<s>[INST] Find every $z$s that fit $\cos(z) = -2$ I couldn't find any, I tried to write $\cos(z)$ as $\cos(x)\cos(iy)-\sin(x)\sin(iy)$ which then gave me
$\cos(x)\cosh(y) - i\sin(x)\sinh(y) = -2$ 
$\sin(x)=0$ so that imaginary part become $0$
now we have to find $\cosh(y) = -2$ which is not true for no $y$.
is it right or i made a mistake in my substitutions? 
 [/INST] Be:
$$
\cos z=\frac{e^{iz}+e^{-iz}}{2}=-2
$$
Then:
\begin{eqnarray}
\frac{e^{iz}+e^{-iz}}{2} &=& -2 \\
e^{iz}+e^{-iz} &=& -4  \\
e^{2iz} + 1 &=& -4e^{iz}  \\
\left(e^{iz}\right)^2 + 4\left(e^{iz}\right) + 1 &=& 0  \\
e^{iz} &=& \frac{-4 \pm \sqrt{16-4}}{2}     \\
e^{iz} &=& -2 \pm \sqrt{3}     \\
\end{eqnarray}
Then since $e^{iz}=\cos(z)+i\sin(z)$, $e^{iz}=e^{i(z+2k\pi)}$ with $k\in\mathbb{Z}$: 
\begin{eqnarray}
e^{i(z+2k_1\pi)} &=& -2 \pm \sqrt{3}     \\
i(z +2k_1\pi) &=& \ln\left(-2 \pm \sqrt{3}\right) \\
z &=& 2k_1\pi - i\ln\left(-2 \pm \sqrt{3}\right) \\
\end{eqnarray}
There are two set solution to differents solutions:
\begin{eqnarray}
z_{+} &=& 2k_1\pi - i\ln\left(-2 + \sqrt{3}\right) \\
z_{-} &=& 2k_1\pi - i\ln\left(-2 - \sqrt{3}\right) \\
\end{eqnarray}
For $z_{-}$:
\begin{eqnarray}
z_{-} &=& 2k_1\pi - i\ln\left(\left(2 + \sqrt{3}\right)e^{-i(1+2k_2)\pi}\right) \\
z_{-} &=& 2k_1\pi - i\left[\ln\left(2 + \sqrt{3}\right)-i(1+2k_2)\pi\right] \\
z_{-} &=& 2k_1\pi - (1+2k_2)\pi - i\ln\left(2 + \sqrt{3}\right) \\
z_{-} &=& (1+2k)\pi - i\ln\left(2 + \sqrt{3}\right) \\
\end{eqnarray}
Then, with $k\in\mathbb{Z}$ the solutions are:
\begin{eqnarray}
z_{+} &=& 2k\pi - i\ln\left(-2 + \sqrt{3}\right) \\
z_{-} &=& (1+2k)\pi - i\ln\left(2 + \sqrt{3}\right) \\
\end{eqnarray}
</s>