1
$\begingroup$

Given 4 points $A,B,C,D\in\mathbb{C}$, is there an inversion, that would map these points to consecutive vertices of a parallelogram?

Here is my take on this. Suppose points $z_1,z_2,z_3,z_4$ are mapped to consecutive vertices of a parallelogram by the inversion $z\mapsto\frac{1}{\overline z}$. Then we have the following relation: $\frac{1}{\overline z_2}-\frac{1}{\overline z_1}=\frac{1}{\overline z_3}-\frac{1}{\overline z_4}$

Not sure of how to proceed.

Thanks.

  • 1
    you should check your "simplification" – 2017-02-17
  • 0
    Why should it be so? Width of parallelogram nearer to pole gets wider after inversion – 2017-02-17
  • 0
    @mercio Thanks, edited. – 2017-02-17
  • 1
    @lanskey Yes, that is always the case. – 2017-02-17
  • 1
    Hint: the general form of an inversion is $z \mapsto c + \frac{r^2}{\bar z - \bar c}\,$. The parallelogram condition will give (in general) a quadratic equation in $c$. – 2017-02-17
  • 1
    @dxiv I get it now. I was trying to disprove the claim, therefore was looking for a relation on the points, but should have looked for the center of inversion instead. – 2017-02-17

1 Answers 1

1

There is a purely geometric proof, but I do not feel like writing it. Instead, one can also argue like this:

Calculate the cross-ratio $$cr(z_1, z_2, z_3, z_4) = \frac{(z_1 - z_2) (z_3-z_4)}{(z_2 - z_3)(z_4-z_1)} = \alpha$$ Take a complex number $a$ such that $a^2 = \alpha$ and pick an arbitrary point $w_1$. Then form the points $w_2 = w_1 + a, \,\,\,\, w_3 = w_1 + a + 1, \,\,\,\, w_4 = w_1 + 1$. Together with $w_1$ they form a parallelogram with edges $a$ and $1$. Then if you calculate the cross ratio $$cr(w_1, w_2, w_3, w_4) = \frac{(w_1 - w_2) (w_3-w_4)}{(w_2 - w_3)(w_4-w_1)} = \frac{(-a) (a)}{(-1)(1)} = a^2 = \alpha = cr(z_1, z_2, z_3, z_4)$$ Since the cross-ratio is conformal (Mobius) invariant, there exists a Mobius transformation (i.e. a complex linear fractional transformation) that maps the four points $z_1, z_2, z_3$ and $z_4$ to the four points $w_1, w_2, w_3$ and $w_4$. It can be constructed by solving the equation

$$cr(w, w_2, w_3, w_4) = cr(z, z_2, z_3, z_4)$$ for $w$ obtaining a linear-fractional map $w = T(z)$ with coefficients formed out of the points $z_2, z_3, z_4$ and $w_2, w_3, w_4$. The cross-ratio condition guarantees that $w_j = T(z_j)$ for $j=1,2,3,4$. Now, take the centroid $G_w$ (the intersection point of the diagonals) of parallelogram $w_1w_2w_3w_4$ and map it back to $G_z = T^{-1}(G_w)$ via the inverse Mobius map $T^{-1}$. Any inversion with respect to a circle centered at $G_z$ and arbitrary radius will map $z_1z_2z_3z_4$ to a parallelogram.