4.3 Algebra of Complex Numbers – Class 11 Mathematics

Introduction

Mathematics evolves when existing systems become insufficient. Real numbers are powerful, but they fail to provide solutions for equations like x² + 1 = 0. To overcome this limitation, mathematicians introduced complex numbers, extending the number system.

Complex numbers may appear abstract at first, but once you understand their algebra, they behave almost exactly like real numbers. In fact, every algebraic property (closure, associativity, distributivity) continues to hold in the world of complex numbers.

This topic – Algebra of Complex Numbers – is not just NCERT syllabus content, but also a critical building block for competitive exams like JEE Main, JEE Advanced, KCET, and COMEDK.

Exam Weightage

• JEE Main: 1–2 questions (~8 marks) frequently test simplification, conjugates, or division.
• KCET/COMEDK: Always 1 direct question (~5 marks), usually on division or simplification.
• Boards (CBSE/State): Appears as a 4-mark short-answer question.

The Standard Form

Every complex number can be expressed as:

z = a + ib

where:

• a → Real part of z (Re(z))
• b → Imaginary part of z (Im(z))
• i → Imaginary unit, defined as √(−1), with i² = −1

Examples:

• 5 + 3i → Re(z) = 5, Im(z) = 3
• −2i → Re(z) = 0, Im(z) = −2
• 7 (a pure real number) → Re(z) = 7, Im(z) = 0

1. Addition and Subtraction of Complex Numbers

If z₁ = a + ib and z₂ = c + id, then:

• z₁ + z₂ = (a + c) + i(b + d)
• z₁ − z₂ = (a − c) + i(b − d)

This is similar to adding vectors component-wise.

Example 1:
(3 + 4i) + (5 − 2i) = (3 + 5) + i(4 − 2) = 8 + 2i

Example 2:
(6 − i) − (4 + 3i) = (6 − 4) + i(−1 − 3) = 2 − 4i

2. Multiplication of Complex Numbers

Multiply binomials and use i² = −1.

z₁z₂ = (a + ib)(c + id) = (ac − bd) + i(ad + bc)

Example 1:
(2 + 3i)(4 − i) = (8 − (−3)) + i(−2 + 12) = 11 + 10i

Example 2 (Squaring a complex number):
(3 + 4i)² = 9 + 24i + 16i² = 9 + 24i − 16 = −7 + 24i

Notice that multiplication mixes real and imaginary parts, and results must always be simplified.

3. Division of Complex Numbers

Division requires eliminating i from the denominator. This is done using the conjugate.

If z₁ = a + ib, z₂ = c + id, then:

z₁ / z₂ = [(a + ib)(c − id)] / (c² + d²)
= [(ac + bd) + i(bc − ad)] / (c² + d²)

Example 1:
(3 + 2i) / (1 − i)
= (3 + 2i)(1 + i) / (1² + (−1)²)
= (3 + 3i + 2i + 2i²) / 2
= (1 + 5i)/2 = 0.5 + 2.5i

Example 2:
(5 − 6i)/(2 + 3i)
= (5 − 6i)(2 − 3i)/(4 + 9)
= (10 − 15i − 12i + 18i²)/13
= (10 − 27i − 18)/13
= (−8 − 27i)/13

4. Conjugates and Division Shortcut

The conjugate of z = a + ib is z̅ = a − ib.

Key property:
z × z̅ = a² + b² = |z|² (always real).

Shortcut for division:
z₁/z₂ = (z₁z̅₂) / |z₂|²

5. Power of i

Core idea

The powers of the imaginary unit i repeat in a cycle of 4:

• i¹ = i
• i² = −1
• i³ = −i
• i⁴ = 1
  …and then it repeats: i⁵ = i, i⁶ = −1, i⁷ = −i, i⁸ = 1, etc.

Rule: To compute iⁿ for any integer n, divide n by 4 and use the remainder r ∈ {0,1,2,3}.

