Solution
= Solution
On $\mathbb C\setminus[-1,1]$, the chosen <branch of a multivalued function> is characterized by $f(z)^2=z^2-1$ and $f(x)>0$ for $x>1$; equivalently $f(z)/z\to1$ as $z\to\infty$. Define $h(z)=-f(-z)$. Then $h(z)^2=z^2-1$ and $h(z)/z\to1$ at infinity, so uniqueness of the normalized <holomorphic square root> gives $h=f$. Therefore \b[$f(-z)=-f(z)$].