The "positive square root" mentioned here is log$|z|$+$i$arg$(z)$ and the "negative square root" is log$|z|$+$i[$arg$(z)+2\pi]$. In the middle of the page I penciled in a minus sign in front of the $i$ because that is what you get if you use the positive square root for the case $\zeta>1$. But then the segments $A$ and $C$ would be mapped onto the lower half plane, and now I don't know what to make of their claim that this function is the inverse of $\sin(z)$. Do they want to use the negative square root instead which does give $i$? But then $B'$ is reversed which messes with the continuity.
In conclusion: I see now now how to explain this. The answer below is right (or at least has the right idea; I think it needs a little more explanation). I have deleted all the previous edits and comments for clarity because there was a lot written that was confusing and wrong. The idea is simply that the branch is chosen so that arg$(1-\zeta^{2})$ is $0$ when $|\zeta|<1$ (or any multiple of $4\pi$ because this corresponds to positive square root.) The question now reduces to figuring out what arg$(1-\zeta^{2})$ is for $\zeta>1$, which is not that hard. If $|z_{0}|<1$ and $z>1$, then as $\zeta$ travels from $z_{0}$ to $z$ in the upper half plane (recall $\zeta$ cannot belong to the two forbidden half lines in the lower half plane), $1-\zeta^{2}$ approaches $1-z^{2}$ from below. Hence arg$(-1)=-\pi$ and this explains why the text gets $i$ and not $-i$.
Best Answer
Note that it says "when $\zeta > 1$".
Consider the complex number $\zeta = x+i\epsilon$, where $x>1\gg\epsilon>0$. Then $1-\zeta^2 = -(x^2-1-\epsilon^2)- 2ix\epsilon$ has a negative real and imaginary part, and so will be mapped to the fourth quadrant near the imaginary axis. In particular, in the limit that $\epsilon\rightarrow 0^+$, $(1-\zeta^2)^{1/2} \rightarrow - i(x^2-1)^{1/2}$, and accordingly $(1-\zeta^2)^{-1/2} \rightarrow i(x^2-1)^{-1/2}$.
For $\zeta < 1$, the opposite is true. Then $(1-\zeta^2)^{-1/2}\rightarrow -i(x^2-1)^{-1/2}$, by a similar argument to the above.