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)

Write your answer in rectangular form when rectangular form is given and in polar form when polar form is given.

Example 5 :

(4 + 4i) (5 - 3i)

Solution :

(4 + 4i) (5 - 3i)

= 20 - 12i + 20i - 12i2

= 20 + 8i - 12(-1)

= 20 + 12 + 8i

= 32 + 8i

Example 6 :

4√2(cos 7π/4 + i sin 7π/4) ⋅ 2(cos π/6 + i sin π/6)

Solution :

4√2(cos 7π/4 + i sin 7π/4) ⋅ 2(cos π/6 + i sin π/6)

= 4√2(2) [cos (7π/4 + π/6) + i sin (7π/4 + π/6)]

= 8√2 [cos (21π + 2π)/4 + i sin (21π + 2π)/4]

= 8√2 [cos (23π/4) + i sin (23π/4)]

Example 7 :

[2√2(cos 7π/6 + i sin 7π/6)] / [6(cos 11π/6 + i sin 11π/6)]

Solution :

= [2√2(cos 7π/6 + i sin 7π/6)] / [6(cos 11π/6 + i sin 11π/6)]

= (2√2/6)[cos (7π/6 - 11π/6) + i sin (7π/6 - 11π/6)]

= (√2/3)[cos (7π - 11π)/6 + i sin (7π - 11π)/6]

= (√2/3)[cos (-4π/6) + i sin (-4π/6)]

= (√2/3)[cos (-2π/3) + i sin (-2π/3)]

Find the product of the complex numbers in polar form. Answer in both polar form and rectangular form.

Example 8 :

𝑧1 = 4(cos 225° + 𝑖sin 225°) and 𝑧2 = 3(cos 90° + 𝑖sin 90°)

Solution :

Given that, 𝑧1 = 4(cos 225° + 𝑖sin 225°)

𝑧2 = 3(cos 90° + 𝑖sin 90°)

𝑧1𝑧2 4(cos 225° + 𝑖sin 225°)  3(cos 90° + 𝑖sin 90°)

= 4(3) [cos (225° + 90°) + i sin  (225° + 90°)]

Polar form :

= 12 [cos 315° + i sin 315°]

Converting into rectangular form :

= 12 [cos (360 - 45) + i sin (360 - 45)]

= 12 [cos 45 + i sin 45]

= 12 [√2/2 + i√2/2]

= 12 (1/2) (√2 + i√2)

= 6 (√2 + i√2)

Example 9 :

𝑧1√2(cos 2π/3 + 𝑖 sin 2π/3) and 𝑧2 = 1/5(cos π/6 + 𝑖sin π/6)

Solution :

Given that, 𝑧1√2(cos 2π/3 + 𝑖 sin 2π/3)

𝑧2 = 1/5(cos π/6 + 𝑖sin π/6)

= [√2(cos 2π/3 + 𝑖 sin 2π/3)] ⋅ [1/5(cos π/6 + 𝑖sin π/6)]

= (√2/5) [cos (2π/3 + π/6) + i sin (2π/3 + π/6)]

= (√2/5) [cos (4π + π)/6 + i sin (4π + π)/6]

Polar form :

= (√2/5) [cos 5π/6 + i sin 5π/6]

Converting into rectangular form, we get

= (√2/5) [cos (π - π/6) + i sin (π - π/6)]

= (√2/5) [cos (π/6) + i sin (π/6)]

= (√2/5) [√3/2 + i (1/2)]

= (√2/5) [(√3 + i)/2]

Multiplying both numerator and denominator by √2

= (1/5) [(√3 + i)/√2]

Rectangular form :

= (√3 + i) / 5√2

Find the quotient of the complex numbers in polar form: 𝑧1/𝑧2 Write the answer in both polar form and rectangular form. 

Example 10 :

𝑧1 = 2(cos210° + 𝑖sin210°) and 𝑧2 = 8(cos60° + 𝑖sin60°)

Solution :

𝑧1 = 2(cos 210° + 𝑖sin 210°) and 𝑧2 = 8(cos 60° + 𝑖sin 60°)

𝑧1 / 𝑧2 = (2/8) cos (cos 210° + 𝑖sin 210°)/(cos 60° + 𝑖sin 60°)

= (1/4)[cos (210 - 60) + i sin (210 - 60)]

= (1/4)[cos 150 + i sin 150]

= (1/4)[cos (180 - 30) + i sin (180 - 30)]

= (1/4)[cos 30 + i sin 30]

= (1/4)[√3/2 + i (1/2)]

= (1/4)(√3 + i)/2)]

= (√3 + i)/8

Example 11 :

𝑧= 2/5 (cos𝜋/2 +𝑖 sin 𝜋/2) and 𝑧= 1/2 (cos 5𝜋/4 + 𝑖 sin 5𝜋/4)

Solution :

𝑧= 2/5 (cos𝜋/2 +𝑖 sin 𝜋/2) and 𝑧= 1/2 (cos 5𝜋/4 + 𝑖 sin 5𝜋/4)

𝑧1 / 𝑧2 = (2/5)/(1/2) (cos (𝜋/2 - 5𝜋/4) + 𝑖 sin (𝜋/2 - 5𝜋/4))

= (1/5) (cos (-3𝜋/4) + 𝑖 sin (-3𝜋/4))

(1/5) (cos (3𝜋/4) - 𝑖 sin (3𝜋/4))

(1/5) (cos (𝜋 - 𝜋/4) - 𝑖 sin (𝜋 - 𝜋/4))

(1/5) (√2/2 - 𝑖 √2/2)

(1/5) (√2- 𝑖 √2)/2)

= (√2- 𝑖 √2)/10

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Digital SAT Math Problems and Solutions (Part - 134)

    Apr 02, 25 12:40 AM

    digitalsatmath143.png
    Digital SAT Math Problems and Solutions (Part - 134)

    Read More

  2. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Apr 02, 25 12:35 AM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More

  3. Digital SAT Math Problems and Solutions (Part 135)

    Apr 02, 25 12:32 AM

    digitalsatmath147.png
    Digital SAT Math Problems and Solutions (Part 135)

    Read More