= Square root of one minus z squared
{title2=$f(z)=\sqrt{1-z^2},\quad f(0)=1$}
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.
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