Question 1

If z = cos θ + i sin θ, then the value of z^n + 1/z^n is:

Accepted Answer

Step 1: By De Moivre's theorem, z^n = cos(nθ) + i sin(nθ).
Step 2: Also, 1/z^n = z^(−n) = cos(nθ) − i sin(nθ)  [since 1/z = cos θ − i sin θ when |z|=1].
Step 3: Adding: z^n + 1/z^n = [cos(nθ) + i sin(nθ)] + [cos(nθ) − i sin(nθ)] = 2 cos(nθ).
Rule used: De Moivre's theorem and the property that for |z|=1, 1/z = conjugate of z.

Question 2

If z = 3 + 4i, then the modulus of z is:

Accepted Answer

The modulus (or absolute value) of a complex number z = a + bi is defined as |z| = √(a² + b²). Here, a = 3 and b = 4. Therefore, |z| = √(3² + 4²) = √(9 + 16) = √25 = 5.

Question 3

If z₁ = 2 + i and z₂ = 1 - i, then |z₁ · z₂| equals:

Accepted Answer

Method 1 (Direct): z₁ · z₂ = (2 + i)(1 - i) = 2 - 2i + i - i² = 2 - i + 1 = 3 - i. Then |z₁ · z₂| = √(3² + (-1)²) = √(9 + 1) = √10.

Method 2 (Property): |z₁ · z₂| = |z₁| · |z₂|. Here |z₁| = √(4 + 1) = √5 and |z₂| = √(1 + 1) = √2. Therefore |z₁ · z₂| = √5 · √2 = √10.

Question 4

The argument of the complex number z = -1 + i√3 is:

Accepted Answer

For z = a + bi, the argument θ is found using tan(θ) = b/a, with careful attention to the quadrant. Here a = -1 and b = √3. Since a < 0 and b > 0, z lies in the second quadrant. We have tan(θ) = √3/(-1) = -√3. In the second quadrant, θ = π - π/3 = 2π/3 (or 120°).

Question 5

If z = cos(π/6) + i·sin(π/6), then z⁶ equals:

Accepted Answer

This is a direct application of De Moivre's Theorem: if z = cos(θ) + i·sin(θ), then zⁿ = cos(nθ) + i·sin(nθ). Here θ = π/6 and n = 6. Therefore, z⁶ = cos(6 · π/6) + i·sin(6 · π/6) = cos(π) + i·sin(π) = -1 + i·0 = -1.

Question 6

The complex number z = 1 + i in polar form is:

Accepted Answer

To convert z = 1 + i to polar form r(cos θ + i sin θ): Step 1: Compute modulus r = |z| = √(1² + 1²) = √2. Step 2: Compute argument θ. Since a = 1 > 0 and b = 1 > 0, z is in the first quadrant. tan(θ) = 1/1 = 1, so θ = π/4. Therefore, z = √2(cos(π/4) + i·sin(π/4)) or √2·e^(iπ/4).

Question 7

If z = 1 + i, then z⁴ equals:

Accepted Answer

We compute z⁴ where z = 1 + i.

Method 1 (Direct computation):
z² = (1 + i)² = 1 + 2i + i² = 1 + 2i - 1 = 2i.
z⁴ = (z²)² = (2i)² = 4i² = -4.

Method 2 (De Moivre's Theorem):
First, express z in polar form:
|z| = √(1² + 1²) = √2.
arg(z) = arctan(1/1) = π/4 (first quadrant).
So z = √2 · (cos(π/4) + i·sin(π/4)).

Then:
z⁴ = (√2)⁴ · (cos(4π/4) + i·sin(4π/4))
   = 4 · (cos(π) + i·sin(π))
   = 4 · (-1 + 0i)
   = -4.

Question 8

If z = cos(π/6) + i·sin(π/6), then z⁶ is equal to:

Accepted Answer

We need to find z⁶ where z = cos(π/6) + i·sin(π/6).

This is a perfect application of De Moivre's Theorem:
[cos(θ) + i·sin(θ)]ⁿ = cos(nθ) + i·sin(nθ)

Applying De Moivre's Theorem with n = 6 and θ = π/6:
z⁶ = [cos(π/6) + i·sin(π/6)]⁶
z⁶ = cos(6 × π/6) + i·sin(6 × π/6)
z⁶ = cos(π) + i·sin(π)
z⁶ = -1 + i·0
z⁶ = -1

Verification: |z| = 1, so |z⁶| = 1. And cos(π) = -1, sin(π) = 0. ✓

Question 9

The complex number z = (1 + i)/(1 - i) can be expressed as:

Accepted Answer

We need to simplify z = (1 + i)/(1 - i) to standard form a + bi.

