Imaginary numbers are a subset of complex numbers that involve the imaginary unit, denoted by i, where i = \(\sqrt{-1}\). These numbers are crucial in mathematics and engineering, especially when dealing with problems that have no real number solutions. Let’s explore imaginary numbers in more detail:

Table of Contents

**Definition of Imaginary Numbers**

An imaginary number is expressed in the form 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 Unit (i):**

The imaginary unit is the fundamental building block of imaginary numbers. It is defined as i = \(\sqrt{-1}\) . The square of ii results in −1, providing a unique solution to equations with no real roots.

**Basic Operations with Imaginary Numbers:**

**Addition and Subtraction:**

Imaginary numbers are added or subtracted by combining their coefficients.

Example: \(3i + 4i = 7i/)

**Multiplication:**

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

Example: \(4i \times 3i = -12\)

**Division:**

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

Example: \(\frac{3i}{2 – i}\)

**Complex Numbers**

Imaginary numbers are part of the broader set of complex numbers, which include both real and imaginary components. A complex number is expressed as \(a + bi\), where aa and b are real numbers. Imaginary numbers are a special case when a = 0.

## Examples on Imaginary Numbers

**1. Example: \(3i + 2i\)**

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

Explanation: Combine the imaginary parts.

**2. Example: \((5 – 2i) – (3 + 4i)\)**

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

Explanation: Subtract the imaginary parts.

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

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

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

**4. Example: \( \frac{3i}{2 – i}\)**

**Solution:** Multiply numerator and denominator by the conjugate of the denominator.

Result: \(\frac{3i(2 + i)}{5} = \frac{6i + 3i^2}{5}\) = \(-\frac{3}{5} + \frac{6}{5}i\)