How to distribute complex numbers?

Complex numbers are used to find the square root of negative numbers. It comprises the real and imaginary parts. The complex number is of the form a + ib where a is the real part and b is the imaginary part. The real part of the complex number is any number and is represented by Re(z) where ‘z’ is any complex number. The imaginary part of the complex number comprises any number multiplied with ‘i’, The ‘i’ stands for iota. The complex numbers cannot be represented on the number line. They are represented on a plane called Argand Plane. For example: Let z = 3 + 5i. Here Re(z) = 3 and Im(z) = 5.

About Iota

The alphabet ‘i’ is known as iota. The value  ‘i’ is √-1. It is basically used to denote imaginary numbers. Values of iota are

1. i² = -1
2. i³ = -i
3. i⁴ = 1

Operations on Complex Numbers

We can perform addition, subtraction, multiplication, division, etc, and can also find conjugate and absolute values of complex numbers. 

• Addition:

Let a + ib and c + id be two complex numbers. The result of the addition operation is that real parts are added separately and imaginary numbers are added separately. The result is (a + c) + i(b + d). 

• Subtraction:

Let a + ib and c + id be two complex numbers. The result of the subtraction operation is that real parts are subtracted separately and imaginary numbers are subtracted separately. The result is (a – c) + i(b – d). 

• Multiplication:

Let a + ib and c + id be two complex numbers. The result of the multiplication operation is that real parts are multiplied together, imaginary parts are multiplied together and real and imaginary parts are also multiplied. The result is (a + ib)×(c+id) = ac + iad +ibc – ad.

• Conjugate:

Conjugate is nothing but changing the sign of operation in the complex number. If a + ib is a complex number than the conjugate is a – ib.

• Division:

Let a + ib and c + id be two complex numbers. We rationalize the denominator by multiplying it with its conjugate. The result is

(a + ib)/(c + id) = (a + ib)×(c – id)/(c² – d²).

• Absolute value:

The absolute value is also known as the modulus. It is the absolute value of the complex number. The absolute value of c+id is given by √c² + d²

Properties of Complex Numbers

Let a + ib, c + id , x+iy be three complex numbers.

Commutative law: In commutative law, the numbers can be swapped without any change in numbers. Commutative law in complex numbers is valid for multiplication and addition. It is

(a + ib) + (c + id) = (c + id) + (a + ib)

Associativity: The associativity of three complex numbers is satisfied under addition and multiplication. It is 

{(a + ib) + (c + id)}+ (x + iy) = (a + ib) + {(c + id) + (x + iy)}

Distributivity: Let  (a + ib), (c + id), (x + iy) be three complex numbers. The distributive Property is 

(a + ib) × {(c+id) + (x + iy)} = {(a + ib)×(c + id)} + {(a + ib)×(x + iy)}

How to Distribute Complex Numbers?

Distributing means dividing. Complex numbers satisfy the distributive property. The property states that if we multiply a complex number by the sum of two complex numbers it will be the same as multiplying the complex number with each complex number separately and then adding the result. Let (a + ib), (c + id) , (x + iy) be three complex numbers. Therefore the distributive Property is (a + ib) × {(c+id) + (x + iy)} = {(a + ib)×(c + id)} + {(a + ib)×(x + iy)}. Let us illustrate it  with the help of an example

Let there be three complex numbers : 2i, 3 + 7i , 7 + 3i

= (2i) × {(3 + 7i) +( 7 + 3i)} 

= 2i × (10 + 10 i) 

= 20i -20 —–(1)

Now let us distribute the 2i individually with the given operands

= (2i) × {(3 + 7i) +( 7 + 3i)} 

= 2i×(3 + 7i) + (2i)×(7 + 3i) 

= 6i -14 + 14i – 6 

= 20i -20    (i² = -1) ——(2)

from eq (1) and (2)

Thus, distributive property is satisfied.

Sample Problems

Problem 1: Compute (2i + 6i )×(4i + 8i -2i)

Solution: 

2i×(4i + 8i -2i) + 6i×(4i + 8i -2i)

= 2i×(10i) + 6i×(10i)

= -20 – 60

= -80

Problem 2: Find the value of  2i×(9 + i/9 + 8i)

Solution:

2i×( 9 + i/9 + 8i)

 = 2i×9 + 2i ×( i/9 + 8i) 

= 18i + 2i ×( 73i/9)

= -146/9 + 18i

Problem 3: Find (4 + 5i)×(9 + 3i)

Solution:

(4 + 5i)×(9 + 3i)

= 4×(9 + 3i) + 5i×(9 + 3i)

=36 + 12i + 45i -15

= 21 +57i

Problem 4: Solve (4 + 5i)×(4 – 5i)

Solution:

(4 + 5i)×(4 – 5i)

= 4×(4 – 5i) + 5i×(4 – 5i)

=16 -20i + 20i +25

=41

Problem 5: Solve (4 + 5i)×(9 – 5i)

Solution:

(4 + 5i)×(9 – 5i)

= 4×(9 – 5i) + 5i×(9 – 5i)

= 36 – 20i + 45i +25

= 41 + 25i