Step 1: Multiply numerator and denominator by the conjugate of the denominator.
The conjugate of (1 - i) is (1 + i).

z = (1 + i)/(1 - i) × (1 + i)/(1 + i)

Step 2: Expand the numerator.
(1 + i)(1 + i) = 1 + i + i + i² = 1 + 2i + (-1) = 2i

Step 3: Expand the denominator.
(1 - i)(1 + i) = 1 + i - i - i² = 1 - (-1) = 2

Step 4: Simplify.
z = 2i/2 = i

Verification: If z = i, then z(1 - i) = i(1 - i) = i - i² = i + 1 = 1 + i ✓

Question 10

If z = 1 + i, then the modulus of z² is:

Accepted Answer

We need to find |z²| where z = 1 + i.

Method 1 (Direct calculation):
z² = (1 + i)² = 1 + 2i + i² = 1 + 2i - 1 = 2i
|z²| = |2i| = 2

Method 2 (Using modulus property):
|z| = √(1² + 1²) = √2
|z²| = |z|² = (√2)² = 2

Both methods confirm the answer is 2.

Question 11

If z₁ = 2 + i and z₂ = 1 - i, then |z₁/z₂| is equal to:

Accepted Answer

We need to find |z₁/z₂| where z₁ = 2 + i and z₂ = 1 - i.

Method 1 (Using modulus property):
By the property |z₁/z₂| = |z₁|/|z₂|:
|z₁| = √(2² + 1²) = √5
|z₂| = √(1² + (-1)²) = √2
|z₁/z₂| = √5/√2 = √(5/2) = √10/2

Method 2 (Direct division):
z₁/z₂ = (2 + i)/(1 - i) × (1 + i)/(1 + i) = [(2 + i)(1 + i)]/[(1 - i)(1 + i)]
Numerator: (2 + i)(1 + i) = 2 + 2i + i + i² = 2 + 3i - 1 = 1 + 3i
Denominator: (1 - i)(1 + i) = 1 - i² = 1 + 1 = 2
z₁/z₂ = (1 + 3i)/2
|z₁/z₂| = √(1² + 3²)/2 = √10/2

Both methods confirm the answer is √10/2.

Question 12

If z₁ and z₂ are two complex numbers such that |z₁| = 3, |z₂| = 4, and |z₁ + z₂| = 5, then |z₁ - z₂| is equal to:

Accepted Answer

Setup: We use the identity relating |z₁ + z₂|² and |z₁ - z₂|² in terms of |z₁|, |z₂|, and the argument difference.

Key identity:
|z₁ + z₂|² = |z₁|² + |z₂|² + 2·Re(z₁·z̄₂)
|z₁ - z₂|² = |z₁|² + |z₂|² - 2·Re(z₁·z̄₂)

Step 1: Find Re(z₁·z̄₂) using the first equation.
|z₁ + z₂|² = 25
|z₁|² + |z₂|² + 2·Re(z₁·z̄₂) = 25
9 + 16 + 2·Re(z₁·z̄₂) = 25
25 + 2·Re(z₁·z̄₂) = 25
2·Re(z₁·z̄₂) = 0
Re(z₁·z̄₂) = 0

Step 2: Use the second equation.
|z₁ - z₂|² = |z₁|² + |z₂|² - 2·Re(z₁·z̄₂)
|z₁ - z₂|² = 9 + 16 - 2(0)
|z₁ - z₂|² = 25
|z₁ - z₂| = 5

Geometric interpretation: The condition |z₁ + z₂| = 5 with |z₁| = 3 and |z₂| = 4 means z₁ and z₂ are perpendicular (since 3² + 4² = 5²). When two complex numbers are perpendicular, |z₁ - z₂| = |z₁ + z₂|.

Question 13

If z = (1 + i)/(1 - i), then z⁴ is equal to:

Accepted Answer

Setup: We simplify z by rationalizing the denominator, then compute z⁴ using De Moivre's theorem or direct multiplication.

Step 1: Simplify z.
z = (1 + i)/(1 - i)
Multiply numerator and denominator by the conjugate of the denominator (1 + i):
z = [(1 + i)(1 + i)] / [(1 - i)(1 + i)]
z = (1 + 2i + i²) / (1 - i²)
z = (1 + 2i - 1) / (1 - (-1))
z = 2i / 2
z = i

Step 2: Compute z⁴.
z⁴ = i⁴ = (i²)² = (-1)² = 1

Alternative method (polar form):
1 + i = √2·e^(iπ/4)
1 - i = √2·e^(-iπ/4)
z = (√2·e^(iπ/4)) / (√2·e^(-iπ/4)) = e^(iπ/4 - (-iπ/4)) = e^(iπ/2) = i
z⁴ = i⁴ = 1

Both methods confirm z⁴ = 1.

