Square root of one minus z squared ($f(z)=\sqrt{1-z^2},f(0)=1$)

On $\mathbb{C}\setminus ((-\infty ,-1]\cup [1,\infty ))$, choose the holomorphic branch $f(z)=\sqrt{1-z}\sqrt{1+z}$, taking each factor's principal square root. This gives $f(x)>0$ for $-1<x<1$. Across the branch cuts, its limits are $f(x+i0)=-i\sqrt{x^2-1}$ and $f(x-i0)=+i\sqrt{x^2-1}$ for $x>1$, and $f(x+i0)=+i\sqrt{x^2-1}$ and $f(x-i0)=-i\sqrt{x^2-1}$ for $x<-1$. The endpoints are branch points; analytic continuation around a single endpoint changes the square-root sign.