Complex Numbers & Quadratic Equations – Chapter 4

1

The value of i² is:

(A) 1

(B) −1

(C) i

(D) −i

✓ ANSWER: (B)

By definition, i = √(−1), so i² = −1.

2

The value of i⁴⁰ is:

(A) i

(B) −1

(C) −i

(D) 1

✓ ANSWER: (D)

40 = 4×10 ⟹ i⁴⁰ = (i⁴)¹⁰ = 1¹⁰ = 1

3

The real part of the complex number z = 3 − 5i is:

(A) −5

(B) 5

(C) 3

(D) −3

✓ ANSWER: (C)

For z = a + ib, Re(z) = a. Here z = 3 − 5i, so Re(z) = 3.

4

If z₁ = 2 + 3i and z₂ = 1 − i, then z₁ + z₂ is:

(A) 3 + 2i

(B) 1 + 4i

(C) 3 − 2i

(D) 1 − 4i

✓ ANSWER: (A)

z₁ + z₂ = (2+1) + (3+(−1))i = 3 + 2i

5

The conjugate of z = 4 − 7i is:

(A) −4 + 7i

(B) 4 + 7i

(C) −4 − 7i

(D) 7 + 4i

✓ ANSWER: (B)

Conjugate of z = a + ib is z̄ = a − ib. Here z̄ = 4 + 7i.

6

The modulus of z = 3 + 4i is:

(A) 7

(B) 1

(C) 5

(D) 25

✓ ANSWER: (C)

|z| = √(3² + 4²) = √(9 + 16) = √25 = 5

7

The value of i⁻³⁵ is:

(A) 1

(B) −1

(C) i

(D) −i

✓ ANSWER: (C)

i⁻³⁵ = 1/i³⁵. Now 35 = 4×8 + 3, so i³⁵ = i³ = −i.
∴ i⁻³⁵ = 1/(−i) = i/1 = i (multiplying by i/i: i/(−i²) = i/1 = i)

8

The product (3 + 5i)(2 + 6i) equals:

(A) 6 + 30i

(B) −24 + 28i

(C) 24 − 28i

(D) 36 + 28i

✓ ANSWER: (B)

(ac − bd) + i(ad + bc) = (3×2 − 5×6) + i(3×6 + 5×2)
= (6 − 30) + i(18 + 10) = −24 + 28i

9

Which property ensures z + 0 = z for every complex number z?

(A) Closure Law

(B) Commutative Law

(C) Additive Identity

(D) Additive Inverse

✓ ANSWER: (C)

The existence of Additive Identity (0 + 0i) ensures z + 0 = z.

10

The multiplicative inverse of z = 2 − 3i is:

(A) (2 + 3i)/13

(B) (2 − 3i)/13

(C) (2 + 3i)/5

(D) −(2 + 3i)/13

✓ ANSWER: (A)

z⁻¹ = z̄/|z|² = (2 + 3i)/(4 + 9) = (2 + 3i)/13

11

The value of i⁹ + i¹⁹ is:

(A) 0

(B) 2i

(C) −2i

(D) 1

✓ ANSWER: (A)

9 = 4×2+1 ⟹ i⁹ = i
19 = 4×4+3 ⟹ i¹⁹ = i³ = −i
i⁹ + i¹⁹ = i + (−i) = 0

12

The additive inverse of z = 5 − 3i is:

(A) 5 + 3i

(B) −5 + 3i

(C) −5 − 3i

(D) 3 − 5i

✓ ANSWER: (B)

Additive inverse of a + ib is −a − ib. Here it is −5 + 3i.

13

If z = a + ib, then z · z̄ equals:

(A) a² − b²

(B) a² + b²

(C) 2ab·i

(D) |z|

✓ ANSWER: (B)

z · z̄ = (a + ib)(a − ib) = a² − (ib)² = a² + b² = |z|²

14

For the quadratic equation ax² + bx + c = 0, complex roots appear when:

(A) b² − 4ac > 0

(B) b² − 4ac = 0

(C) b² − 4ac < 0

(D) b² = 4ac + 1

✓ ANSWER: (C)

When discriminant D = b² − 4ac < 0, the equation has no real roots — it has two complex conjugate roots.

15

The modulus of z = 2 − 5i is:

(A) √21

(B) √29

(C) 7

(D) 3

✓ ANSWER: (B)

|z| = √(2² + (−5)²) = √(4 + 25) = √29

16

The value of i¹ + i² + i³ + i⁴ is:

(A) 2

(B) −2

(C) 0

(D) 1

✓ ANSWER: (C)

i + (−1) + (−i) + 1 = 0

17

The plane used to represent complex numbers geometrically is called:

(A) Cartesian Plane

(B) Argand Plane

(C) Polar Plane

(D) Complex Grid

✓ ANSWER: (B)

The plane having a complex number assigned to each point is called the Argand Plane (or complex plane).

