Find the algebraic branch points of w=√((1-z^2)(1-k^2z^2))…
Find the algebraic branch points of $w=\sqrt{(1-z^2)(1-k^2z^2)}$, $0<k<1$, and use branch cuts to define single-valued branches.
Explanation of Method
Locate the zeros of the radicand (each contributes a simple branch point), then check the behavior at infinity by substituting $\zeta=1/z$. Since the radicand is a degree-$4$ (even-degree) polynomial in $z$, the point at infinity turns out to be regular, not a branch point — a key difference from the odd-degree case seen in $\sqrt{z-a}$.
Step 1 · Zeros of the radicand
$$ (1-z^2)(1-k^2z^2)=0 \ \Longrightarrow\ z=\pm1 \quad\text{or}\quad z=\pm\frac1k $$
Since $0<k<1$, these four points are ordered on the real axis: $\ -\dfrac1k<-1<1<\dfrac1k$.
Step 2 · Confirm each is a branch point
Near $z=1$, write $z=1+\varepsilon e^{i\phi}$: $\ 1-z^2\approx-2\varepsilon e^{i\phi}$, while $1-k^2z^2$ stays near the nonzero value $1-k^2\neq0$. As $\phi:0\to2\pi$, $\arg(1-z^2)$ changes by $2\pi$, so $\Delta[\arg w]=\frac12(2\pi)=\pi$ — $w$ does not return to its value, confirming $z=1$ is a branch point. The same argument applies at $z=-1,\ \pm\frac1k$.
Step 3 · Behavior at infinity
Substitute $\zeta=1/z$. The radicand becomes:
$$ (1-z^2)(1-k^2z^2)=\frac{k^2-(1+k^2)\zeta^2+\zeta^4}{\zeta^4} $$
As $\zeta\to0$, the numerator $\to k^2\neq0$ — a nonzero constant, so the numerator's square root is single-valued (analytic, non-branching) near $\zeta=0$. Therefore:
$$ w=\frac{\sqrt{k^2-(1+k^2)\zeta^2+\zeta^4}}{\zeta^2}\ \sim\ \frac{k}{\zeta^2}=kz^2 \quad(\zeta\to0) $$
a single-valued pole of order $2$ — not a branch point. (A quick check: on a giant loop of radius $R\to\infty$, the radicand $\sim-k^2z^4$, so $\arg w$ changes by $\frac12(8\pi)=4\pi$, a multiple of $2\pi$ — confirming $w$ returns to its value after one large loop.)
Step 4 · Choosing branch cuts
With exactly four finite branch points and none at infinity, pair them with two cuts along the real axis:
$$ \left[-\frac1k,-1\right] \quad\text{and}\quad [1,\tfrac1k] $$
On the cut plane $\mathbb C\setminus\left(\left[-\frac1k,-1\right]\cup\left[1,\frac1k\right]\right)$, the function is single-valued.
Convention on the real axis
For $z\in(-1,1)$, both factors $1-z^2$ and $1-k^2z^2$ are positive, so the natural (principal) branch gives $w=\sqrt{(1-z^2)(1-k^2z^2)}>0$ there.
Remark · connection to elliptic integrals
This function appears in $\displaystyle\int\frac{dz}{\sqrt{(1-z^2)(1-k^2z^2)}}$, the elliptic integral of the first kind with modulus $k$; the two branch cuts here correspond to the two "sheets" that glue into a torus (genus-$1$ Riemann surface) once the point at infinity — being regular, not a branch point — is included as an ordinary point on each sheet.
Result
Because the radicand has even degree $4$, all four zeros pair up without needing infinity as a fifth branch point — unlike odd-degree radicands such as $\sqrt{z-a}$, where infinity is always a branch point.