How to show $\arcsin{x} = \frac{\pi}{2} + i \ln{(x+\sqrt{x^2-1})}$? Is the following identity correct
$\arcsin{x} = \frac{\pi}{2} + i \ln{(x+\sqrt{x^2-1})}?$
Here, $x < 1$. How can we show that it is true? One way to see it is by differentiating, since
$\frac{d}{dx} (LHS) = \frac{d}{dx} (RHS).$ 
Thanks.
 A: One way to see why it is correct (but not to derive it) is to plug back into the definition and grind it through. Recall $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}.$$
Lets plug in $x=\frac{\pi}{2}+i\ln\left(z+\sqrt{z^{2}-1}\right).$ Then we have $$\sin(x)=\frac{1}{2i}\left(\exp\left(\frac{\pi i}{2}+-\ln\left(z+\sqrt{z^{2}-1}\right)\right)-\exp\left(\frac{-i\pi}{2}+\ln\left(z+\sqrt{z^{2}-1}\right)\right)\right)$$
$$=\frac{1}{2}\left(\exp\left(-\ln\left(z+\sqrt{z^{2}-1}\right)\right)+\exp\left(\frac{-i\pi}{2}+\ln\left(z+\sqrt{z^{2}-1}\right)\right)\right)$$
$$=\frac{1}{2}\left(\frac{1}{z+\sqrt{z^{2}-1}}+z+\sqrt{z^{2}-1}\right)=z.$$
The last equality follows from rationalizing the denominator, and the cancellations that follow.
Hope that helps,
A: Using $\sin(t) = \frac{e^{it} - e^{-it}}{2i}$ we see that if $t$ satisfies $\sin(t) = x$ and $iw = e^{-it}$, we have $x = \frac{w + 1/w}{2}$ so $2 x w = w^2 + 1$.  The solutions of this quadratic equation for $w$ are $w = x + \sqrt{x^2-1}$ (for both branches of the square root).  Now 
$w = e^{-it -i \pi/2}$, i.e. $-it - i \pi/2 = \log w$ (for one of the branches of log), or $t = \pi/2 + i \log(x + \sqrt{x^2-1})$.  Now the question is, if $t = \arcsin(x)$ (presumably using the principal branch of arcsin), what are the correct branches of the square root and log?  For $x = 0$,
$\arcsin(0) = 0 = \pi/2 + i \log(0+\sqrt{-1})$ using the principal branches of the square root and log, and other choices would give us different multiples of $\pi$.  However, $\pi/2 + i \log(x + \sqrt{x^2-1})$ (using the principal branches) has branch cuts in different places than the principal branch of $\arcsin(x)$, so some caution must be used.  It looks to me like the formula (using principal branches) is valid for real $x$ with $-1 \le x \le 1$, for all imaginary $x$, and when $\Re(x) \Im(x) > 0$. 
A: Write $x=cos(\theta)$ for some $\theta$, and notice that
$$\sqrt{x^2-1} = i \sqrt{1-x^2} = i \operatorname{sin}(\theta)$$
So the right hand side is:
$$\frac{\pi}{2} + i \operatorname{ln}(\operatorname{cos}{\theta} + i \operatorname{sin}{\theta})$$
But $\operatorname{ln}(\operatorname{cos}{\theta} + i \operatorname{sin}{\theta})$ is $i\theta + 2\pi k$ for some $k$. So the right hand side is:
$$ \frac{\pi}{2} - \theta - 2 \pi k$$
But the left hand side is:
$$\operatorname{arcsin}(\operatorname{cos}(\theta))$$
