Description of how to calculate quadratic equations with complex numbers

## Quadratic Equations and Complex Numbers

This article describes an important application of complex numbers, the solution of quadratic equations. An example of the solvability of a quadratic equation has already been described in the article Complex Numbers, namely $$x^2 = -1$$.

Here we describe, with complex numbers all quadratic equations are solvable.

We come back to the equation $$x^2= -1$$. This quadratic equation is not solvable in the domain of real numbers; but in the world of complex numbers.

The equation $$z^2= -1$$ has the complex solution $$z = i$$. The unknown $$z$$ instead of $$x$$ indicates that a complex solution is allowed.

Obviously, the following solution is also correct

$$(-i)^2=(-1)^2·i^2=1·(-1)=-1$$

So $$-i$$ is a solution of the equation $$z^2 = -1$$.

The quadratic equation $$z^2 = -1$$ has two solutions, namely the two conjugate complex numbers $$z_1 = i$$ and $$z_2 = -i$$.

As next example, we solve the equation $$z^2 = -2$$. We write

$$z^2=2·(-1)$$

The two conjugate complex solutions are

$$z_1=\sqrt{2i}$$   and   $$z_2=-\sqrt{2i}$$

The equation $$z^2=-2=2·(-1)$$ is equivalent to the equation

$$\displaystyle\frac{z^2}{2}=\left(\frac{z}{\sqrt{2}}\right)^2=-1$$ with the two solutions   $$\displaystyle\frac{z_1}{\sqrt{2}}=i$$   und   $$\displaystyle\frac{z_2}{\sqrt{2}}=-1$$
That is $$z_1=\sqrt{2i}$$   and   $$z_2=-\sqrt{2i}$$.

The quadratic equations discussed in the previous section, in which $$z$$ occurred only as $$z^2$$, were easily solved.In this section we will treat a quadratic equation of the form   $$az^2+bz^2 +c = 0$$   with   $$a,b,c$$   real and   $$a\not =0$$.

As an example we calculate the following quadratic equation

$$z^2-6z+13=0$$

We try to apply the solution method from the previous section to this equation. To write the left side of the equation as a square, we use the method of quadratic addition. In our example, this looks like this.

$$0=z^2-6z+13=z^2-2·3z+3^2-3^2+13=(z-3)^2-9+13=(z-3)^2+4)$$

The equation   $$z^2-6z+13=0$$   is equivalent to the equation $$(z-3)^2+4=0$$. The reshaped equation looks no less complicated than before, but it can be solved like the equations discussed in the previous section

The equation $$(z-3)^2+4=0$$ is equivalent to $$(z-3)^2 = -4=4·(-1)$$

If we take the root of it we get $$z-3=2i$$ or $$z-3 = -2i$$

The solutions are $$z_1 = 3+2i$$ and $$z_2=3-2i$$

## General solution formula for quadratic equations

We again use the general quadratic equation of the form

$$az^2+bz+c=0$$

A division by $$a$$ results

$$\displaystyle z^2+\frac{b}{a}·z+\frac{c}{a}=0$$

The summand

$$\displaystyle\frac{b}{a}·z=2·\frac{b}{2a}·z$$

is taken as the product of a binomial formula and the quadratic addition

$$\displaystyle +\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2$$

This results in

$$\displaystyle 0=z^2+\frac{b}{a}·z+\frac{c}{a}=z^2+2·\frac{b}{2a}·z+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a}=\left(z+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}$$
$$\displaystyle = \left(z+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{4ac}{4a·a}=\left(z+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}$$

The equation $$az^2+bz+c=0$$ is thus equivalent to the equation

$$\displaystyle \left(z+\frac{b}{2a}\right)^2 = \frac{b^2-4ac}{(2a)^2}$$

Whether the number on the right is positive or negative depends on whether it is $$b^2-4ac$$ positive or negative. The denominator $$(2a)^2$$ is always positive. The term $$D=b^2-4ac$$ is the discriminant of the quadratic equation. It therefore depends on the sign of the discriminant whether the quadratic equation has a real or not a real solution.

To solve the equation, we set $$D$$ for $$b^2-4ac$$ and write

$$\displaystyle \left(z+\frac{b}{2a}\right)^2=\frac{D}{(2a)^2}$$

For the equation we get the two solutions

$$\displaystyle z_1+\frac{b}{2a}= \frac{ω}{2a}$$   and   $$\displaystyle z_2+\frac{b}{2a}= \frac{-ω}{2a}$$

also   $$\displaystyle z_1+\frac{-b}{2a}+ \frac{ω}{2a}$$   and   $$\displaystyle z_2+\frac{-b}{2a}- \frac{ω}{2a}$$

We write the short solution

$$\displaystyle z_{1,2}\frac{-b±ω}{2a}$$

## Summary

The solution of the quadratic equation $$az^2+bz+c=0$$ with $$a,b,c$$ real and $$a ≠0$$ is

$$\displaystyle z_{1,2}\frac{-b±ω}{2a}$$

The complex number ω is $$ω^2=D=b^2-4ac$$

Depending on the sign of the discriminant $$D$$, the solutions are as follows

Ist $$D>0$$, then are $$z_1$$ und $$z_2$$ real.

Ist $$D=0$$, then are $$z_1=z_2$$ real.

Ist $$D < 0$$, then are $$z_1$$ and $$z_2$$ conjugated complex to each other.