Proving $\sec^{-1}(z) = -i\log\left[ \frac{1}{z} + \left(\frac{1}{z^2} - 1 \right)^{1/2}\right]$ [closed]

Prove that $$\sec^{-1}(z) = -i\log\left[ \frac{1}{z} + \left(\frac{1}{z^2} - 1 \right)^{1/2}\right]$$

Here are some hints: \begin{align} \frac{e^{iz}+e^{-iz}}{2}=\frac{\cos z+i\sin z+\cos(-z)+i\sin(-z)}{2}=\cos z. \end{align} Thus $y(z):=\sec z=\frac{2}{e^{iz}+e^{-iz}}$. Solve for $z$ in terms of $y$ by factoring out $e^{-iz}$ and writing $\alpha=e^{iz}$. We end up solving a quadratic.

$$\sec(x) = z\implies \color{blue}{x = \sec^{-1}z} \implies z = \frac 2{e^{ix} + e^{-ix}}$$ considering princ. value only (Now, we are interested in finding x)

let $$ u = e^{ix} \implies z = \frac 2{u + \frac1u}$$ $$u = \frac {1+ (1-z^2)^{\frac12}}{z}$$ $$\ln u = \ln (e^{ix}) =ix=\ln \left(\frac {1+ (1-z^2)^{1/2}}{z}\right)$$ $$x = \frac1i......$$