How to find all solutions of $\cos(z) = 0$, where $z\in\mathbb{C}$?

Question

I am stuck on finding all solutions of the equation $\cos(z) = 0$, where $z\in\mathbb{C}$.

I found this proof, however, I cannot figure out the logic behind the last few steps. \begin{align} \cos(z) &= \dfrac{e^{iz}+e^{-iz}}{z} \\\\ &= 0 \\\\ e^{iz}+e^{-iz} &= 0 \\\\ e^{i(x+iy)} &= -e^{-i(x+iy)} \\\\ e^{ix}e^{-y} &= -e^{ix}e^y \\\\ e^{ix}e^{-y} &= e^{i\pi}e^{-ix}e^y \\\\ e^{-y}e^{ix} &= e^ye^{i(\pi-x)} \\\\ |e^{-y}e^{ix}| &= |e^ye^{i(\pi-x)}| \end{align}

This is where I get a bit confused. It follows from the modulus that, [ e^{-y} = e^y ] but why? Are we just shaving off the imaginary component? Or is it because $e^{-y}$ is real so it acts as the radius in $re^{i\theta}$?

So we have \begin{align} -y &= y \\\\ y &= 0 \\\\ e^{ix} &= e^{i(\pi-x)} \end{align}

Therefore we have from this that, $x - (\pi - x)= 2\pi k, \forall k\in\mathbb{Z}$... but why is that? Where is the $2\pi k$ coming from? I understand it has to do with the unit circle... is $i = 2\pi k$?

From that result we have \begin{align} 2x &= \pi + 2\pi k \\\\ x &= \dfrac{\pi}{2} + \pi k \\\\ z &= \dfrac{\pi}{2} + \pi k, \forall k\in\mathbb{Z} \end{align}

When I tried this problem on my own I did, \begin{align} \cos(z) &= \dfrac{e^{iz}+e^{-iz}}{2} \\\\ &= 0 \\\\ e^{iz} + e^{-iz} &= 0 \\\\ e^{iz} &= -e^{-iz} \\\\ e^{iz} &= e^{i\pi}e^{-iz} \\\\ e^{iz} &= e^{i(\pi-z)} \\\\ z &= \pi - z \\\\ 2z &= \pi \\\\ z &= \dfrac{\pi}{2} \end{align}

But this does not result in the infinite solutions it should result in.

Answer 1

Hint for your own attempt: Just because two exponentials are equal, that doesn't mean the exponents are equal.

Answer 2

To understand the proof that you found, this properties might be useful:

$\bullet$ For $x\in\mathbb{R}$, you have $e^{ix}=\cos x+i\sin x$, in particular $|e^{ix}|=1$.

$\bullet$ If $x\in\mathbb{R}$ and $e^{ix}=1$, then this means $\cos x+i\sin x=1$, comparing real and imaginary parts, this implies that $\cos x=1$ and $\sin x=0$, this happens when $x=2\pi k$ for some $k\in\mathbb{Z}$.

Answer 3

Maybe a faster trick is $$e^{iz} - e^{-iz} = e^{-iz}(e^{2iz} - 1) = 0$$

Then the right factor has to be zero, since the left factor can not.