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Question Number 86476 by DuDono last updated on 28/Mar/20

$$\mathrm{show}\mathrm{that}$$$${\sin }^{-1}\alpha =-i\ln (\alpha \pm \sqrt{{\alpha }^2-1})-\frac{\pi }{2}$$

Commented by MJS last updated on 28/Mar/20

$$\mathrm{the}\mathrm{path}\mathrm{is}\mathrm{this}:$$$$y=\sin x$$$$y=\frac{{\mathrm{e}}^{\mathrm{i}x}-{\mathrm{e}}^{-\mathrm{i}x}}{2i}$$$$2\mathrm{i}y={\mathrm{e}}^{\mathrm{i}x}-\frac{1}{{\mathrm{e}}^{\mathrm{i}x}}$$$$\mathrm{let}{\mathrm{e}}^{\mathrm{i}x}=t\Rightarrow x=2n\pi -\mathrm{i}\ln t$$$$2\mathrm{i}y=t-\frac{1}{t}\Rightarrow t=\mathrm{i}y\pm \sqrt{1-y^2}=\mathrm{i}(y\pm \sqrt{y^2-1})$$$$\mathrm{now}\mathrm{insert}\mathrm{into}x=2n\pi -\mathrm{i}\ln t$$$$...$$

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