Find all complex solutions of $\sin(z)=1$ Find all complex solutions of $\sin(z)=1$.
How would I go about this?

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 $k\in\mathbb{Z}.$
