| The determinant is 0 iff the points are collinear or cocircular.
Let me use "z" as an abbreviation for the column z1-z2-z3-z4, etc.
Let z = x + iy where x and y are real (columns). By subtracting the third
column from the second, pulling out a factor of 2, subtracting the (new) second
column from the third, and pulling out a factor of i, we see that the old
determinant is 0 iff the determinant |x^2+y^2 x y 1| is.
Now if all 4 points are identical, the determinant is zero. If not, pick two
that are distinct; WLOG let these be z1 and z2. The points on the line
determined by z1 and z2 are exactly those of the form a*z1 + (1-a)*z2; so if the
4 points are collinear, we can subtract multiples of the top two rows from the
bottom two rows so as to make the 2x3 rectangle in the lower right corner all 0.
So the determinant is zero in this case.
If no two points are collinear, then z1, z2, and z3 determine a unique circle
which passes through them; call the center of this circle z0, and let its radius
be r>0. Subtracting and adding multiples of the last three columns from the
first one, the determinant |x^2+y^2 x y 1| = |(x-x0)^2+(y-y0)^2 x y 1|, which is
clearly zero whenever z4 lies on the circle (since the first column is a
multiple of the last one). If z4 doesn't lie on the circle, subtract r^2 times
the last column from the first one; the resulting determinant has only one
nonzero entry in the first column, so it is zero iff the 3x3 determinant |x y 1|
is zero. But it is easy to see that the three rows are linearly independent if
z1,z2,z3 are not collinear, so the determinant is zero iff z1-z4 are collinear
or cocircular, QED.
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| My analysis has an error at the point where it says "if no 2 points are
collinear..." which is a braino in any case. Try this instead: if all 4 points
are collinear, we can proceed as indicated. Otherwise, some 3 are not
collinear, WLOG they can be taken to be z1,z2, and z3, and they determine a
unique circle; and again we can proceed as indicated.
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