PRODUCT OF TWO COMPLEX NUMBERS IN POLAR FORM

Here we are going to see how to find the product of two complex numbers in polar form.

Case 1 :

Case 2 :

By using one of the above methods, we may find the product of two or more complex numbers.

In case we have power for any complex numbers written polar form,  we have bring down the power using Demoiver's theorem and multiply.

Let us look into some example problems based on the above concept.

Example 1 :

Find the product of following complex numbers 

(cos 2θ + i sin 2θ) (cos 3θ + i sin 3θ)

Solution :

=  (cos 2θ + i sin 2θ) (cos 3θ + i sin 3θ)

=  cos (2θ + 3θ) + i sin (2θ + 3θ)

=  cos 5θ + i sin 5θ

Example 2 :

Find the product of following complex numbers 

(cos 4θ + i sin 4θ)^-6 (cos θ + i sin θ)⁸

Solution :

=  (cos 4θ + i sin 4θ)^-6 (cos θ + i sin θ)⁸

First we have to bring down the power using Demoiver's theorem.

So,

=  (cos (-24θ) + i sin (-24θ)) (cos 8θ + i sin 8θ)

=  cos (-24θ + 8θ) + i sin (-24θ + 8θ)

=  cos (-16θ) + i sin (-16θ)

=  cos 16θ - i sin 16θ

Example 3 :

Find the product of following complex numbers 

(cos 4θ + i sin 4θ)¹² (cos 5θ - i sin 5θ)^-6

Solution :

=  (cos 4θ + i sin 4θ)¹² (cos 5θ - i sin 5θ)^-6

=  (cos 4θ + i sin 4θ)¹² (cos (-5θ) + i sin (-5θ))^-6

=  (cos 12(4θ) + i sin 12(4θ)) (cos -6(-5θ) + i sin -6(-5θ))

=  (cos 48θ + i sin 48θ) (cos 30θ + i sin 30θ)

=  cos (48+30)θ +  i sin (48+30)θ

=  cos 78θ +  i sin 78θ 

Example 4 :

Find the product of following complex numbers 

(cos α + i sin α)³/(sin β + i cos β)⁴

Solution :

=  (cos α + i sin α)³/(sin β + i cos β)⁴

The denominator is not in the polar form. In order to simplify this, first we have to convert the denominator into polar form.

(sin β + i cos β)⁴  =  [cos (90 - β) + i sin (90 - β)]⁴

=  (cos α + i sin α)³/[cos (90 - β) + i sin (90 - β)]⁴

=  (cos 3α + i sin 3α)/cos 4(90 - β) + i sin 4(90 - β)

=  cos (3α - (360 - 4β)) + i sin (3α - (360 - 4β))

=  cos (3α - 360 + 4β) + i sin (3α - 360 + 4β)

=  cos ((3α + 4β) - 360) + i sin ((3α + 4β) - 360)

=  cos (360 - (3α + 4β)) + i sin (360 - (3α + 4β))

=  cos (3α + 4β) + i sin (3α + 4β)

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