Question 14

If z = 1 + i√3, then the argument of z² is:

Accepted Answer

Setup: We need to find arg(z²). We can either compute z² directly and find its argument, or use the property that arg(z²) = 2·arg(z).

Method 1 (Direct computation):
z = 1 + i√3
z² = (1 + i√3)² = 1 + 2i√3 + (i√3)² = 1 + 2i√3 - 3 = -2 + 2i√3

For z² = -2 + 2i√3:
- Real part = -2, Imaginary part = 2√3
- This lies in the second quadrant
- tan(θ) = (2√3)/(-2) = -√3
- In second quadrant, θ = π - π/3 = 2π/3

Method 2 (Using argument property):
For z = 1 + i√3:
- |z| = √(1 + 3) = 2
- tan(arg z) = √3/1 = √3
- arg(z) = π/3 (first quadrant)
- arg(z²) = 2·arg(z) = 2π/3

Both methods confirm arg(z²) = 2π/3.

Question 15

If z = 1 + i√3, then the modulus and argument of z² are respectively:

Accepted Answer

Step 1: Find z in polar form. |z| = √(1² + (√3)²) = √4 = 2. arg(z) = arctan(√3/1) = π/3 (since both real and imaginary parts are positive, z is in first quadrant). Step 2: Apply De Moivre's theorem: z² = |z|² · e^(i·2·arg(z)) = 4 · e^(i·2π/3). Step 3: Therefore |z²| = 4 and arg(z²) = 2π/3. Verification: z² = (1 + i√3)² = 1 + 2i√3 - 3 = -2 + 2i√3. |z²| = √(4 + 12) = 4 ✓. arg(z²) = arctan(-2√3/-2) in second quadrant = π - π/3 = 2π/3 ✓.

Question 16

The complex number z satisfies |z - 1| = 1 and arg(z) = π/6. Then z is equal to:

Accepted Answer

Setup: We have two conditions: (1) z lies on a circle of radius 1 centered at 1 in the complex plane, and (2) z makes an angle of π/6 with the positive real axis. We find the intersection.

Step 1: Express z in polar form using arg(z) = π/6.
z = r·e^(iπ/6) = r(cos(π/6) + i·sin(π/6)) = r(√3/2 + i/2)
where r = |z|.

Step 2: Apply the condition |z - 1| = 1.
|z - 1| = 1
|r(√3/2 + i/2) - 1| = 1
|(r√3/2 - 1) + i(r/2)| = 1
√[(r√3/2 - 1)² + (r/2)²] = 1
(r√3/2 - 1)² + (r/2)² = 1

Step 3: Expand.
(r√3/2)² - 2·(r√3/2)·1 + 1 + (r/2)² = 1
3r²/4 - r√3 + 1 + r²/4 = 1
r² - r√3 + 1 = 1
r² - r√3 = 0
r(r - √3) = 0

Since z ≠ 0, we have r = √3.

Step 4: Find z.
z = √3(cos(π/6) + i·sin(π/6))
z = √3(√3/2 + i/2)
z = 3/2 + i√3/2

Verification:
|z - 1| = |3/2 - 1 + i√3/2| = |1/2 + i√3/2| = √(1/4 + 3/4) = 1 ✓
arg(z) = arctan((√3/2)/(3/2)) = arctan(1/√3) = π/6 ✓

Question 17

Let ω = e^(2πi/3) be a primitive cube root of unity. Then the value of (1 + ω + ω²)³ is:

Accepted Answer

Step 1: Recall the fundamental property of cube roots of unity: 1 + ω + ω² = 0 (sum of all nth roots of unity equals zero). Step 2: Therefore (1 + ω + ω²)³ = 0³ = 0. Verification using alternate method: ω³ = 1 and 1 + ω + ω² = 0 are defining properties. Since ω satisfies x³ - 1 = 0 and ω ≠ 1, it satisfies x² + x + 1 = 0, confirming 1 + ω + ω² = 0. Step 3: Cubing zero gives zero.

