How to multiply complex numbers in polar form ?
Let z1 = r1(cos θ1 + i sin θ1 ) 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.
z1 . z2 = r1r2[cos (θ1 + θ2) + i sin (θ1 + θ2)]
Find the product of z1 and z2.
Example 1 :
z1 = 7(cos 25˚ + i sin 25˚)
z2 = 2(cos 130˚ + i sin 130˚)
Solution :
By using the z1 . z2 formula, we get
z1 . z2 = (7 . 2)[cos (25˚ + 130˚) + i sin (25˚ + 130˚)]
z1 . 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 . z2 formula, we get
z1 . z2 = (√2 . (1/2))[cos (118˚ - 19˚) + i sin (118˚ - 19˚)]
z1 . 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 :
z1 . z2 = (5 . 3)[cos (π/4 + 5π/3) + i sin (π/4 + 5π/3)]
Taking the least common multiple, we get
z1 . z2 = 15[cos ((3π + 20π)/12) + i sin ((3π + 20π)/12)]
z1 . 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 :
z1 . z2 = (√3 . 1/3)[cos (3π/4 + π/6) + i sin (3π/4 + π/6)]
Taking the least common multiple, we get
z1 . z2 = (1/√3)[cos ((9π + 2π)/12) + i sin ((9π + 2π)/12)]
z1 . 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)]