Investigate the change of w=√(1-z^2) as z traces a closed…

Investigate the change of $w=\sqrt{1-z^2}$ as $z$ traces a closed Jordan curve $c$ in the cases: $\ 1^\circ\ c$ encloses one of $\pm1$; $\ 2^\circ\ c$ encloses both $\pm1$; $\ 3^\circ\ \pm1$ lie outside $c$.

Case $1^\circ$: $c$ encloses exactly one of $\pm1$

Say $c$ encloses $z=1$ but not $z=-1$: then $\arg(1-z)$ changes by $2\pi$ over the loop while $\arg(1+z)$ changes by $0$ (its zero $z=-1$ is outside $c$):

$$ \Delta[\arg(1-z^2)]=2\pi+0=2\pi \qquad\Longrightarrow\qquad \Delta[\arg w]=\frac12(2\pi)=\pi $$

A change of $\pi$ in argument means $w\to w\,e^{i\pi}=-w$: the function switches sign. (The case where $c$ encloses only $z=-1$ is identical by symmetry.) Two circuits are needed to return to the original value.

Case $2^\circ$: $c$ encloses both $\pm1$

Now both factors' arguments increase by $2\pi$:

$$ \Delta[\arg(1-z^2)]=2\pi+2\pi=4\pi \qquad\Longrightarrow\qquad \Delta[\arg w]=\frac12(4\pi)=2\pi $$

A full $2\pi$ change returns $w$ exactly to its starting value: $w\to w\,e^{i2\pi}=w$. The function is single-valued on any contour enclosing both branch points together.

Case $3^\circ$: $\pm1$ both lie outside $c$

Neither factor's argument changes:

$$ \Delta[\arg(1-z^2)]=0 \qquad\Longrightarrow\qquad \Delta[\arg w]=0 $$

$w$ trivially returns to its original value, as expected in any simply connected region avoiding both branch points.

General principle

For $w=\sqrt{f(z)}$ with $f$ having simple zeros at finitely many points: a contour enclosing an odd number of these zeros flips the sign of $w$; a contour enclosing an even number (including zero) returns $w$ unchanged.