Description of how to calculate quadratic equations with 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 \(z_2=3-2i\)

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

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

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.