Class 11 Maths Chapter 4: Complex Numbers and Quadratic Equations MCQs with answers

Answer: (c) −8

For z = a + ib the imaginary part is the real number b, not ib. Here b = −8, so Im z = −8. Writing −8i takes ib for b; −√5 is Re z.

Answer: (a) −4

Multiply above and below by the conjugate 1 + 2i: (2 + ki)(1 + 2i)/5 = [(2 − 2k) + (4 + k)i]/5. It is purely real when its imaginary part is 0, so 4 + k = 0 and k = −4. Taking 4 drops the sign of i² = −1; 1 comes from setting the real part to zero instead.

Answer: (a) i⁶

6 = 4 × 1 + 2, so i⁶ = i² = −1. Since 8 and −4 are multiples of 4, i⁸ = i⁻⁴ = 1; and 9 = 4 × 2 + 1 gives i⁹ = i.

Answer: (d) −1 + i

Every four consecutive powers of i add to 0, so the first 48 terms add to 0. What is left is i⁴⁹ + i⁵⁰ = i + (−1) = −1 + i. The answer 0 forgets that 50 is not a multiple of 4.

Answer: (b) −2 + 3i

z₂ = z₁ − (z₁ − z₂) = (4 − 7i) + (−6 + 10i) = −2 + 3i. Adding instead of subtracting gives 10 − 17i; 2 − 3i is −z₂.

Answer: (c) 3

z₁z₂ = (8 + 3) + i(−2 + 12) = 11 + 10i, so Re (z₁z₂) = 11, while Re z₁ · Re z₂ = 8. The difference is 3, which is −Im z₁ · Im z₂ = −(3)(−1). The answer 0 assumes Re (z₁z₂) = Re z₁ Re z₂, which is false.

Answer: (b) −12 + 5i

|−12 + 5i| = √(144 + 25) = √169 = 13. The others give √89, √170 and √145. 13 − i has real part 13 but its modulus is larger than 13.

Answer: (c) third quadrant

z̄ = −3 − 2i is the point (−3, −2), the mirror image of (−3, 2) in the real axis; both coordinates are negative, so it lies in the third quadrant. The second quadrant is where z itself lies.

Answer: (d) 1 + 2i

z is the inverse of z⁻¹: z = 5/(1 − 2i) = 5(1 + 2i)/(1² + 2²) = 1 + 2i. Check: (1 + 2i)(1 − 2i)/5 = 5/5 = 1. The option 1 − 2i forgets to take the conjugate.

Answer: (a) −12

√(−16) = 4i and √(−9) = 3i, so the product is 12i² = −12. Writing √((−16)(−9)) = √144 = 12 uses √a × √b = √(ab), which fails when both a and b are negative.