• If r = 0 → iⁿ = 1
• If r = 1 → iⁿ = i
• If r = 2 → iⁿ = −1
• If r = 3 → iⁿ = −i

Negative powers: i⁻ⁿ = 1 / iⁿ. Use the cycle first, then take reciprocal.
Examples:

• i⁻¹ = 1/i = −i (since 1/i = −i)
• i⁻² = 1/i² = 1/(−1) = −1
• i⁻³ = 1/i³ = 1/(−i) = i
• i⁻⁴ = 1

Quick examples

1. i²⁰²⁵ → 2025 ÷ 4 leaves remainder 1 ⇒ i²⁰²⁵ = i
2. i²⁰²⁶ → remainder 2 ⇒ i²⁰²⁶ = −1
3. i¹⁰⁰ → remainder 0 ⇒ i¹⁰⁰ = 1
4. i⁻²⁷ → first find i²⁷ = i³ (remainder 3) = −i, so i⁻²⁷ = 1/(−i) = i

Mixed expressions

• (1 + i)⁴ = (1 + i)² · (1 + i)² = (1 + 2i + i²)² = (2i)² = −4
• i⁵ + i⁶ + i⁷ + i⁸ = i + (−1) + (−i) + 1 = 0

Exam pointers (JEE/KCET/COMEDK)

• Reduce exponent mod 4 first.
• For sums like iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³, the result is always 0 (complete cycle).
• For negative exponents, convert using 1/i = −i.

Common pitfalls

• Forgetting the 4-cycle.
• Treating i like a real base (e.g., using real-number exponent rules blindly).
• Mishandling 1/i (remember: 1/i = −i).

6. The square roots of a negative real number

Core idea

For any positive real number a > 0:

• √(−a) = i √a (principal square root)
• The two square roots of the negative real number −a are ± i √a

Examples

• √(−9) = 3i; the two roots are ± 3i
• √(−48) = √(16·3) with a negative sign → 4√3 · i, so roots are ± 4√3 i
• Solve x² + 25 = 0 → x = ± 5i

Principal vs. both roots

• The radical sign √ denotes the principal (chosen) value: √(−a) = i√a (with the positive √a).
• When solving equations like x² = −a, write both roots: x = ± i√a.

Connection to quadratics

• For ax² + bx + c = 0 with discriminant D = b² − 4ac < 0, roots are:
  x = [−b ± √(−D)] / (2a) = [−b ± i √(D′)] / (2a), where D′ = −D > 0.
  Example: x² + 2x + 5 = 0 → D = 4 − 20 = −16 → x = (−2 ± 4i)/2 = −1 ± 2i.

Exam pointers

• Pull out the minus sign before taking square root: √(−a) = i √a, not −√a.
• In simplification problems, express final answers as a + ib.
• In MCQs, check sign conventions carefully (principal value vs. both roots).

Common pitfalls

• Writing √(−a) = −√a (incorrect).
• Missing the ± when solving x² = −a.
• Dropping i in simplification (e.g., √(−49) = 7, wrong; correct is 7i).

7. Identities

These are the core algebraic identities you’ll repeatedly use. Assume z = a + ib, w = c + id.

Conjugate and modulus identities

• Conjugate: z̅ = a − ib
• Modulus: |z| = √(a² + b²)
• Product with conjugate: z · z̅ = |z|² = a² + b² (real, ≥ 0)
• Conjugate of sum: (z + w)̅ = z̅ + w̅
• Conjugate of product: (z w)̅ = z̅ w̅
• Conjugate of quotient: (z / w)̅ = z̅ / w̅ (provided w ≠ 0)
• Modulus of product: |z w| = |z| · |w|
• Modulus of quotient: |z / w| = |z| / |w| (w ≠ 0)

Proof sketch for z · z̅ = |z|²:
(a + ib)(a − ib) = a² − iab + iab − i² b² = a² + b².

Real and imaginary parts via conjugate

• Re(z) = (z + z̅)/2
• Im(z) = (z − z̅) / (2i)

Example: For z = 3 − 4i:

• z̅ = 3 + 4i
• Re(z) = (z + z̅)/2 = (3 − 4i + 3 + 4i)/2 = 3
• Im(z) = (z − z̅)/(2i) = (−8i)/(2i) = −4

Algebraic identities (sum/product/difference)

• (a + ib)(a − ib) = a² + b²
• (z + w)(z̅ + w̅) = (z + w)(overline{z + w}) = |z + w|² (always real)
• |z + w|² = |z|² + |w|² + 2 Re(z w̅)
  (very useful in geometry/inequalities)

Triangle inequality (preview; formal in modulus section)

• |z + w| ≤ |z| + |w|
• |z − w| ≥ ||z| − |w||

These are typically emphasized in 4.4 (Modulus), but it’s good to know them here since they appear alongside identities.