18

For z = x + iy, its conjugate z̄ is represented in the Argand plane as the reflection of z about the:

(A) Imaginary axis

(B) Real axis

(C) Origin

(D) Line y = x

✓ ANSWER: (B)

z = (x, y) and z̄ = (x, −y). The point (x, −y) is the mirror image of (x, y) about the real axis.

19

Which of the following is NOT true for complex numbers z₁ and z₂?

(A) |z₁z₂| = |z₁||z₂|

(B) z₁z₂̄ = z̄₁z̄₂

(C) √(z₁·z₂) = √z₁·√z₂ (when both negative)

(D) z₁ + z₂̄ = z̄₁ + z̄₂

✓ ANSWER: (C)

√a × √b ≠ √(ab) when both a and b are negative real numbers. This is a well-known caveat from the chapter.

20

The value of i⁴ᵏ⁺³ for any integer k is:

(A) 1

(B) i

(C) −1

(D) −i

✓ ANSWER: (D)

i⁴ᵏ = 1, i⁴ᵏ⁺¹ = i, i⁴ᵏ⁺² = −1, i⁴ᵏ⁺³ = −i

21

The roots of x² + 4 = 0 are:

(A) ±2

(B) ±2i

(C) ±4i

(D) No roots

✓ ANSWER: (B)

x² = −4 ⟹ x = ±√(−4) = ±2√(−1) = ±2i

22

For |z₁z₂|, which identity holds?

(A) |z₁| + |z₂|

(B) |z₁| − |z₂|

(C) |z₁| · |z₂|

(D) |z₁|² + |z₂|²

✓ ANSWER: (C)

Standard property: |z₁z₂| = |z₁| · |z₂|

23

The value of (1 − i)⁴ is:

(A) 4

(B) −4

(C) 4i

(D) −4i

✓ ANSWER: (A)

(1−i)² = 1 − 2i + i² = 1 − 2i − 1 = −2i
[(1−i)²]² = (−2i)² = 4i² = 4(−1) = −4
Note: Correct answer is (B) −4.

24

If 4x + i(3x − y) = 3 + i(−6), the value of y is:

(A) 33/4

(B) −33/4

(C) 3/4

(D) 6

✓ ANSWER: (A)

Equating real parts: 4x = 3 ⟹ x = 3/4
Equating imaginary parts: 3x − y = −6
3(3/4) − y = −6 ⟹ 9/4 − y = −6 ⟹ y = 9/4 + 6 = 33/4

25

The roots of x² + x + 1 = 0 are:

(A) Real and equal

(B) Real and distinct

(C) Complex conjugates

(D) Zero

✓ ANSWER: (C)

D = 1² − 4(1)(1) = 1 − 4 = −3 < 0
Roots = (−1 ± √(−3))/2 = (−1 ± i√3)/2 — these are complex conjugates.

NUM N1

Express (5 − 3i)³ in the form a + ib.

✓ SOLUTION

Using (a−b)³ = a³ − 3a²b + 3ab² − b³ with a=5, b=3i:
= 125 − 3(25)(3i) + 3(5)(9i²) − 27i³
= 125 − 225i + 135(−1) − 27(−i)
= 125 − 225i − 135 + 27i
= −10 − 198i

NUM N2

Express (5 + √2i)/(1 − √2i) in the form a + ib.

✓ SOLUTION

Multiply numerator and denominator by conjugate (1 + √2i):
= (5 + √2i)(1 + √2i) / [(1)² + (√2)²]
= (5 + 5√2i + √2i + 2i²) / (1 + 2)
= (5 + 6√2i − 2) / 3
= (3 + 6√2i) / 3 = 1 + 2√2i

NUM N3

Find the multiplicative inverse of z = 4 − 3i.

✓ SOLUTION

z̄ = 4 + 3i, |z|² = 4² + 3² = 16 + 9 = 25
z⁻¹ = z̄/|z|² = (4 + 3i)/25 = 4/25 + 3i/25

NUM N4

If 4x + i(3x − y) = 3 + i(−6), find x and y.

✓ SOLUTION

Equating real parts: 4x = 3 ⟹ x = 3/4
Equating imaginary parts: 3x − y = −6
3(3/4) − y = −6 ⟹ 9/4 + 6 = y ⟹ y = 33/4

NUM N5

Solve the equation x² + 2 = 0 over complex numbers.

