomes
$$
w^2-2iw-1=0
$$
A: As egreg suggested, $\sin(z) = \frac{1}{2i} \left( e^{iz} - e^{-iz} \right)$ for all $z \in \mathbb{C}$. So,
$$ \Big( \sin(z) = 1 \Big) \Leftrightarrow \Big( e^{2iz} - 2ie^{iz} - 1 = 0 \Big) $$
So $z \in \mathbb{C}$ is such that $\sin(z) = 1$ if and only if $e^{iz}$ is a root of the polynomial $X^{2} - 2iX -1 \in \mathbb{C}[X]$. Since $X^{2} - 2iX -1 = (X-i)^{2}$, we get : $e^{iz} = i = e^{i\frac{\pi}{2}}$.
$$ 
\begin{eqnarray*}
\Big( e^{iz} = e^{i\frac{\pi}{2}} \Big) & \Leftrightarrow & \Big( e^{i(z-\frac{\pi}{2})} = 1 \Big) \\
 & \Leftrightarrow & z - \frac{\pi}{2} \in 2\pi \mathbb{Z} \\
\end{eqnarray*}
$$
