FIND THE PRODUCT OF COMPLEX NUMBERS IN POLAR FORM

Find the product of the complex numbers. Leave answers in polar form.

Problem 1 :

z₁ = 6(cos 20° + i sin 20°)

z₂ = 5(cos 50° + i sin 50°)

Solution:

By using z₁ ⋅ z₂ formula,

z₁ ⋅ z₂ = r₁r₂[cos (θ₁ + θ₂) + i sin (θ₁ + θ₂)]

z₁ ⋅ z₂ = (6 ⋅ 5)[cos(20° + 50°) + i sin (20° + 50°)]

z₁ ⋅ z₂ = 30(cos 70° + i sin 70°)

Problem 2 :

z₁ = 4(cos 15° + i sin 15°)

z₂ = 7(cos 25° + i sin 25°)

Solution:

By using z₁ ⋅ z₂ formula,

z₁ ⋅ z₂ = r₁r₂[cos (θ₁ + θ₂) + i sin (θ₁ + θ₂)]

z₁ ⋅ z₂ = (4 ⋅ 7)[cos(15° + 25°) + i sin (15° + 25°)]

z₁ ⋅ z₂ = 28(cos 40° + i sin 40°)

Problem 3 :

Solution:

By using z₁ ⋅ z₂ formula,

z₁ ⋅ z₂ = r₁r₂[cos (θ₁ + θ₂) + i sin (θ₁ + θ₂)]

Problem 4 :

Solution:

By using z₁ ⋅ z₂ formula,

z₁ ⋅ z₂ = r₁r₂[cos (θ₁ + θ₂) + i sin (θ₁ + θ₂)]

Problem 5 :

Solution:

By using z₁ ⋅ z₂ formula,

z₁ ⋅ z₂ = r₁r₂[cos (θ₁ + θ₂) + i sin (θ₁ + θ₂)]

Problem 6 :

Solution:

By using z₁ ⋅ z₂ formula,

z₁ ⋅ z₂ = r₁r₂[cos (θ₁ + θ₂) + i sin (θ₁ + θ₂)]

Problem 7 :

z₁ = 1 + i

z₂ = -1 + i

Solution:

z₁ = 1 + i

x + iy = r(cos θ + i sin θ)

r = √(1² + 1²)

r = √2

1 + i = √2 cos θ  + √2 i sin θ 

The angle lies in the first quadrant.

θ = 45° (or) π/4

z₂ = -1 + i

r = √(-1)² + 1²)

r = √2

-1 + i = √2 cos θ  + √2 i sin θ 

θ lies in second quadrant.

θ = π - α

α = 45°

θ = π - π/4

θ = 3π/4

Problem 8 :

z₁ = 1 + i

z₂ = 2 + 2i

Solution:

z₁ = 1 + i

x + iy = r(cos θ + i sin θ)

r = √(1² + 1²)

r = √2

1 + i = √2 cos θ  + √2 i sin θ 

The angle lies in the first quadrant.

θ = 45° (or) π/4

z₂ = 2 + 2i

r = √(2)² + (2)²)

r = √8 

2 + 2i = √8 (cos θ  + i sin θ) 

θ lies in first quadrant.

θ = π/4