Ten multiple-choice questions on complex numbers in polar form: modulus and argument, conjugates, polar conversion, products, De Moivre’s theorem, distance and midpoint. Choose the one best answer, then read the explanation to learn which mistake each wrong option comes from.
PrecalculusCommon Core: N-CN.B.4 · N-CN.B.5 · N-CN.B.6 · N-CN.A.3
🎯 Complex Numbers in Polar Form – Precalculus math quiz
Question 1
Find \(|-6+8i|\).
A
\(14\)
B
\(2\)
C
\(10\)
D
\(100\)
Question 2
What is the argument of the complex number shown?A point in the complex plane.
A
\(45^{\circ}\)
B
\(135^{\circ}\)
C
\(225^{\circ}\)
D
\(315^{\circ}\)
Question 3
What is the conjugate of \(4-7i\)?
A
\(4+7i\)
B
\(-4+7i\)
C
\(-4-7i\)
D
\(7-4i\)
Question 4
Evaluate \((3+2i)(3-2i)\).
A
\(5\)
B
\(9\)
C
\(13\)
D
\(-13\)
Question 5
Which is the polar form of \(-1+i\)?
A
\(\sqrt{2}\left(\cos 45^{\circ}+i\sin 45^{\circ}\right)\)
B
\(2\left(\cos 135^{\circ}+i\sin 135^{\circ}\right)\)
C
\(\sqrt{2}\left(\cos 225^{\circ}+i\sin 225^{\circ}\right)\)
D
\(\sqrt{2}\left(\cos 135^{\circ}+i\sin 135^{\circ}\right)\)
Question 6
Find \(2\left(\cos 30^{\circ}+i\sin 30^{\circ}\right)\cdot5\left(\cos 60^{\circ}+i\sin 60^{\circ}\right)\).
A
\(7\left(\cos 90^{\circ}+i\sin 90^{\circ}\right)\)
B
\(10\left(\cos 30^{\circ}+i\sin 30^{\circ}\right)\)
C
\(10\left(\cos 90^{\circ}+i\sin 90^{\circ}\right)\)
D
\(10\left(\cos 1800^{\circ}+i\sin 1800^{\circ}\right)\)
Question 7
Use De Moivre's theorem to evaluate \((1-i)^{6}\).
A
\(8i\)
B
\(-8i\)
C
\(8\)
D
\(-8\)
Question 8
What is the distance between the two complex numbers shown?Two complex numbers.
A
\(7\)
B
\(25\)
C
\(1\)
D
\(5\)
Question 9
Find the midpoint of \(-4+i\) and \(6+5i\).
A
\(2+6i\)
B
\(5+2i\)
C
\(-5-2i\)
D
\(1+3i\)
Question 10
The figure shows \(3+i\). Which number is \(i(3+i)\), the point rotated \(90^{\circ}\) counterclockwise?The point 3 + i and a rotated point.
A
\(1-3i\)
B
\(-1+3i\)
C
\(-1-3i\)
D
\(1+3i\)
🎉 Quiz complete!
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Remember:
- Modulus |a + bi| = √(a² + b²); the conjugate a − bi satisfies z · z̄ = |z|².
- Polar form: z = r(cos θ + i sin θ), with a = r cos θ and b = r sin θ.
- Product: multiply moduli, add arguments. Quotient: divide moduli, subtract arguments.
- De Moivre: [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ).
- Distance between z₁ and z₂ is |z₁ − z₂|; the midpoint is (z₁ + z₂)/2.
See every chapter on the Precalculus math page.