✓ SOLUTION

x² = −2 ⟹ x = ±√(−2) = ±√2 · i
∴ x = √2i or x = −√2i

NUM N6

Solve: x² + x + 1 = 0 over complex numbers.

✓ SOLUTION

D = b² − 4ac = 1 − 4 = −3
x = (−1 ± √(−3)) / 2 = (−1 ± √3·i) / 2
∴ x = (−1 + i√3)/2 or (−1 − i√3)/2

NUM N7

Express i⁻³⁹ in the form a + ib.

✓ SOLUTION

39 = 4×9 + 3 ⟹ i³⁹ = i³ = −i
i⁻³⁹ = 1/(−i) = 1/(−i) × (i/i) = i/(−i²) = i/1 = 0 + i

NUM N8

Find the conjugate of [(3 − 2i)(2 + 3i)] / [(1 + 2i)(2 − i)].

✓ SOLUTION

Numerator: (3−2i)(2+3i) = 6+9i−4i−6i² = 6+5i+6 = 12+5i
Denominator: (1+2i)(2−i) = 2−i+4i−2i² = 2+3i+2 = 4+3i
Now: (12+5i)/(4+3i) × (4−3i)/(4−3i)
= (48−36i+20i−15i²) / (16+9)
= (48+15 − 16i) / 25 = (63 − 16i)/25
Conjugate = 63/25 + 16i/25

NUM N9

Express (3(7 + 7i) + i(7 + 7i)) in the form a + ib.

✓ SOLUTION

= 21 + 21i + 7i + 7i²
= 21 + 28i − 7
= 14 + 28i

NUM N10

Solve the quadratic equation: x² − 2x + 5 = 0.

✓ SOLUTION

D = (−2)² − 4(1)(5) = 4 − 20 = −16
x = (2 ± √(−16)) / 2 = (2 ± 4i) / 2
∴ x = 1 + 2i or 1 − 2i

NUM N11

Find the real numbers x and y if (x − iy)(3 + 5i) is the conjugate of −6 − 24i.

✓ SOLUTION

Conjugate of (−6 − 24i) = −6 + 24i
So (x − iy)(3 + 5i) = −6 + 24i
(3x + 5y) + i(5x − 3y) = −6 + 24i
Equating: 3x + 5y = −6 ... (1)
5x − 3y = 24 ... (2)
From (1): 3x + 5y = −6; From (2)×5 + (1)×3: 25x − 15y + 9x + 15y = 120 − 18 ⟹ 34x = 102 ⟹ x = 3
Sub in (1): 9 + 5y = −6 ⟹ y = −3

NUM N12

Find the modulus of (1+i)/(1−i) − (1−i)/(1+i).

✓ SOLUTION

(1+i)/(1−i) = (1+i)²/[(1−i)(1+i)] = (1+2i−1)/2 = 2i/2 = i
(1−i)/(1+i) = (1−i)²/2 = (1−2i−1)/2 = −2i/2 = −i
Expression = i − (−i) = 2i
Modulus = |2i| = 2

NUM N13

Prove that Re(z₁z₂) = Re z₁ · Re z₂ − Im z₁ · Im z₂.

✓ SOLUTION

Let z₁ = a + ib, z₂ = c + id
z₁z₂ = (ac − bd) + i(ad + bc)
∴ Re(z₁z₂) = ac − bd
Re z₁ · Re z₂ = a·c, Im z₁ · Im z₂ = b·d
∴ Re z₁ · Re z₂ − Im z₁ · Im z₂ = ac − bd = Re(z₁z₂) ✓

NUM N14

Solve: 21x² − 28x + 10 = 0 over complex numbers.

✓ SOLUTION

D = (−28)² − 4(21)(10) = 784 − 840 = −56
x = (28 ± √(−56)) / (2×21) = (28 ± 2√14·i) / 42
= (14 ± √14·i) / 21
∴ x = (14 + i√14)/21 or (14 − i√14)/21

NUM N15

If ((1+i)/(1−i))ᵐ = 1, find the least positive integral value of m.

✓ SOLUTION

(1+i)/(1−i) × (1+i)/(1+i) = (1+i)²/(1+1) = (1+2i−1)/2 = 2i/2 = i
So the equation becomes iᵐ = 1.
iᵐ = 1 when m is a multiple of 4.
Least positive integral value: m = 4

Chapter 4: Complex Numbers & Quadratic Equations

Multiple Choice Questions

Numerical / Descriptive Questions

Ready to Crack AAI ATC Exam? 🚀