I am trying to solve $ \cosh(z)=-5, \ \forall z\in\mathbb{C}$.
My attempt:
\begin{align} \cosh(z)&=-5 \\ \frac{e^z+e^{-z}}{2}&=-5 \\ (e^z)^2+10e^z+1&=0 \\ e^z&=\frac{-10+ \sqrt{96}}{2} \\ e^z&=-5+ \sqrt{6} \\ \implies z&=\pm\log(-5+2\sqrt{6}) \ \ \ \text{(log is mulitvalued)} \end{align}
The answer provided is $$\pm\log(5+2\sqrt{6})+(2k+1)\pi i \ \ k\in\mathbb{Z}$$ My questions is, how is
$$\pm\log(-5+2\sqrt{6}) \equiv\pm\log(5+2\sqrt{6})+(2k+1)\pi i \ \ k\in\mathbb{Z} \ \ \text{(where log is mulitvalued)}$$
Note that $-5 \pm \sqrt 6$ is a real number so Arg$( -5 \pm \sqrt 6 )=0$