What are the zeros of $\cos z$? My attempt $:$
I know that $\cos z = \frac {e^{iz} + e^{-iz}} {2}$. So $\cos z = 0$ if and only if $e^{2iz} = -1$. Let $z = x + iy$. Then we have $e^{-2y + i 2x} = -1$ i.e. $e^{-2y} \cos 2x = -1$ and $e^{-2y} \sin 2x = 0$. $\implies$ $y = 0$ and $x = \frac {(2n - 1) \pi} {2}$, $n \in \mathbb N$. Hence the solutions are given by $z = \frac {(2n - 1) \pi} {2}$, $n \in \mathbb N$ which are precisely the solutions of $\cos x = 0$ when $x \in \mathbb R$.
Is it correct at all? Please check it.
Thank you in advance. 
 A: Yes, but to me it is somewhat circular (since you have to know the zeroes of $\sin z$).
Perhaps a better way (since it doesn't involve determining $x$ and $y$ separately) is to use the periodicity of the exponential (it has period $2\pi i$). Since you know that $e^{\pi i}=-1$, and since $e^r=e^s$ if and only if $r-s$ is an integral multiple of the period, you can write
$$e^{2iz} = -1$$
$$e^{2iz} = e^{\pi i}$$
so
$$2iz-\pi i = 2\pi i k$$
$$2iz = (2k+1)\pi i$$
$$z = (2k+1)\frac{\pi}{2}$$
for some integer $k$
A: Yes, that's fine, although you want all integers. A more natural way is via the cute identity
$$ \lvert \cos{(x+iy)} \rvert^2 = \cos^2{x} + \sinh^2{y}, $$
which is easy to prove with some trigonometric identities. Then the result is obvious, since $\sinh{y} \neq 0$ if $y \neq 0$ for real $y$.
A: Well, it's almost right : you should define $n \in \mathbb{Z}$ and not $n \in \mathbb{N}$.
Plus, simply for curiosity, what are the arguments behind your implication sign?
A: 
My attempt $:$
I know that $\cos z = \frac {e^{iz} + e^{-iz}} {2}$. So $\cos z = 0$
  if and only if $e^{2iz} = -1$. Let $z = x + iy$. Then we have $e^{-2y+ i2x}
= -1$ i.e. $e^{-2y} \cos 2x = -1$ and $e^{-2y} \sin 2x = 0$.

That is correct.

$\implies$ $y = 0$ and $x = \frac {(2n - 1) \pi} {2}$,

This step is correct but not clear to follow. You should rather say $e^{-2y}\sin(2x)=0 \implies \sin(2x)=0$ as $e^{-2y}\neq 0$. Hence, $x=\frac{\pi}{2}k$ in which $k\in \mathbb{Z}$. 
If you plug this into the first eqaution $e^{-2y} \cos 2x = -1$ you will obtain $e^{-2y}(-1)^k=-1 \implies e^{-2y}=(-1)^{k+1}$. 
Can you complete it from here? 
A: Hint:)
A simple approach is for $e^{2iz}=-1$ we see $2iz=\ln|-1|+i(-\pi+2k\pi)$ so $\color{blue}{z=\dfrac{2k-1}{2}\pi}$ means that the zeros of $w=\cos z$ are reals.
