A question in my complex analysis book (Gamelin's "Complex Analysis", question I.8.7) asks me to prove that $e^{iz} = \cos(z) \pm \sqrt{\cos^2(z) - 1}$. Using the identity $\cos(z) = \frac{e^{iz} + e^{-iz}}{2}$ and the quadratic formula this is easy enough:
$\begin{align*} 2\cos(z) &= e^{iz} + e^{-iz} \\ 2\cos(z) e^{iz} &= e^{2iz} + 1 \\ 0 &= e^{2iz} - 2\cos(z) e^{iz} + 1 \\ e^{iz} &= \frac{2\cos(z) \pm \sqrt{4\cos^2(z) - 4}}{2} \\ e^{iz} &= \cos(z) \pm \sqrt{\cos^2(z) - 1} \end{align*}$
But now I am trying to interpret this result.
- Does this imply that $e^{iz}$ is a multi-valued function, one value for the $+$ branch and one value for the $-$ branch?
- Since $\cos$ is an even function (i.e. $\cos(z) = \cos(-z)$ for all $z$), $$\cos(z) \pm \sqrt{\cos^2(z) - 1} = \cos(-z) \pm \sqrt{\cos^2(-z) - 1},$$ correct? But $e^{iz} \neq e^{-iz}$ in general.
Help on either or both bullets would be appreciated.