ZE
ZESTEXAM

CDS Complex Numbers

Study Material — 1 PYQs (2019–2019) · Concept Notes · Shortcuts

CDS Complex Numbers is a frequently tested subtopic — 1 previous year questions from 2019–2019 papers are included below with concept notes, key rules and shortcut tricks.

1 PYQs
2019–2019
38 Practice
MCQs
10 Key Points
to remember
Free
no login needed
Take Free Mock →Full Practice Set
Also for:NDAAgniveerCAPFAFCAT
PYQs
1
Practice
38
Key Points
10
Access
Free
Previous Year Questions

CDS Complex Numbers — Past Exam Questions

1 questions from actual CDS papers · all shown free · click option to reveal solution

Exam Q 12019Previous Year Pattern

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

Concept Notes

Complex Numbers— Rules & Concept

1.Core Concept

Read this first — the foundation of the topic

Complex numbers are numbers that contain both real and imaginary parts. They were created to solve equations like x² + 1 = 0, which has no real solution. A complex number is written as z = a + bi, where 'a' is the real part, 'b' is the imaginary part, and 'i' is the imaginary unit where i² = -1.

2.Key Rules

Core rules you must know cold

The real part is denoted as Re(z) = a, and the imaginary part as Im(z) = b. Two complex numbers are equal only if their real parts are equal AND their imaginary parts are equal. The conjugate of z = a + bi is z* = a - bi. The modulus (absolute value) of z is |z| = √(a² + b²).

3.Formula Block

Memorise — at least one formula appears in every paper
• Addition: (a + bi) + (c + di) = (a + c) + (b + d)i
• Subtraction: (a + bi) - (c + di) = (a - c) + (b - d)i
• Multiplication: (a + bi)(c + di) = (ac - bd) + (ad + bc)i
• Division: (a + bi)/(c + di) = [(a + bi)(c - di)]/[c² + d²]
• Powers of i: i¹ = i, i² = -1, i³ = -i, i⁴ = 1, then cycle repeats

4.Exam Patterns

What examiners ask — read before attempting PYQs
Shortcut for Powers of i

Divide the power by 4. The remainder determines the answer: remainder 0 → 1, remainder 1 → i, remainder 2 → -1, remainder 3 → -i

Example

i²³ = i³ = -i (since 23 ÷ 4 gives remainder 3).

5.Worked Example

Solve this step-by-step before moving on
1
Step 1

Add real parts: 3 + 2 = 5

2
Step 2

Add imaginary parts: 4i + (-7i) = -3i

3
Step 3

Result = 5 - 3i

4
Step 4

Modulus = √(5² + (-3)²) = √(25 + 9) = √34 Worked Example 2: Multiply (1 + 2i)(3 - i).

1
Step 1

Use FOIL method: (1)(3) + (1)(-i) + (2i)(3) + (2i)(-i)

2
Step 2

= 3 - i + 6i - 2i²

3
Step 3

Since i² = -1, we get: 3 - i + 6i - 2(-1)

4
Step 4

= 3 - i + 6i + 2 = 5 + 5i Division Shortcut: To divide complex numbers, multiply both numerator and denominator by the conjugate of the denominator. This eliminates the imaginary part from the denominator. Most

6.Exam Traps

Common mistakes students make — avoid these

Students forget that i² = -1 when expanding products. They often write i² = i or leave it as i². Always substitute i² = -1 immediately after multiplication.

Another frequent error is incorrectly finding the conjugate - remember to change only the sign of the imaginary part, not the real part.

Key Points to Remember

  • Complex number format: z = a + bi where a is real part, b is imaginary part
  • Imaginary unit: i² = -1, i³ = -i, i⁴ = 1, then pattern repeats every 4 powers
  • Modulus formula: |z| = √(a² + b²) for z = a + bi
  • Conjugate of a + bi is a - bi (change sign of imaginary part only)
  • For division, multiply by conjugate of denominator to rationalize
  • Two complex numbers equal only if real parts equal AND imaginary parts equal
  • Powers of i shortcut: divide power by 4, use remainder (0→1, 1→i, 2→-1, 3→-i)
  • Multiplication: (a + bi)(c + di) = (ac - bd) + (ad + bc)i
  • Addition/Subtraction: combine like terms (real with real, imaginary with imaginary)
  • Quadratic with negative discriminant gives complex roots: x = [-b ± √(b²-4ac)]/2a

Exam-Specific Tips

  • i² = -1 is the fundamental definition of imaginary unit
  • i⁴ = 1, making powers of i cycle every 4 terms
  • For z = a + bi, modulus |z| = √(a² + b²)
  • Conjugate of z = a + bi is z* = a - bi
  • Re(z₁ + z₂) = Re(z₁) + Re(z₂) and Im(z₁ + z₂) = Im(z₁) + Im(z₂)
  • |z|² = z × z* = a² + b² for any complex number z
  • If z₁ = z₂, then a₁ = a₂ and b₁ = b₂
  • For pure imaginary number bi, modulus = |b|
Practice MCQs

Complex Numbers — Practice Questions

38graded MCQs · easy to hard · full solution & trap analysis · showing 20 of 38

All MCQs →
Practice 1easy

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

Practice 2easy

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

Practice 3easy

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

Practice 4easy

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

Practice 5easy

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

Practice 6easy

If z = 1 + i, then z⁴ equals:

Practice 7easy

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

Practice 8easy

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

Practice 9easy

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

Practice 10easy

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

Practice 11medium

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

Practice 12medium

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

Practice 13medium

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

Practice 14medium

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

Practice 15medium

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

Practice 16medium

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

Practice 17medium

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

Practice 18medium

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

Practice 19medium

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

Practice 20medium

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

18 more practice questions in the Study Panel

Difficulty-graded, bookmarkable, with timed mode. Free account — no credit card.

Create Free Account →Browse Questions

60-Second Revision — Complex Numbers

  • Remember: i² = -1 always, substitute immediately after multiplication
  • Formula: |a + bi| = √(a² + b²) for modulus calculations
  • Shortcut: i^n pattern repeats every 4 - divide power by 4, use remainder
  • Division trick: multiply by conjugate of denominator
  • Trap: Don't forget to change i² to -1 when expanding products
  • Conjugate: Only change sign of imaginary part, keep real part same
  • Equality: Complex numbers equal only if both parts separately equal
←Sets, Relations & FunctionsQuadratic Equations & Polynomials→
Studied the notes? Now test yourself
See how Complex Numbers appears in the real CDS paper
Full timed mock · Instant All-India percentile · Free
Start Free Mock →Practice Panel
✓ Free forever for basic prep✓ No app download✓ Real exam-pattern questions
Test Complex Numbers under exam conditions
Free CDS mock · instant rank · no login
Free Mock →