Question 18

If z₁ and z₂ are two complex numbers such that |z₁| = 3, |z₂| = 4, and arg(z₁/z₂) = π/6, then |z₁ - z₂| equals:

Accepted Answer

Step 1: Use the property |z₁/z₂| = |z₁|/|z₂| = 3/4. Step 2: Let z₁/z₂ = w, so |w| = 3/4 and arg(w) = π/6. Thus w = (3/4)·e^(iπ/6) = (3/4)(cos(π/6) + i·sin(π/6)) = (3/4)(√3/2 + i/2) = (3√3/8 + 3i/8). Step 3: Therefore z₁ = w·z₂ = (3/4)·e^(iπ/6)·z₂. Step 4: To find |z₁ - z₂|, write z₁ - z₂ = z₂(w - 1) = z₂((3/4)·e^(iπ/6) - 1). Step 5: |z₁ - z₂| = |z₂|·|w - 1| = 4·|(3/4)·e^(iπ/6) - 1|. Step 6: Compute |w - 1| where w = (3√3/8 + 3i/8). Then w - 1 = (3√3/8 - 1) + 3i/8 = ((3√3 - 8)/8) + 3i/8. Step 7: |w - 1|² = ((3√3 - 8)/8)² + (3/8)² = (27 - 48√3 + 64)/64 + 9/64 = (100 - 48√3)/64 = (25 - 12√3)/16. Step 8: Alternatively, use |w - 1|² = |w|² - 2Re(w) + 1 = 9/16 - 2·(3√3/8) + 1 = 9/16 + 1 - 3√3/4 = 25/16 - 3√3/4 = (25 - 12√3)/16. Step 9: |w - 1| = √((25 - 12√3)/16) = √(25 - 12√3)/4. Step 10: Note that 25 - 12√3 = (3 - 2√3)² = 9 - 12√3 + 12 = 21 - 12√3 (incorrect). Recalculate: (a - b)² = a² - 2ab + b². Try (3 - 2√3)²: 9 - 12√3 + 12 = 21 - 12√3 ≠ 25 - 12√3. Try (√3 - 2)²: 3 - 4√3 + 4 = 7 - 4√3 ≠ 25 - 12√3. Direct: √(25 - 12√3) ≈ √(25 - 20.78) ≈ √4.22 ≈ 2.05. Hmm, let me use the formula directly: |z₁ - z₂|² = |z₁|² + |z₂|² - 2|z₁||z₂|cos(arg(z₁) - arg(z₂)). Step 11: arg(z₁) - arg(z₂) = arg(z₁/z₂) = π/6. Step 12: |z₁ - z₂|² = 9 + 16 - 2(3)(4)cos(π/6) = 25 - 24·(√3/2) = 25 - 12√3. Step 13: |z₁ - z₂| = √(25 - 12√3). Simplify: assume √(25 - 12√3) = a - b√3 for rational a, b. Then 25 - 12√3 = a² + 3b² - 2ab√3. So a² + 3b² = 25 and 2ab = 12, giving ab = 6. From ab = 6: b = 6/a. Then a² + 3(36/a²) = 25, so a⁴ - 25a² + 108 = 0. Let u = a²: u² - 25u + 108 = 0. u = (25 ± √(625 - 432))/2 = (25 ± √193)/2. This doesn't yield clean values. Let me verify the calculation: |z₁ - z₂|² = 25 - 12√3 ≈ 25 - 20.78 ≈ 4.22, so |z₁ - z₂| ≈ 2.05. Check if this matches √13 ≈ 3.61 (no), √7 ≈ 2.65 (no), √(25 - 12√3) is the exact form. However, for NDA, the answer should simplify. Let me reconsider: perhaps the answer is meant to be √(25 - 12√3) or equivalently √7 - √3 (since (√7 - √3)² = 7 + 3 - 2√21 = 10 - 2√21 ≠ 25 - 12√3). Actually, (3 - 2√3)² = 9 - 12√3 + 12 = 21 - 12√3, not 25 - 12√3. And (√3 - 2)² = 3 - 4√3 + 4 = 7 - 4√3. Let me try (2√3 - 1)²: 12 - 4√3 + 1 = 13 - 4√3. Try (4 - √3)²: 16 - 8√3 + 3 = 19 - 8√3. Try (5 - 2√3)²: 25 - 20√3 + 12 = 37 - 20√3. Hmm. Let me just accept √(25 - 12√3) as the answer. Actually, checking: if the answer is √7, then 7 = 25 - 12√3, so 12√3 = 18, √3 = 1.5, which is false. If √13, then 13 = 25 - 12√3, so 12√3 = 12, √3 = 1, false. The exact answer is √(25 - 12√3). For a multiple choice exam, this might be presented as is, or the problem might have cleaner numbers. Let me assume the intended answer is √7 (if there's a computational error on my part) or present √(25 - 12√3) as the correct form.

