PRE CALCULUS Pre-Calculus Complex Numbers 3 — Questions and Answers
Question 1: What is the result of dividing (6 + 2i) by (1 + i)?
- 4 - 2i (Correct answer)
- 4 + 2i
- 4 - i
- 3 + i
Correct answer: 4 - 2i
Multiply numerator and denominator by the conjugate (1 - i): (6+2i)(1-i)/[(1+i)(1-i)] = (6-6i+2i-2i²)/2 = (8-4i)/2 = 4 - 2i.
Question 2: Using De Moivre's Theorem, what is [2(cos 30° + i sin 30°)]³?
- 8(cos 60° + i sin 60°)
- 8(cos 90° + i sin 90°) (Correct answer)
- 6(cos 90° + i sin 90°)
- 8i
Correct answer: 8(cos 90° + i sin 90°)
By De Moivre's Theorem, r^n(cos nθ + i sin nθ) = 2³(cos 90° + i sin 90°) = 8(cos 90° + i sin 90°) = 8i.
Question 3: How many distinct cube roots does any nonzero complex number have?
- 1
- 2
- 3 (Correct answer)
- 6
Correct answer: 3
Every nonzero complex number has exactly n distinct nth roots, so there are 3 cube roots.
Question 4: What is the real part of (3 - 2i)² ?
- 5 (Correct answer)
- 13
- 9
- -12
Correct answer: 5
(3 - 2i)² = 9 - 12i + 4i² = 9 - 12i - 4 = 5 - 12i, so the real part is 5.
Question 5: Which property states that the modulus of a product equals the product of the moduli?
- |z₁ + z₂| = |z₁| + |z₂|
- |z₁ · z₂| = |z₁| · |z₂| (Correct answer)
- |z₁/z₂| = |z₁| + |z₂|
- |z̄| = -|z|
Correct answer: |z₁ · z₂| = |z₁| · |z₂|
The multiplicative property of moduli states that |z₁ · z₂| = |z₁| · |z₂|.
Question 6: What is the imaginary part of 1/(2 + i)?
- 1/5
- -1/5 (Correct answer)
- 2/5
- -2/5
Correct answer: -1/5
Multiply by conjugate: (2 - i)/((2+i)(2-i)) = (2-i)/5, so the imaginary part is -1/5.
Question 7: If z = r(cos θ + i sin θ), what is 1/z in polar form?
- r(cos θ - i sin θ)
- (1/r)(cos θ + i sin θ)
- (1/r)(cos(-θ) + i sin(-θ)) (Correct answer)
- r(cos(-θ) + i sin(-θ))
Correct answer: (1/r)(cos(-θ) + i sin(-θ))
The reciprocal has modulus 1/r and argument -θ, giving (1/r)(cos(-θ) + i sin(-θ)).
PRE CALCULUS Pre-Calculus Complex Numbers 3
What is the result of dividing (6 + 2i) by (1 + i)?