A proposition of trigonometry, or complex numbers

Prop. If \(0,1,z,w\) all lie on a circle centered at \(1/2\) then \(zw\) is the orthogonal projection of \(0\) onto the line through \(z\) and \(w\).

Proof. If \(z,w\) to lie on that circle, then \[(z-\frac12)(\overline{z}-\frac12)=\frac14,\text{ which simplifies to }z\overline{z}=\frac{z+\overline{z}}2\] \[(w-\frac12)(\overline{w}-\frac12)=\frac14,\text{ which simplifies to }w\overline{w}=\frac{w+\overline{w}}2.\] Introducing names \(z=a+bi,w=c+di\) these equations turn into \[a^2+b^2=a,\quad c^2+d^2=c.\] Two things need to be shown: \((1)\) \(zw\) lies on the line through \(z\) and \(w\), and \((2)\) \(zw\) is perpendicular to \(z-w\).

\((1)\) Verify that the following determinant is zero. \begin{align*} \begin{vmatrix}1&a&b\\1&c&d\\1&ac-bd&ad+bc\end{vmatrix}&=acd+bc^2-acd+bd^2-a^2d-abc+abc-b^2d+ad-bc\\ &=b(c^2+d^2)-(a^2+b^2)d+ad-bc=bc-ad+ad-bc=0. \end{align*} \((2)\) Verify that \(zw/(z-w)\) is imaginary, or equivalently that \(zw(\overline z-\overline w)\) is imaginary. \begin{align*} zw(\overline z-\overline w)+\overline{zw(\overline z-\overline w)}&=z\overline zw-zw\overline w+z\overline z\overline w-\overline zw\overline w=z\overline z(w+\overline w)-(z+\overline z)w\overline w\\ &=\frac{z+\overline z}2(w+\overline w)-(z+\overline z)\frac{w+\overline w}2=\frac12(z+\overline z)(w+\overline w)(1-1)=0. \end{align*}

Extremely rational cubic polynomials

By an extremely rational cubic polynomial, I mean a cubic polynomial \(p\) such that both \(p\) and \(p'\) are products of rational linear factors. A parametrization will now be derived.

Seek to satisfy \(p=(a+x)(b+x)(c+x),\ p'=3(d+x)(e+x)\) as polynomials.
\begin{align*}
0=p'(-e)&=ab+ac+bc-2(a+b+c)e+3e^2 \\
(3e-a-b-c)^2&=(a+b+c)^2-3(ab+ac+bc) \\
&=a^2+b^2+c^2-ab-ac-bc
\end{align*}We thus need to parametrize triples \((a,b,c)\) such that the right-hand side is a square.

By the cubic number system, I mean triples \((a,b,c)\) with the following operations.
\begin{align*}
(a,b,c)+(d,e,f)&\equiv(a+d,b+e,c+f) \\
(a,b,c)(d,e,f)&\equiv(ad+bf+ce,ae+bd+cf,af+be+cd) \\
Q(a,b,c)&\equiv a^2+b^2+c^2-ab-ac-bc
\end{align*}This system is a close sibling of the usual complex number system. Its relevance to us is in the last definition. It is the expression we want to make into a square. Verify the identity \(Q(zw)=Q(z)Q(w)\) to realize that our task can be fulfilled by setting \((a,b,c)\equiv(u,v,w)^2\). Normally, one would use fractions \(a,b,c,u,v,w\) in this number system, but we can just as well use complex parameters \(u,v,w\).

Here is the parametrization.
\begin{align*}
p&=(a+x)(b+x)(c+x),\quad p'=3(d+x)(e+x),\quad p''=6(f+x) \\
a&=u^2+2vw,\quad b=w^2+2uv,\quad c=v^2+2uw\\
d,e&=\frac13((u+v+w)^2\pm Q(u,v,w)),\quad f=\frac13(u+v+w)^2\end{align*}
Here is an example.
\begin{align*}
(u,v,w)&=(1,2,7) \\
(a,b,c)&=(1^2+2\times 2\times 7,\ 7^2+2\times 1\times 2,\ 2^2+2\times 1\times 7)=(29,53,18) \\
p&=(29+x)(53+x)(18+x)=27666+3013x+100x^2+x^3 \\
Q(u,v,w)&=1^2+2^2+7^2-1\times 2-1\times 7-2\times 7=31 \\
Q(a,b,c)&=29\times 29+53\times 53+18\times 18-29\times 53-29\times 18-53\times 18=961 \\
d,e&=\frac13({10}^2\pm 31)=\frac{131}3,23\\
p'&=3(\frac{131}3+x)(23+x)=(131+3x)(69+x)=3013+200x+3x^2\\
f&=\frac{100}3,\quad p''=6(\frac{100}3+x)=200+6x

\end{align*}
An example involving complex parameters \(u,v,w\) is best given visually and interactively.