How to find all the solutions to cos$(z)=0$ 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 $k\in \mathbb{Z}$
 A: Logarithms are treacherous. It is safer and easier to note that we are looking at the equation $e^{iz}=-e^{-iz}$.  Let $z=a+ib$, where $a$ and $b$ are real. Then the norm of $e^{i(a+ib)}$ is $e^{-b}$, and the norm of $-e^{-iz}$ is $e^{b}$. If $b \ne 0$, the norms don't match. 
So $z$ has to be real, and we are in familiar territory.
A: No. The part where you seem to be assuming $\log\mathrm e^{\mathrm iz}=\mathrm i\mathrm e^{-y}x$ is wrong. Rather, (the principal value of) that logarithm is $\log\mathrm e^{\mathrm iz}=\mathrm iz$. (The natural logarithm is the inverse of the exponential function.)
You also got a factor of $2$ wrong in the arguments somewhere, presumably in trying to resolve the square. The possible arguments for numbers whose square is $-1$ are $\pi\left(\frac12+k\right)$, without the $2$.
Then by comparing real and imaginary parts you can conclude that $x=\pi\left(\frac12+k\right)$, $y=0$.
A: since you know that $cos(z) = \frac{e^{iz} + e^{-iz}}{2}$. So  $\frac{e^{iz} + e^{-iz}}{2}=0$ thus ${e^{iz} + e^{-iz}} = 0$ suppose you let $u = e^{iz}$. You now have $u + \frac{1}{u} = 0$, then $u^2 +1 =0$; hence $(u-i) (u+i) = 0$.  Therefore $u=i$ or $u=-i$. we now substitute back to get $e^{iz}=i$ or $e^{iz}=-i$, by taking natural log from both side we see that $e^{iz}=-i$ can not be a solution thus $e^{iz}=i$ is a solution. Hence $z= -iln(i)$. so the value that satisfied this equation is when $z= -iln(i)$ 
