Before we introduce imaginary and complex numbers, let us understand the types of numbers we may have already stumbled across.

**Recall:** A set is a collection of objects. We use curly braces to define what is in a set. If \(A\) is a set and \(x\) is one of the objects of \(A\), then we write \(x \in A\), and is read as “\(x\) is an element of \(A\)”.

We have learnt about the following sets:

- \(\mathbb{N} = \{0,1,2,3,\dots\} \) = the set of
**natural numbers**(or counting numbers) - \(\mathbb{Z} = \{\dots, -3,-2,-1,0,1,2,3,\dots\} \) = the set of
**integers** - \(\mathbb{Z}^+ = \{1,2,3,\dots\} \) = the set of
**positive****integers** - \(\mathbb{Z}^- = \{\dots, -3,-2,-1\}\) = the set of
**negative****integers** - \(\mathbb{Q} = \left\{ \frac{p}{q} \text{ such that } p,q \in \mathbb{Z}\text{ and } q \ne 0\right\} \) = the set of
**rational numbers**, or the set of numbers which can be expressed as a fraction of two integers, where the denominator is non-zero (e.g. \( \frac{1}{3}\),\(-3\), \(0.1729\), \(- \frac{999}{1000} \)) - \(\mathbb{R}\) = the set of
**real numbers**(which includes the**irrational numbers**e.g. \( \pi\), \(e\), \(\sqrt{5}\))

**Question to think about:** What is a real number?^{1}

### Why do we need a new type of number?

#### Allowing us to solve equations

When we first started learning about numbers, we “invented” new types of numbers to allow us to give solutions to different equations.

We first learnt about the **positive integers**, allowing us to give solutions to equations such as $$x-1=0.$$

We were then introduced to **negative integers**, allowing us to give solutions to equations such as $$x+1=0.$$

We were then introduced to **rational numbers**, allowing us to give solutions to equations such as $$2x+1=0.$$

And then came **irrational numbers**, allowing us to give solutions to equations such as $$x^2-2=0.$$

But what about an equation like $$x^2 + 4 = 0?$$

#### Sum and product of roots of quadratic equations

**Recall: **The formulas (called Vieta’s formulas) for finding the sum and product of roots \(\alpha\) and \(\beta\) of quadratic equations of the form \(ax^2 + bx + c = 0 \): $$\alpha + \beta = -\frac{b}{a} \quad \alpha\beta = \frac{c}{a}.$$

**Recall: **You may have also learnt the quadratic formula for solving quadratic equations: $$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a},$$ which has real solutions if and only if the discriminant \(b^2-4ac\) is non-negative.

If the discriminant is negative, its square root is undefined and so there are no real number solutions to the quadratic equation. However, we notice that we can still find the sum and product of roots. For example, for the equation \(x^2 + x + 1 = 0 \), the discriminant is \(-3\), and if we try to solve it using the quadratic formula we have $$ x = \frac{-1 \pm \sqrt{-3}}{2},$$ which doesn’t seem to make sense because we cannot take the square root of \( -3 \). But by Vieta’s formulas, the sum of roots is \(-1\) and the product of roots is \(1\). Isn’t it strange that the roots don’t exist, yet the sum and product of the roots exist?