Question: As seen in underlined part of below attached picture: 
$±\sqrt{1-z^2}$ is implied by $\sqrt{1-z^2}$ But how? what this means?
Why they considered only positive root?
My attempt: let's try by considering negative root!
Say $e^{iw}=iz-\sqrt{1-z^2}$ then, By taking natural logarithm on both sides we get,
$iw=\ln(iz-\sqrt{1-z^2})$
$w=\frac{1}{i}\ln(iz-\sqrt{1-z^2})$
Hence if we put $z=0$, then we get,
$w=\frac{1}{i}\ln(-1)$
as logarithm function is not defined for negative numbers, hence we see that,$w=\sin^{-1}z$ will not be defined at $z=0$
But we have $sin^{-1}(0)=0$ hence we must consider, positive root!
Is am I correct?
Further, is considering negative root is wrong? What we get by considering negative root?
Please help...