Imaginary numbers, a fascinating subset of complex numbers, introduce us to a realm where creativity meets mathematical precision. Rooted in the imaginary unit ii—where \(i^2 = -1\) —these numbers play a pivotal role in solving equations that defy real-number solutions. In this exploration, we will dive into the definition, basic operations, applications, and graphical representation of imaginary numbers.

## Definition of Imaginary Numbers

An imaginary number is expressed as bi, where b is a real number and i is the imaginary unit. The general form of an imaginary number is xi, where x is any real number. Imaginary numbers are a subset of complex numbers, specifically those with a real part of 0.

## Imaginary Unit (i):

The imaginary unit ii is the cornerstone of imaginary numbers, defined as i = \(\sqrt{-1}\). Its unique property of \(i^2 = -1\) provides a solution to equations that would otherwise have no real roots.

### Basic Operations with Imaginary Numbers:

#### Addition and Subtraction:

Imaginary numbers are added or subtracted by combining their coefficients.

Example: \(2i + 3i = 5i/)

#### Multiplication:

Multiplying imaginary numbers involves leveraging the fact that \(i^2 = -1\)

Example: \(4i \times 2i = -8\)

#### Division:

Division requires multiplying the numerator and denominator by the conjugate of the denominator.

Example: \(\frac{3i}{2 – i}\)

### Complex Numbers:

Imaginary numbers find their place within the broader set of complex numbers, expressed as \(a + bi\), where a and b are real numbers. Imaginary numbers occur as a special case when the real part (a) is zero.

## Examples on Imaginary Numbers

**1. Example: \(2i + 4i\)**

**Solution:** \(2i + 4i = 6i\)

Explanation: Combine the imaginary parts.

**2. Example: \((4 – 2i) – (2 + 4i)\)**

**Solution:** \((5 – 2i) – (3 + 4i) = 2 – 6i\)

(4−2i)−(2+4i)=4−2i−2−4i

Combine the real parts: 4 – 2 = 2

Combine the imaginary parts: (-2i – 4i) = -6

So, (4 – 2i) – (2 + 4i) = 2−6i

**3. Example: \((2i) \times (4 – i)\)**

**Solution:** \((2i) \times (4 – i) = 8i – 2i^2 = 8i + 2\)

Explanation: Apply the distributive property and \(i^2 = -1\)