# Quadratics with Complex Roots, Lill’s Circle and the Hyperbola

I already have a few papers that show how Lill’s method can be used to solve quadratic equations. In my paper “Lill’s Method and Graphical Solutions to Quadratic Equations” I showed 2 methods that can solve quadratics with real roots and 1 method for solving quadratics with complex roots. That paper also mentions that the complex roots of a quadratic equation are the inverse of each other with respect to Lill’s circle. In my post “Lill’s Method and Geometric Solutions to Quadratic Equations With Complex Roots” I show how to obtain the complex roots of a second degree polynomial using tangents to Lill’s circle.

I also have a few posts that show how hyperbolas and Lill’s method can be used to solve cubic equations (my cubic posts link). In this post I want to show how hyperbolas are connected to the solutions of quadratic equations and Lill’s circle. I will focus only on quadratics with complex roots.

### Systems of Equations, Hyperbola and Lill’s Circle

Let’s have a quadratic with complex roots p(x)=az2 +bz+c, where z is a complex variable. If we let z=x+iy, where i is the imaginary number, then p(x)=a(x+iy)2 +b(x+iy)+c. If we multiply the terms (and let p(x)=0 to get the roots) we get:

a(x2 +2ixy+i2 y2 )+bx+iby+c=0

We know that i2 =-1 so the equation can be written as:

ax2 +2aixy-ay2 +bx+iby+c=0

We have terms that are real and terms that are imaginary since they are multiplied by i. We can make a system of 2 equations, one equation for the real part and one equation for the imaginary part:

Real: ax2 -ay2 +bx+c=0 (rectangular hyperbola)

Imaginary: 2axy+by=0 (describes the x-axis and a line perpendicular to the x-axis)

The intersection of these 2 equations give the roots of p(x). The real part is a rectangular hyperbola with the center at the point (-b/2a,0). The asymptotes of the rectangular hyperbola make 45 degree angles with the x-axis . The imaginary part describes the x-axis and the perpendicular to the x-axis at (-b/2a,0). So the imaginary part describes 2 lines perpendicular to each other.

The Lill circle representation of the monic version of p(x)= z2 +(b/a)z+c/a intersects the rectangular hyperbola ax2 -ay2 +bx+c=0 at the same points it intersects the line parralel to the x-axis and that passes through the center of Lill’s circle . I will not prove the property, but I will show a random example.

### Example

Let’s use p(x)=2z2 +1.4z+3. The roots of the equation can be obtained using Wolfram Alpha. The lines and the curve generated by the 2 equations  2 x^(2)-2 y^(2)+1.4 x+3=0 and 2*2 x y+1.4 y=0 can be seen below. You can also see the Wolfram Alpha result.

Now we can superimpose the Lill method representation of the monic polynomial z2 +0.7z+1.5. P0P1 =1, P1P2=0.7 and P2P3=1.5 . The Lill circle has the center at point C and passes trough P0 and P3. We can see that the line parallel to the x-axis and that passes through the center C intersects Lill’s circle of the quadratic at the points A and B. A and B are also the points of intersection between the Lill circle and the hyperbola 2 x^(2)-2 y^(2)+1.4 x+3=0.

### Final Notes

To describe a hyperbola we need 5 points (GeoGebra has a drawing function that requires 5 points for a general conic). We know that points A and B are on the hyperbola. We can reflect the points A and B about the x-axis to obtain 2 additional points on the hyperbola. Below we can see the center H of the rectangular hyperbola 2 x^(2)-2 y^(2)+1.4 x+3=0 and the 2 asymptotes that make 45 degrees with the x-axis.

To obtain the fifth point we can use the “Hyperbola Secant Theorem” that I mentioned in other posts. The theorem says :  On any secant of an hyperbola, the segments between the curve and the asymptotes are equal. We can draw a secant through a known point on the hyperbola and we can use the theorem to obtain a new point on the hyperbola.

To make the things more complicated, we can try to work with quadratics that have complex coefficients. If the quadratic has complex coefficients, I believe that the equation of the imaginary terms will also describe a hyperbola, and this hyperbola passes through the center C of Lill circle. We already see that the line described by 2axy+by=0 passes through C . In our example the line HC is the line that passes through Lill’s circle center C and the point (-b/2a,0) or (-0.35,0).

I want to add that I don’t believe that the properties discussed in this post were discussed anywhere else. The methods for solving quadratics discussed in the previous post and the paper mentioned in the beginning of this post are much easier to apply and they use only compass and straightedge constructions. However, the beauty of this method is that it connects the parabola, circle and the hyperbola in an elegant manner. This method proves again that Lill’s method can be connected to various interesting properties.