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Put $A=\wp (w)$ and $B={\wp }^{\prime }(w)$. At $z=0$, the first column has $\wp (z)=z^{-2}+O(z^2)$ and ${\wp }^{\prime }(z)=-2z^{-3}+O(z)$, whereas the third has $\wp (-z-w)=A+Bz+O(z^2)$. Expanding the determinant, the coefficient multiplying ${\wp }^{\prime }(z)$ is $\wp (-z-w)-A=O(z)$, so the apparent cubic pole reduces to order at most two. The only other possible pole is at $z=-w$, and $f(-z-w)=-f(z)$ gives the same bound there. Thus, if $f$ is nonconstant, its total pole order and valency are at most four. A constant case is handled by finding a zero below.

Column coincidences give zeros at $z=w$, at $z=-2w$, and at the four solutions of $2z=-w$ on $\mathbb{C}/\Lambda$. The latter are distinct because $\Lambda /2\Lambda$ has four elements. If $3w\notin \Lambda$, the first two points are distinct from each other and from these four solutions.

One exceptional pole issue needs care: when $2w\in \Lambda$, the first two prospective zeros are $w$ and $0$, the possible pole locations. They are actually removable zeros. Here $B=0$, and, with $C={\wp }^{\prime \prime }(w)$, evenness about this half-period givesThe determinant becomes $(A-\wp (z)){\wp }^{\prime }(-z-w)+{\wp }^{\prime }(z)(\wp (-z-w)-A)$. Its two $C/z$ contributions cancel and it is $O(z)$; reflection transfers this conclusion to $z=w$. Thus there really are six distinct zeros also in this case.

If $3w\in \Lambda$ but $w\notin \Lambda$, then $z=w=-2w$ is one of the four solutions of $2z=-w$. All three columns coincide there, with no pole. Both the determinant and its first derivative vanish, since each term in the derivative still has two identical columns. Hence there are four distinct zeros, at least one multiple, giving at least five zeros counted with multiplicity.

For a nonconstant elliptic function, zeros and poles in a period parallelogram have equal total multiplicity, by the argument principle. Either count contradicts total pole order at most four. A constant with a zero is zero, soThis is the elliptic collinearity determinant. Consequently, when $z_1+z_2+z_3\in \Lambda$, the three vectors $(1,\wp (z_i),{\wp }^{\prime }(z_i))$ have zero determinant. The three points on the Weierstrass equation of an elliptic curve are therefore collinear in ${\mathbb{C}}^2$.