Class 11 Maths Chapter 4: Complex Numbers and Quadratic Equations worksheet with answer key

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: (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: (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 + 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.

Let z = a + ib. Then iz = −b + ia, and z̄ = a − ib. Equating parts: a = −b and −b = a, so b = −a and z = a − ia. Then |z| = √(a² + a²) = |a|√2 = √2, so a = ±1. Hence z = 1 − i or z = −1 + i.

• writes iz = −b + ia and equates with a − ib to get b = −a (1)
• uses |z| = √2 to get a = ±1: z = 1 − i or −1 + i (1)

√(−4) = 2i and √(−25) = 5i, so √a × √b = 10i² = −10. But ab = 100 and √(ab) = 10. Since −10 ≠ 10, √a × √b ≠ √(ab) when a and b are both negative.

• finds √a × √b = (2i)(5i) = −10 (1)
• finds √(ab) = 10 and concludes they differ (1)

z = (2 + i)(1 + i)/((1 − i)(1 + i)) = (2 + 2i + i − 1)/2 = 1/2 + (3/2)i. The point (1/2, 3/2) has both coordinates positive, so it is in the first quadrant. Its distance from O is |z| = √(1/4 + 9/4) = √10/2.

• z = 1/2 + (3/2)i (1)
• first quadrant and distance √10/2 (1)

Taking moduli, |x + iy|² = |7 + 24i|, so x² + y² = √(49 + 576) = 25. Expanding, (x + iy)² = (x² − y²) + 2xyi, so x² − y² = 7 and 2xy = 24. Adding and subtracting with x² + y² = 25: x² = 16, y² = 9. Since xy = 12 > 0, x and y have the same sign: (x, y) = (4, 3) or (−4, −3).

• x² + y² = |7 + 24i| = 25 (1)
• x² − y² = 7, 2xy = 24, so x² = 16, y² = 9 (1)
• uses xy > 0: (4, 3) or (−4, −3) (1)

z̄₁ z₂ is the conjugate of z₁ z̄₂ (the conjugate of a product is the product of the conjugates). So the sum is w + w̄ = 2 Re w with w = z₁ z̄₂, which is real. w = (1 + 2i)(3 + i) = 3 + i + 6i + 2i² = 1 + 7i, so the sum is 2 × 1 = 2. Check: z̄₁ z₂ = (1 − 2i)(3 − i) = 1 − 7i, and (1 + 7i) + (1 − 7i) = 2.

• notes z̄₁ z₂ is the conjugate of z₁ z̄₂, so the sum is 2 Re (z₁ z̄₂), real (1)
• finds z₁ z̄₂ = 1 + 7i (1)
• the sum is 2 (1)

z₁ + z₂ = 4 and z₁z₂ = |z₁|² = 5. So 4³ = z₁³ + z₂³ + 3 × 5 × 4, that is 64 = z₁³ + z₂³ + 60, and z₁³ + z₂³ = 4. It is real because z₂ = z̄₁, so z₂³ is the conjugate of z₁³ and the sum is 2 Re (z₁³). (Check: z₁³ = 2 + 11i.)

• z₁ + z₂ = 4 and z₁z₂ = 5 (1)
• substitutes in the identity: z₁³ + z₂³ = 4 (1)
• reason: z₂³ is the conjugate of z₁³, so the sum is 2 Re z₁³ (1)