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Complex Numbers Polar Form Test Answer Key with Scoring – Precalculus

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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers

Precalculus Math Test Answer Keys – Mathbot, the Math Made Easy robot coach, with free printable test answer keys for Precalculus

Use the scoring notes to award partial credit. Most questions give a point for the correct modulus or angle and a point for the final form. Exact values are required wherever the question asks for them.

Standards: N-CN.B.4, N-CN.B.5, N-CN.B.6, N-CN.A.3 · Level: Grade 12 Precalculus

Question 2 Rectangular to Polar (2 points)

(a) r = √(8 + 8) = 4.

(b) The reference angle is 45° and the number is in Quadrant II, so θ = 135°, and the polar form is 4(cos 135° + i sin 135°).

Answer: r = 4; 4(cos 135° + i sin 135°)

Scoring: 1 point for each part.

Question 3 Polar to Rectangular (2 points)

cos 210° = −√32 and sin 210° = −12. So a = 10·(−√32) = −5√3 and b = 10·(−12) = −5.

Answer: −5√3 − 5i

Scoring: 1 point for the two values from the unit circle, 1 point for the result.

Question 4 Product and Quotient (2 points)

(a) 6 · 3 = 18 and 100° + 40° = 140°, so 18(cos 140° + i sin 140°).

(b) 6 ÷ 3 = 2 and 100° − 40° = 60°, so 2(cos 60° + i sin 60°).

Answer: (a) 18(cos 140° + i sin 140°); (b) 2(cos 60° + i sin 60°)

Scoring: 1 point for each part.

Question 5 Distance and Midpoint (2 points)

Difference: 5 − 8i. Distance = √(25 + 64) = √89 ≈ 9.43.

Midpoint = 1 + 6i2 = 12 + 3i.

Answer: √89 and 12 + 3i

Scoring: 1 point for the distance, 1 point for the midpoint.

Question 6 Using De Moivre (3 points)

(a) 35 = 243 and 5 · 12° = 60°, so 243(cos 60° + i sin 60°).

(b) √3 − i = 2(cos(−30°) + i sin(−30°)). The seventh power is 27(cos(−210°) + i sin(−210°)) = 128(cos 150° + i sin 150°) = 128(−√32 + 12i) = −64√3 + 64i.

Answer: (a) 243(cos 60° + i sin 60°); (b) −64√3 + 64i

Scoring: (a) 1 point. (b) 1 point for the polar form of √3 − i and 1 point for the correct power and final rectangular form.

Question 7 Cube Roots of −8 (3 points)

-3-2-1123-3-2-1123ReIm1 + √3 i−21 − √3 i
The three cube roots of −8 form an equilateral triangle on the circle of radius 2.

−8 = 8(cos 180° + i sin 180°). The cube roots have modulus 2 and angles (180° + 360°k)/3 = 60°, 180°, 300°.

2(cos 60° + i sin 60°) = 1 + √3i; 2(cos 180° + i sin 180°) = −2; 2(cos 300° + i sin 300°) = 1 − √3i.

Answer: 1 + √3i, −2, 1 − √3i

Scoring: 1 point for the modulus and angles, 2 points for the three correct roots (1 point if only two).

Question 8 Scaling and Turning (2 points)

(a) 3i(5 + 2i) = 15i + 6i2 = −6 + 15i.

(b) |P| = √29 and |Q| = √(36 + 225) = √261 = 3√29. Q is three times as far from the origin as P, and it is P turned 90° counterclockwise.

Answer: Q = −6 + 15i; |Q| = 3|P|

Scoring: 1 point for each part.

Question 9 Modulus of a Product (2 points)

zw = 3 − 12i + i − 4i2 = 7 − 11i.

|zw| = √(49 + 121) = √170. Also |z| = √10 and |w| = √17, and √10 · √17 = √170.

Answer: zw = 7 − 11i and both sides equal √170

Scoring: 1 point for the product, 1 point for the comparison of moduli.

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