# Dual plot for complex roots of quadratic equation

$x^2 - \sqrt 3 x + 1/2 =0 \tag{1}$

can be plotted on $x$- axis as its parabola intersection at $(\sqrt 3/2 \pm 1/2,0).$

In an improvization I assign $x$-axis as real $x$-axis and $y$-axis as imaginary axis. For plotting complex roots it is possible to plot roots of quadratic equations on same $x$- but different $y$- as a dual representation (convenient) mode converting the Plot to an Argand/Gauss diagram ?

That is, is there some geometric way to represent complex roots $( \sqrt 3 /2, \pm\, i/2)$ of

$$x^2 -\sqrt 3 x +1 =0$$ as well on such dual diagram?

What geometric constructions may be necessary to make any new line cut the parabola or its morph at the complex roots? The situation shown: • @John Thanks for edit. – Narasimham May 23 '15 at 7:10

I am not sure if I understood well your question. So forgive me if the following answer does not satisfy you.

Anyway, if you have an equation like $$x^2-6x+10=0$$ not having real solutions, you can still plot the function $$f(x)=x^2-6x+10.$$ See the blue curve below. (The coefficient belonging to $x^2$ will be $1$ during this argumentation.) The fact that the original equation does not have real solutions is equivalent to the fact that $f$ does not intersect with the $x$ axis. Mirror the plot of $f$ over the line tangent to its minimum being at $m=3$. The resulting (red) curve will have intersections with the $x$ axis (at $3\pm 1$ or $m \pm 1$ this time). The complex solutions of the original equation are $3\pm i.$ (Where the multiplier of $i$ is $1$. Note that the intersection points are at $m\pm 1$.)

This is true in general: Suppose that $f(x)=x^2+bx+c$ does not intersect the $x$ axis and that its minimum is taken at $m$. Mirror $f$ over its tangent line at $m$. The resulting function $g(x)=2f(m)-f(x)$ will intersect the $x$ axis at points, say $m+u$ and $m-u$. The complex solutions of $x^2+bx+c=0$ are then $$m \pm iu.$$

That is, the complex roots can be constructed without having to use the solution formula.

Consider the equation given in the OP:

$$x^2 -\sqrt 3 x +1 =0.$$

Here the blue line represents the function $x^2 -\sqrt 3 x +1 =0$ whose minimum is at; $m=\frac{\sqrt{3}}{2}.$ The purple line is the mirror image whose intersection points with the $x$ axis are at $\frac{\sqrt{3}}{2}\pm\frac{1}{2}$ so the complex roots are

$$\frac{\sqrt{3}}{2}\pm i\frac{1}{2}.$$

• Thanks for your answer. It is what I wanted. The simple question of complex root representation stayed with me for quite a long time! – Narasimham 8 mins ago edit – Narasimham May 25 '15 at 12:07

In answering my question towards generality I take essential input is from zoli about placement of roots in this familiar situation.

When discriminant $\Delta =(b^2- 4 ac)$ < 0 a second degree equation $a x^2 + b x + c = 0$ has two complex roots. When $\Delta >0,$ two real roots. Their components are respectively

$$\alpha = \frac {-b}{2a} , \beta = \frac {\sqrt { 4 a c -b^2}}{2 a}$$

These are half sum and half difference (real part) of complex roots respectively.

When the graph of second degree equation the parabola is reflected with respect to tangent to at extreme point where the derivative vanishes it supplies two real roots on intersection with x-axis. Parabolas

$$y_1= a x^2 + b x + c , y_2= 2 ( c - \frac {b^2}{4 a} ) -y_1.$$

share a common extremum contact point as shown. The four roots can be placed on ends of horizontal and vertical diameters of a Circle whose diameter equals difference of roots. The circle can be conveniently defined in the Argand/Gauss complex plane used to represent complex numbers. Improvised black dots are simply rotated through $\pi/2$ to push them back into the appropriate complex plane.

If real roots of $y_2$ are given, complex roots for $y_1$ can be found also but serves no useful purpose now.

One can say that every second degree polynomial of complex roots is associated with its image or conjugate with real roots by reflection at extreme point tangents and vice-versa.