DIVISION OF COMPLEX NUMBERS IN POLAR FORM

Find the quotient z1/z2 of the complex numbers. Leave answers in polar form. Express the argument as an angle between 0° and 360°.

Problem 1 :

z1 = 20(cos 75° + i sin 75°)

z2 = 4(cos 25° + i sin 25°)

Solution:

By using the z1/z2 formula,

z1z2=r1r2[cos (𝜃1-𝜃2)+i sin (𝜃1-𝜃2)]z1z2=204[cos (75°-25°)+i sin (75°-25°)]z1z2=5 (cos 50° + i sin 50°)

Problem 2 :

z1 = 50(cos 80° + i sin 80°)

z2 = 10(cos 20° + i sin 20°)

Solution:

By using the z1/z2 formula,

z1z2=r1r2[cos (𝜃1-𝜃2)+i sin (𝜃1-𝜃2)]z1z2=5010[cos (80°-20°)+i sin (80°-20°)]z1z2=5 (cos 60° + i sin 60°)

Problem 3 :

z1=3cos 𝜋5+i sin 𝜋5z2=4cos 𝜋10+i sin 𝜋10

Solution:

By using the z1/z2 formula,

z1z2=r1r2[cos (𝜃1-𝜃2)+i sin (𝜃1-𝜃2)]z1z2=34cos 𝜋5-𝜋10+i sin 𝜋5-𝜋10Taking the least common multiple, we getz1z2=34cos 2𝜋-𝜋10+i sin 2𝜋-𝜋10z1z2=34cos 𝜋10+i sin 𝜋10

Problem 4 :

z1=3cos 5𝜋18+i sin 5𝜋18z2=10cos 𝜋16+i sin 𝜋16

Solution:

By using the z1/z2 formula,

z1z2=r1r2[cos (𝜃1-𝜃2)+i sin (𝜃1-𝜃2)]z1z2=310cos 5𝜋18-𝜋16+i sin 5𝜋18-𝜋16Taking the least common multiple, we getz1z2=310cos 40𝜋-9𝜋144+i sin 40𝜋-9𝜋144z1z2=310cos 31𝜋144+i sin 31𝜋144

Problem 5 :

z1 = cos 80° + i sin 80°

z2 = cos 200° + i sin 200°

Solution:

By using the z1/z2 formula,

z1z2=r1r2[cos (𝜃1-𝜃2)+i sin (𝜃1-𝜃2)]z1z2=[cos (80°-200°)+i sin (80°-200°)]z1z2=[cos (-120°)+i sin (-120°)]

Problem 6 :

z1 = cos 70° + i sin 70°

z2 = cos 230° + i sin 230°

Solution:

By using the z1/z2 formula,

z1z2=r1r2[cos (𝜃1-𝜃2)+i sin (𝜃1-𝜃2)]z1z2=[cos (70°-230°)+i sin (70°-230°)]z1z2=[cos (-160°)+i sin (-160°)]

Problem 7 :

z1 = 2 + 2i

z2 = 1 + i

Solution:

z1 = 2 + 2i

x + iy = r(cos θ + i sin θ)

r = √(22 + 22)

r = √8

2 + 2i = √8 cos θ  + √8 i sin θ 

√8 cos θ = 2

cos 𝜃=28cos 𝜃=222cos 𝜃=12

√8 sin θ = 2

sin 𝜃=28sin 𝜃=222sin 𝜃=12

The angle lies in the first quadrant.

θ = 45° (or) π/4

2+2i=8 cos 𝜋4+i sin 𝜋4

z2 = 1 + i

r = √(12 + 12)

r = √2

1 + i = √2 cos θ  + √2 i sin θ 

√2 cos θ = 1

cos θ = 1/√2

√2 sin θ = 1

sin θ = 1/√2

θ lies in first quadrant.

θ = 45° (or) π/4

1+i=2 cos 𝜋4+i sin 𝜋4z1z2=r1r2[cos (𝜃1-𝜃2)+i sin (𝜃1-𝜃2)]z1z2=82cos 𝜋4-𝜋4+i sin𝜋4-𝜋4z1z2=2[cos (0)+i sin (0)]

Problem 8 :

z1 = 2 - 2i

z2 = 1 - i

Solution:

z1 = 2 - 2i

x + iy = r(cos θ + i sin θ)

r = √(22 + (-2)2)

r = √8

2 - 2i = √8 cos θ  + √8 i sin θ 

√8 cos θ = 2

cos θ = 1/√2

√8 sin θ = -2

sin θ = -1/√2

The angle lies in the fourth quadrant.

θ = 2π - α

α = 45°

θ = 2π - π/4

θ = 7π/4

2-2i=8cos 7𝜋4+i sin7𝜋4

z2 = 1 - i

r = √12 + (-1)2)

r = √2

1 - i = √2 cos θ  + √2 i sin θ 

√2 cos θ = 1

cos θ = 1/√2

√2 sin θ = -1

sin θ = -1/√2

θ lies in the fourth quadrant.

θ = 2π - α

α = 45°

θ = 2π - π/4

θ = 7π/4

1-i=2 cos 7𝜋4+i sin 7𝜋4z1z2=r1r2[cos (𝜃1-𝜃2)+i sin (𝜃1-𝜃2)]z1z2=82cos 7𝜋4-7𝜋4+i sin7𝜋4-7𝜋4z1z2=2[cos (0)+i sin (0)]

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