Let G=\mathbb{C}\backslash \{x \in \mathbb{R} | |x|\ge 1 \}. We want to find a holomorphic function $f$ so that $f(0)=i,\qquad\text{and}\qquad (f(z))^2 = z^2 - 1 \text{ for all }z\in G.$
VVVs work:
$(f(z))^{2} = z^{2}-1 = (z-1)(z+1)$
let l be a logarithmic function, then $f_{2} = \sqrt{z-1} = exp(\frac{1}{2}l(z-1))$ for z\in \mathbb{C} \backslash \{ x \in \mathbb{R} | x \ge + 1\} and $f_{1} = \sqrt{z+1}=exp(\frac{1}{2}l(z+1))$ for $z\in \mathbb{C}\backslash \{x \in \mathbb{R}| x\le -1\}$
and one chooses: $l= \frac{1}{2}log(x^{2}+y^{2})+iarctan(y/x)$
and this gives : $f_{1}f_{2} = \sqrt{z^{2}-1}$
attempt 2
Directly one sees also that $f(z) = \pm \sqrt{(z^{2}-1)}$ and with the condition $f(0)=i$ it follows that $f(z) = \sqrt{z^{2}-1}$
Is VVVs work correct ?