Using $$\sin z = \frac{e^{iz}-e^{-iz}}{2i}$$ Prove $$\sin^{-1} z =\frac{1}{i}\ln(iz+\sqrt{1-z^2}) $$
Attempted solution: Let $\sin z = u$ and $e^{iz} = v$. \begin{align*}& 2iu = v - \frac{1}{v}\\&v^2 - 2iuv - 1 = 0\\&v = \frac{2iu \pm \sqrt{-4u^2 + 4}}{2} = iu \pm \sqrt{1-u^2}\\&e^{iz} = iu \pm \sqrt{1-u^2}\\&iz = \ln\left(iu \pm \sqrt{1-u^2}\right)\\&z = \frac{1}{i}\ln\left(iu + \sqrt{1-u^2}\right)\end{align*}
But I'm looking for $\sin^{-1}z$ not $\sin^{-1}u$, so I'm not sure where to go from here.