# Butterfly in Complex Numbers

### Proof

Let $A, B, C, D$ be points on the unit circle and $M$ the intersection of $AB$ and $CD$.The equation of the points $Z$ on the secant line through $A$ and $B$ is

$z+ab\,\overline{z}=a+b.$

The lowercase letters refer to the complex numbers representing the corresponding points.Since $M$ is a point of $AB$ and $CD$ we have

$\displaystyle b=\frac{m-a}{1-a\,\overline{m}}$

and

$\displaystyle d=\frac{m-c}{1-c\,\overline{m}}.$

Let the secant $PQ$ be orthogonal to $OM$, hence tangent to the circle around $O$ through $M$. For a point $Z$ an $PQ$ we have therefore the equation

$z\,\overline{m}+\overline{z}\,m=2m\overline{m}.$

The intersection $X$ of $PQ$ and $AD$ can be obtained as a common solution of the equation of $PQ$ and the equation $z+ad\overline{z}=a+d$ of the secant line throught $A$ and $D$

$\displaystyle x=\frac{a\,m+d\,m-2ad\,m\overline{m}}{m-ad\,\overline{m}}.$

Insert the expression for $d$ from above and write $x$ as

$\displaystyle x=\frac{m(-a+c-m-ac\,\overline{m}+2a\,m\overline{m})} {-m-ac\,\overline{m}+a\,m\overline{m}+c\,m\overline{m}}.$

By the same argument, but inserting the expression for $b$ , we get for the intersection $Y$ of $PQ$ and $BC$ simply by interchanging $a$ and $c$

$\displaystyle y=\frac{m(-c+a-m-ac\,\overline{m}+2c\,m\overline{m})} {-m-ac\,\overline{m}+a\,m\overline{m}+c\,m\overline{m}}.$

The sum of $x$ and $y$ is indeed $2m$ and $M$ is the midpoint of $XY$.