Finding all values of $z\in\mathbb{C}$ such that $\cosh(z)=-1$

Question

I am trying to solve $\cosh(z)=-1$.

So far, I have: \begin{align} \cosh(z)&=-1 \\ \frac{e^z+e^{-z}}{2}&=-1 \\ (e^z+1)^2&=0 \\ e^z&=-1 \ (\text{multiplicity 2}) \\ \implies z&=-\text{Log}(-1)+(2k+1)\pi i, \ \ k\in\mathbb{Z} \\ &=-(\ln|-1|+i\text{Arg}(-1))+(2k+1)\pi i \\ &=2k\pi i \end{align} But I know this is incorrect, as $\cosh((2k+1)\pi i)=-1$. Where have I gone so wrong?