← Back to the Complex Numbers Polar Form worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: N-CN.B.4, N-CN.B.5, N-CN.B.6, N-CN.A.3 · Level: Grade 12 Precalculus
Problem 1 Modulus of a Complex Number
|z| = √(22 + (−5)2) = √(4 + 25) = √29 ≈ 5.39.
Answer: √29 ≈ 5.39
Problem 2 Conjugate and Product
z̄ = 3 + 7i. Then z·z̄ = 32 + 72 = 9 + 49 = 58, and |z|2 = 9 + 49 = 58, so they are equal.
Answer: z̄ = 3 + 7i and z·z̄ = 58 = |z|2
Problem 3 Reading a Point
The point is 6 units left and 1 unit up, so z = −6 + i. Its conjugate is −6 − i. The modulus is √(36 + 1) = √37.
Answer: z = −6 + i, z̄ = −6 − i, |z| = √37
Problem 4 Modulus and Argument
r = √(25 + 25) = 5√2. The reference angle is tan⁻¹(5 ÷ 5) = 45°, and the number is in Quadrant II, so θ = 180° − 45° = 135°.
Answer: r = 5√2, θ = 135°
Problem 5 Distance and Midpoint
Difference: (4 − (−3)) + (1 − 5)i = 7 − 4i. Distance = √(49 + 16) = √65 ≈ 8.06.
Midpoint = (−3 + 4) + (5 + 1)i2 = 12 + 3i.
Answer: PQ = √65 and the midpoint is 12 + 3i
Problem 6 Polar to Rectangular
cos 60° = 12 and sin 60° = √32. So 6·12 = 3 and 6·√32 = 3√3.
Answer: 3 + 3√3i
Problem 7 Rectangular to Polar
r = √(4 + 12) = 4. The reference angle is tan⁻¹(2√3 ÷ 2) = tan⁻¹(√3) = 60°. In Quadrant IV, θ = 360° − 60° = 300°.
Answer: 4(cos 300° + i sin 300°)
Problem 8 Multiplying in Polar Form
Multiply moduli: 4 · 5 = 20. Add angles: 25° + 35° = 60°. Polar form: 20(cos 60° + i sin 60°).
Rectangular: 20·12 = 10 and 20·√32 = 10√3.
Answer: 20(cos 60° + i sin 60°) = 10 + 10√3i
Problem 9 Dividing in Polar Form
Divide moduli: 18 ÷ 6 = 3. Subtract angles: 200° − 80° = 120°. The quotient is 3(cos 120° + i sin 120°).
cos 120° = −12 and sin 120° = √32, so the result is −32 + 3√32i.
Answer: −32 + 3√32i
Problem 10 Product via Polar Form
z₁: r = √2, θ = 45°. z₂: r = √(1 + 3) = 2, θ = 60°.
Product: modulus √2 · 2 = 2√2, argument 45° + 60° = 105°.
Check: z₁z₂ = (1 − √3) + (1 + √3)i, and its modulus squared is (1 − √3)2 + (1 + √3)2 = 4 − 2√3 + 4 + 2√3 = 8, so the modulus is 2√2.
Answer: modulus 2√2, argument 105°
Problem 11 A Tenth Power
1 − i = √2(cos(−45°) + i sin(−45°)). The tenth power has modulus (√2)10 = 32 and angle 10 · (−45°) = −450°, which is coterminal with −90°.
So the result is 32(cos(−90°) + i sin(−90°)) = 32(0 − i) = −32i.
Check: (1 − i)2 = −2i, and (−2i)5 = −32i5 = −32i.
Answer: −32i
Problem 12 Cubing in Polar Form
The modulus becomes 23 = 8 and the angle becomes 3 · 20° = 60°. So the result is 8(cos 60° + i sin 60°) = 8·12 + 8·√32i.
Answer: 4 + 4√3i
Problem 13 A Sixth Power
r = √(3 + 1) = 2 and θ = tan⁻¹(1√3) = 30°. Then (√3 + i)6 = 26(cos 180° + i sin 180°) = 64(−1).
Answer: −64
Problem 14 A Fifth Power
r = √2 and θ = 135° (Quadrant II). Then (−1 + i)5 = (√2)5(cos 675° + i sin 675°). Since (√2)5 = 4√2 and 675° − 360° = 315°, this is 4√2(cos 315° + i sin 315°) = 4√2(√22 − √22i) = 4 − 4i.
Check: (−1 + i)2 = −2i, (−1 + i)4 = −4, and −4(−1 + i) = 4 − 4i.
Answer: 4 − 4i
Problem 15 Cube Roots of 64
Write 64 = 64(cos 0° + i sin 0°). The cube roots have modulus ∛64 = 4 and angles (0° + 360°k)/3 for k = 0, 1, 2, which are 0°, 120° and 240°.
k = 0: 4. k = 1: 4(−12 + √32i) = −2 + 2√3i. k = 2: −2 − 2√3i.
Answer: 4, −2 + 2√3i, −2 − 2√3i
Problem 16 Trail Distance
B − A = (5 − (−2)) + (−3 − 6)i = 7 − 9i. Distance = √(49 + 81) = √130 ≈ 11.40 miles.
Midpoint = (−2 + 5) + (6 − 3)i2 = 1.5 + 1.5i.
Answer: about 11.40 miles; rest stop at 1.5 + 1.5i
Problem 17 Robot Rotation
(a) 2i(3 + 2i) = 6i + 4i2 = −4 + 6i.
(b) (2i)4 = 16i4 = 16, so the position is 16(3 + 2i) = 48 + 32i.
Check: four 90° turns make a full turn and the distance is multiplied by 24 = 16.
Answer: (a) −4 + 6i; (b) 48 + 32i
Problem 18 Alternating Current
Divide moduli: 48 ÷ 6 = 8. Subtract angles: 70° − 25° = 45°. In polar form the current is 8(cos 45° + i sin 45°) amps.
Rectangular: 8·√22 = 4√2 ≈ 5.66, so the current is about 5.66 + 5.66i.
Answer: 8(cos 45° + i sin 45°) ≈ 5.66 + 5.66i amps
Problem 19 Sixth Roots of Unity
(a) By De Moivre, z6 = cos 360° + i sin 360° = 1.
(b) The roots are cos θ + i sin θ for θ = 0°, 60°, 120°, 180°, 240°, 300°: 1, 12 + √32i, −12 + √32i, −1, −12 − √32i, 12 − √32i.
(c) They sit 60° apart on the unit circle, so they form a regular hexagon whose side length equals 1.
Answer: a regular hexagon with side length 1
Problem 20 A Quotient of Powers
1 − √3i = 2(cos(−60°) + i sin(−60°)), so its fifth power is 32(cos(−300°) + i sin(−300°)) = 32∠60°.
(1 + i)4 = (√2)4∠180° = 4∠180° = −4.
Quotient: 8∠(60° − 180°) = 8∠(−120°) = 8(−12 − √32i) = −4 − 4√3i.
Answer: −4 − 4√3i
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