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Algebraic Operations on Complex Numbers | Class 11 Maths
• Last Updated : 04 Dec, 2020

A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation i² = −1. For example, 5+6i is a complex number, where 5 is a real number and 6i is an imaginary number. Therefore, the combination of both the real number and imaginary number is a complex number. There can be four types of algebraic operations on complex numbers which are mentioned below. The four operations on the complex numbers include:

• Addition
• Subtraction
• Multiplication
• Division

### Addition of Complex Numbers

To add two complex numbers, just add the corresponding real and imaginary parts.

(a + bi) + (c + di) = (a + c) + (b + d)i

Examples:

• (7 + 8i) + (6 + 3i)  = (7 + 6) + (8 + 3)i = 13 + 11i
• (2 + 5i) + (13 + 7i) = (2 + 13) + (7 + 5)i = 15 + 12i
• (-3 – 6i) + (-4 + 14i) = (-3 – 4) + (-6 + 14)i = -7 + 8i
• (4 – 3i ) + ( 6 + 3i) = (4+6) + (-3+3)i = 10
• (6 + 11i) + (4 + 3i) = (4 + 6) + (11 + 3)i = 10 + 14i

### Subtraction of Complex Numbers

To subtract two complex numbers, just subtract the corresponding real and imaginary parts.

(a + bi) − (c + di) = (a − c) + (b − d)i

Examples:

• (6 + 8i)  –  (3 + 4i) = (6 – 3) + (8 – 4)i = 3 + 4i
• (7 + 15i) – (2 + 5i) = (7 – 2) + (15 – 5)i = 5 + 10i
• (-3 + 5i) – (6 + 9i) = (-3 – 6) + (5 – 9)i = -9 – 4i
• (14 – 3i) – (-7 + 2i) = (14 – (-7)) + (-3 – 2)i = 21 – 5i
• (-2 + 6i) – (4 + 13i) = (-2 – 4) + (6 – 13)i = -6 – 7i

### Multiplication of Two Complex Numbers

Multiplication of two complex numbers is the same as the multiplication of two binomials. Let us suppose that we have to multiply a + bi and c + di. We will multiply them term by term.

(a + bi) ∗ (c + di) = (a + bi) ∗ c + (a + bi) ∗ di

= (a ∗ c + (b ∗ c)i)+((a ∗ d)i + b ∗ d ∗ −1)

= (a ∗ c − b ∗ d + i(b ∗ c + a ∗ d))

Example 1:  Multiply (1 + 4i) and (3 + 5i).

(1 + 4i) ∗ (3 + 5i) = (3 + 12i) + (5i + 20i2)

= 3 + 17i − 20

= −17 + 17i

Note: Multiplication of complex numbers with real numbers or purely imaginary can be done in the same manner.

Example 2: Multiply 5 and (4 + 7i).

5 ∗ (4+7i) can be viewed as (5 + 0i) ∗ (4 + 7i)

= 5 ∗ (4 + 7i)

= 20 + 35i

Example 3: Multiply 3i and (2 + 6i).

3i ∗ (2 + 6i) can be viewed as (0 + 3i) ∗ (2 + 6i)

= 3i ∗ (2 + 6i)

= 6i + 18i2

= 6i − 18

= −18 + 6i

Example 4: Multiply (5 + 3i)  and  (3 + 4i).

(5+3i) ∗ (3+4i) = (5 + 3i) ∗ 3 + (5 + 3i) ∗ 4i

= (15 + 9i) + (20i + 12i2)

= (15 − 12) + (20 + 9)i

= 3 + 29i

### Review of Complex Numbers Addition, Subtraction and Multiplication

1. (a + bi) + (c + di) = (a + c) + (b + d)i
2. (a + bi) − (c + di) = (a − c) + (b − d)i
3. (a + bi) ∗ (c + di) = ((a ∗ c − b ∗ d) + (b ∗ c + a ∗ d)i)

## Conjugate of a Complex Number

In any two complex numbers, if only the sign of the imaginary part differs then, they are known as a complex conjugate of each other. Thus conjugate of a complex number a + bi would be a – bi. ### What’s the use of a complex conjugate? Thus we can observe that multiplying a complex number with its conjugate gives us a real number. Thus the division of complex numbers is possible by multiplying both numerator and denominator with the complex conjugate of the denominator.

### Examples of Complex Conjugates ### Properties of Complex Conjugates

Property 1: Property 2: Property 3: Property 4: Property 5: ### Division of Two Complex Numbers

Division of complex numbers is done by multiplying both numerator and denominator with the complex conjugate of the denominator. Example 1: Example 2: Example 3: Example 4: Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

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