PRODUCT OF COMPLEX NUMBERS IN POLAR FORM

How to multiply complex numbers in polar form ? 

Let z1  =  r1(cos θ+ i sin θ) and z2  =  r2(cos θ2 + i sin θ2 ) be two complex numbers in the polar form.

We can use the formula given below to find the product of two complex numbers in the polar form.

z. z2  =  r1r2[cos (θ1 + θ2) + i sin (θ1 + θ2)]

Find the product of z1 and z2.

Example 1 :

z=  7(cos 25˚ + i sin 25˚)

z2  =  2(cos 130˚ + i sin 130˚)

Solution :

By using the z. z2 formula, we get

z1 . z=  (7 . 2)[cos (25˚ + 130˚) + i sin (25˚ + 130˚)]

z. z2  =  14(cos 155˚ + i sin 155˚)

Example 2 :

z1  =  √2(cos 118˚ + i sin 118˚)

z2  =  0.5(cos (-19˚) + i sin (-19˚)

Solution :

By using the z1 . zformula, we get

z. z2  =  (√2 . (1/2))[cos (118˚ - 19˚) + i sin (118˚ - 19˚)]

z. z2  =  (1/√2)(cos 99˚ + i sin 99˚)

Example 3 :

z1  =  5(cos π/4 + i sin π/4)

z2  =  3(cos 5π/4 + i sin 5π/4)

Solution :

z. z2  =  (5 . 3)[cos (π/4 + 5π/3) + i sin (π/4 + 5π/3)]

Taking the least common multiple, we get

z. z2  =  15[cos ((3π + 20π)/12) + i sin ((3π + 20π)/12)]

z. z2  =  15[cos (23π/12) + i sin (23π/12)]

Example 4 :

z1  =  √3(cos 3π/4 + i sin 3π/4)

z2  =  1/3(cos π/6 + i sin π/6)

Solution :

z. z2  =  (√3 . 1/3)[cos (3π/4 + π/6) + i sin (3π/4 + π/6)]

Taking the least common multiple, we get

z. z2  =  (1/√3)[cos ((9π + 2π)/12) + i sin ((9π + 2π)/12)]

z. z2  =  (1/√3)[cos (11π/12) + i sin (11π/12)

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