Questions: Operations with Complex Numbers

Question 1

What is the result of (2 + 3i)(1 − i)?

• A. 5 + i
• B. 2 − 2i
• C. 2 + i
• D. −1 + 5i

Answer: (A)

Using FOIL: (2 + 3i)(1 − i) = 2 − 2i + 3i − 3i². Replacing i² with −1 gives 2 − 2i + 3i + 3 = 5 + i. Option B is the classic error of leaving i² unreplaced — if a student treats i² as just i, the last term becomes −3i and they get 2 − 2i. The substitution i² = −1 is the single rule that makes complex multiplication work; skipping it produces a result that still contains i², which is not simplified.

Question 2

What is (3 + i) ÷ (1 + 2i)?

• A. 1 − i
• B. 3 − i
• C. (1 + 7i)/5
• D. (5 − 5i)/3

Answer: (A)

Multiply numerator and denominator by the conjugate (1 − 2i): numerator = (3 + i)(1 − 2i) = 3 − 6i + i − 2i² = 3 − 5i + 2 = 5 − 5i; denominator = 1² + 2² = 5. Result: (5 − 5i)/5 = 1 − i. Option C results from multiplying by (1 + 2i)/(1 + 2i) instead of the conjugate — the imaginary terms in the denominator add rather than cancel, so the denominator remains complex.

Question 3

The product of any complex number and its conjugate is always a real number.

• T. True
• F. False

Answer: True

(a + bi)(a − bi) = a² − (bi)² = a² − b²i² = a² + b². The imaginary parts cancel exactly because the conjugate has the opposite sign on the imaginary component. This is the difference-of-squares pattern applied to complex numbers, and it is precisely why multiplying by the conjugate is the standard strategy for dividing complex numbers.

Question 4

The modulus of 3 + 4i is 7.

• T. True
• F. False

Answer: False

The modulus |a + bi| = √(a² + b²), not a + b. For 3 + 4i: |3 + 4i| = √(9 + 16) = √25 = 5. The common error is adding the real and imaginary parts directly (3 + 4 = 7), but the modulus is the Pythagorean distance from the origin to the point (3, 4) in the complex plane — the two components are legs of a right triangle, and the modulus is the hypotenuse.

Question 5

Why does multiplying both the numerator and denominator of a complex fraction by the conjugate of the denominator allow you to simplify the result?

Model answer: Multiplying a complex number by its conjugate eliminates the imaginary part: (a + bi)(a − bi) = a² + b², a real number. Applying this to the denominator converts the divisor from a complex number to a real number, making ordinary division straightforward. Multiplying both numerator and denominator by the same expression preserves the value of the fraction (you are multiplying by 1 in the form conjugate/conjugate), so only the form changes.

The key is the identity (a + bi)(a − bi) = a² + b². Once the denominator is real, you can divide the numerator's real and imaginary parts separately by that real number. This is not a trick but a direct application of the difference-of-squares pattern, which guarantees that the cross terms (involving i) always cancel when you multiply conjugate pairs.