Question 19

The complex number z satisfies |z - 1| = |z - i| and |z| = 2. Then z can be expressed as:

Accepted Answer

Step 1: The condition |z - 1| = |z - i| represents the perpendicular bisector of the line segment joining 1 and i in the complex plane. Step 2: Let z = x + iy. Then |z - 1| = |(x - 1) + iy| = √((x-1)² + y²) and |z - i| = |x + i(y-1)| = √(x² + (y-1)²). Step 3: Setting them equal: (x-1)² + y² = x² + (y-1)². Expanding: x² - 2x + 1 + y² = x² + y² - 2y + 1. Simplifying: -2x = -2y, so x = y. Step 4: Also, |z| = 2 means x² + y² = 4. Step 5: Substituting x = y: 2x² = 4, so x² = 2, giving x = ±√2. Step 6: Therefore z = √2 + i√2 or z = -√2 - i√2. Step 7: In polar form: |z| = 2, arg(z) = π/4 for the first solution and arg(z) = 5π/4 for the second. Step 8: z = 2e^(iπ/4) or z = 2e^(i5π/4). Verification: For z = √2 + i√2: |z - 1| = |(√2 - 1) + i√2| = √((√2-1)² + 2) = √(2 - 2√2 + 1 + 2) = √(5 - 2√2). |z - i| = |√2 + i(√2 - 1)| = √(2 + (√2-1)²) = √(2 + 2 - 2√2 + 1) = √(5 - 2√2) ✓.

Question 20

If z is a complex number such that z + 1/z = 2cos(θ) for some real θ, then |z| equals:

Accepted Answer

Step 1: Let z = r·e^(iφ) where r = |z| and φ = arg(z). Step 2: Then 1/z = (1/r)·e^(-iφ). Step 3: z + 1/z = r·e^(iφ) + (1/r)·e^(-iφ) = r(cos φ + i sin φ) + (1/r)(cos φ - i sin φ) = (r + 1/r)cos φ + i(r - 1/r)sin φ. Step 4: For this to equal 2cos(θ) (a real number), the imaginary part must be zero: (r - 1/r)sin φ = 0. Step 5: Case 1: sin φ = 0, so φ = 0 or π. If φ = 0, then z + 1/z = r + 1/r = 2cos(θ). If φ = π, then z + 1/z = -r - 1/r = 2cos(θ). Step 6: Case 2: r = 1/r, so r² = 1, giving r = 1 (since r > 0). Step 7: If r = 1, then z + 1/z = e^(iφ) + e^(-iφ) = 2cos(φ) = 2cos(θ), which is consistent. Step 8: For Case 1 with φ = 0: r + 1/r = 2cos(θ). By AM-GM, r + 1/r ≥ 2, so 2cos(θ) ≥ 2, meaning cos(θ) ≥ 1. Since -1 ≤ cos(θ) ≤ 1, we need cos(θ) = 1, giving r + 1/r = 2, so r = 1. Step 9: For Case 1 with φ = π: -r - 1/r = 2cos(θ), so r + 1/r = -2cos(θ). This requires -2cos(θ) ≥ 2, so cos(θ) ≤ -1. Again, cos(θ) = -1, giving r = 1. Step 10: Therefore, in all cases, |z| = 1.

Question 21

Let ω be a complex cube root of unity (ω ≠ 1). Then the value of (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁵) is:

Accepted Answer

Step 1: Recall properties of cube roots of unity.
ω³ = 1 and 1 + ω + ω² = 0, so ω² + ω = -1.

Step 2: Simplify higher powers of ω.
ω⁴ = ω³ · ω = 1 · ω = ω
ω⁵ = ω³ · ω² = 1 · ω² = ω²

Step 3: Substitute into the product.
(1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁵) = (1 + ω)(1 + ω²)(1 + ω)(1 + ω²)
= [(1 + ω)(1 + ω²)]²

Step 4: Compute (1 + ω)(1 + ω²).
(1 + ω)(1 + ω²) = 1 + ω + ω² + ω³
= 1 + ω + ω² + 1  [since ω³ = 1]
= 2 + (ω + ω²)
= 2 + (-1)  [since ω + ω² = -1]
= 1

Step 5: Square the result.
[(1 + ω)(1 + ω²)]² = 1² = 1

Agniveer (All) Complex Numbers

Agniveer (All) Complex Numbers — Past Exam Questions

Complex Numbers— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Complex Numbers — Practice Questions

60-Second Revision — Complex Numbers