Complex numbers: Introduction

We review how complex numbers are introduced, starting from a simple linear equation. We also review the derivation of the quadratic formula.

Consider the following equation:

$$3x=12.$$

The coefficient of $x$ is 3, and the right-hand side is 12. 3 and 12 are natural numbers: $3,12\in \mathbb{N}$. Solving this for $x$, we have

$$x=4\in \mathbb{N}.$$

So, the solution is obtained within $\mathbb{N}$.

Next, consider

$$2x+4=0.$$

Again we have $2,4\in \mathbb{N}$. But the solution

$$x=-2\in \mathbb{Z}$$

is not found in $\mathbb{N}$, but in $\mathbb{Z}$.

Next, consider

$$2x-3=0$$

where $2,-3\in \mathbb{Z}$. The solution is

$$x=\frac{3}{2}\in \mathbb{Q}$$

which is not in $\mathbb{Z}$.

In general, consider

$$ax+b=0,a,b\in \mathbb{Q},a\ne 0.$$

Then, the solution is always found within $\mathbb{Q}$:

$$x=-\frac{b}{a}\in \mathbb{Q}.$$

As long as all the coefficients are in $\mathbb{Q}$ and the equations are linear, we can always find the solution, if it exists, within $\mathbb{Q}$. The algebra that has grown out of linear equations is called linear algebra. Of course, we can use $\mathbb{R}$ in place of $\mathbb{Q}$. We will study linear algebra to some extent in later posts.

Consider the following non-linear equation:

$$x^2=2.$$

Note $1,2\in \mathbb{Q}$. Its solutions are

$$x=\pm \sqrt{2}\in \mathbb{R}.$$

When we deal with non-linear (in this case, quadratic) equations, we cannot stay in $\mathbb{Q}$, but need (at least) $\mathbb{R}$. In fact, we need more.

Let us define a function $f:\mathbb{R}\to \mathbb{R}$ by

$$f(x)=2x^2-5x+1.$$

The solutions of the equation $f(x)=0$ are the elements of

$$\{\alpha \mid \alpha \in \mathbb{R},f(\alpha )=0\}.$$

The solutions of an equation are sometimes called roots.

Theorem (Solutions of the quadratic equation)

Consider the equation

$$a{x}^2+bx+c=0$$

where $a,b,c\in \mathbb{R}$ and $a\ne 0$. This equation has no solutions in $\mathbb{R}$ if $b^2-4ac<0$. The real solutions are given by the formula

$$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

if $b^2-4ac\ge 0$.

Proof. Suppose that $\alpha \in \mathbb{R}$ is a solution, then $a{\alpha }^2+b\alpha +c=0$. Since $a\ne 0$, we can divide both sides by $a$ to have

$${\alpha }^2+\frac{b}{a}\alpha =-\frac{c}{a}.$$

Complete the square on the left-hand side to get

$${(\alpha +\frac{b}{2a})}^2-{\left(\frac{b}{2a}\right)}^2=-\frac{c}{a}$$

which can be rearranged into

$${(\alpha +\frac{b}{2a})}^2=\frac{b^2-4ac}{4a^2}.$$

If the right-hand side is negative, we have a contradiction since the square of a real number on the left-hand side is non-negative. In this case, our initial assumption that there is a solution in $\mathbb{R}$ is false. If the right-hand side is non-negative, we have

$${(\alpha +\frac{b}{2a})}^2={\left(\frac{\sqrt{b^2-4ac}}{2a}\right)}^2.$$

Note that in $\mathbb{R}$, $u^2=v^2$ if and only if $u=\pm v$. Thus we have

$$\alpha +\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}$$

so

$$\begin{array}{}\text{(eq:qsol)}& \alpha =\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$$

and $b^2-4ac\ge 0$.

Finally, retracing the argument in reverse, we can show that these $\alpha$'s are indeed the solutions. ■

Remark. The last paragraph is necessary because we assumed that $\alpha$ was a solution, but this assumption may well be false. From a false assumption, anything can be ``proved'' to be true. Therefore, after we have obtained Eq. (eq:qsol), we need to verify that this is indeed the solution (i.e., the assumption is true). □

What happens if $b^2-4ac<0$? Since there are no real numbers such that their square is negative, we do not have real solutions in this case. Nevertheless, we introduce a new ``number'' $i=\sqrt{-1}$ called the imaginary unit. Formally,  the square of the imaginary unit is -1: $i^2=(\sqrt{-1}{)}^2=-1$ (hence, it cannot be real!) Using the imaginary unit, we define complex numbers such as $a+bi$ with $a,b\in \mathbb{R}$. In the complex number $a+bi$, $a$ is called the real part, and $b$ is called the imaginary part. A real multiple of $i=\sqrt{-1}$, such as the $b$ in $bi$, is called an imaginary number. However, there are a bunch of questions arising. Is this new number system consistent? What does $a+bi$ (addition between real and imaginary numbers) mean at all? We will make the notion of complex numbers more precise in the next post by constructing them from scratch.