Let $f(z)$ be the principal branch of $z^{1/2}$ on $\mathbb{C}\setminus (-\infty,0]$, i.e. we insist $\theta=\arg(z)\in(-\pi,\pi)$ and define $f(z)=\sqrt{|z|}e^{i\theta/2}$.
Let $g(z)$ be an analytic branch of $(-(z+1)(z-1))^{1/2}$ on $\mathbb{C}\setminus [-1,1]$ with $g(2i)=\sqrt{5}$, e.g. I insisted $\theta_1=\arg(z-1)\in(-\pi,\pi]$ and $\theta_{-1}=\arg(z+1)\in(-\pi,\pi]$ and defined $$g(z)=-i\sqrt{|z+1|}e^{i\theta_{-1}/2}\sqrt{|z-1|}e^{i\theta_{1}/2}$$ Note $g(z)=-if(z+1)f(z-1)$ for $z\notin(-\infty,1]$.
Show $$g(1/z)=-\frac{i}{z}f(1-z^2) \;\text{ for }0<|z|<1$$
I cannot seem to obtain this final result. I understand why the restrictions on $z$ given are necessary for the terms $g(1/z)$, $f(1-z^2)$ to be defined but beyond that I have made little headway. Any help would be appreciated!
Thanks to Maxim's comment for this answer:
Let $D=\mathbb{C}\setminus [-1,1]$. Note $g$ is an analytic branch of $(1-z^2)^{1/2}$ on $D$.
Consider $h(z)=-izf(1-1/z^2)$. $h$ is (obviously) another branch of $(1-z^2)^{1/2}$, and is well-defined on $D$ since $z\in[-1,1]\iff 1-1/z^2\in (-\infty,0]$. But note $h(2i)=(-i)(2i)f(5/4)=\sqrt{5}=g(2i)$, and since there are exactly two branches of a square root (on a given domain), equality at a point implies equality at all points. That is, $g(z)=h(z) \forall z\in D$.
In particular then we have $g(z)=-izf(1-1/z^2)$ for $|z|>1$ and hence $g(1/z)=-\frac{i}{z}f(1-z^2)$ for $0<|z|<1$.