How would you go about finding all the solutions to $\cos(z)=0$, where $z\in \mathbb{C}$?
I have $\cos(z)=0 \implies (e^{iz})^2=-1 \implies \text{Log}(e^{iz})=\text{Log}(e^{i\pi(1/2+2k)})\qquad\text{ where }k\in \mathbb{Z}$
$\implies ie^{-y}x=i\pi(1/2+2k)$
$\implies z= e^{y}\pi(1/2+2k)+iy \qquad\forall k\in \mathbb{Z}\;\text{ and }\;\forall y\in \mathbb{R}$
Is this correct?
Thanks!
EDIT:
$e^{2iz}=e^{i\pi(1+2k)} \Rightarrow z=\pi/2+k\pi$ where $k\in \mathbb{Z}$