Division identity (conjugate method)

• z / w = (z w̅) / |w|², w ≠ 0
  Use this to avoid mistakes in division; it ensures the denominator becomes real.

Worked examples

1. Show (z + w)̅ = z̅ + w̅.
   LHS = (a + ib + c + id)̅ = (a + c) − i(b + d) = (a − ib) + (c − id) = z̅ + w̅.

2. Compute |(3 + 4i)(1 − 2i)|.
   |3 + 4i| = 5, |1 − 2i| = √5 → product modulus = 5√5.

3. If z = 2 − 5i, find Re(z²) and Im(z²).
   z² = (2 − 5i)² = 4 − 20i + 25i² = −21 − 20i.
   Re(z²) = −21, Im(z²) = −20.

4. If z = 1 + i and w = 1 − i, compute Re(z/w).
   z/w = (1 + i)/(1 − i) = (1 + i)(1 + i)/(1 + 1) = (1 + 2i + i²)/2 = (2i)/2 = i.
   Re(z/w) = 0.

Exam pointers

• For “prove” type questions, start from the definition (a + ib form) and simplify.
• When you see |z + w|², expand as (z + w)(z̅ + w̅) and use z z̅ = |z|².
• Convert Re(zw̅) and Im(zw̅) using the formulas above if needed.

Common pitfalls

• Using (z̅)² = (z²)̅ is true, but students sometimes mix this with |z|²; note that |z|² = z z̅, not z² or (z̅)².
• Forgetting to take conjugate of both numerator and denominator in a quotient when needed.
• Writing |z| = a + ib (modulus is always a non-negative real number).

5. Properties of Operations

Closure

For any z₁, z₂ ∈ C,

• z₁ + z₂, z₁ − z₂, z₁z₂, and z₁/z₂ (z₂ ≠ 0) are also complex numbers.

Commutativity

• z₁ + z₂ = z₂ + z₁
• z₁z₂ = z₂z₁

Associativity

• (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
• (z₁z₂)z₃ = z₁(z₂z₃)

Distributivity

• z₁(z₂ + z₃) = z₁z₂ + z₁z₃

Identity Elements

• Additive identity: 0 + 0i
• Multiplicative identity: 1 + 0i

Inverse Elements

• Additive inverse: −z
• Multiplicative inverse: 1/z (z ≠ 0)

Mini Practice Set (with answers)

1. Compute i³⁰⁰³.
2. Simplify i⁻²⁹ + i¹¹.
3. Find the two square roots of −64.
4. Solve x² + 36 = 0.
5. If z = 4 − 3i, compute z̅, |z|, and z · z̅.
6. Prove (z w)̅ = z̅ w̅ using z = a + ib, w = c + id.
7. Show Re(z) = (z + z̅)/2 and Im(z) = (z − z̅)/(2i) for z = 7 − 2i.

Answers

1. 3003 ÷ 4 → remainder 3 ⇒ i³⁰⁰³ = −i
   
2. i⁻²⁹ = 1/i²⁹ = 1/i¹ (since 29 ≡ 1 mod 4) = −i; and i¹¹ = i³ (since 11 ≡ 3 mod 4) = −i ⇒ sum = −i + (−i) = −2i
   
3. ± 8i
   
4. x = ± 6i
   
5. z̅ = 4 + 3i; |z| = √(16 + 9) = 5; z · z̅ = 25
   
6. (a + ib)(c + id) all overbar = (ac − bd) − i(ad + bc) = (a − ib)(c − id) = z̅ w̅
   
7. z = 7 − 2i, z̅ = 7 + 2i → (z + z̅)/2 = 7 = Re(z); (z − z̅)/(2i) = (−4i)/(2i) = −2 = Im(z)
   

Where these show up in exams

• Power of i: Fast simplifications (JEE/KCET single-step MCQs).
• Square roots of negative reals: Quadratic equations with D < 0; standard board/KCET questions.
• Identities: Prove/verify style questions, transform expressions to a + ib, and modulus-conjugate manipulations (JEE Main frequently).

Step-by-Step NCERT Examples

Example 1: Simplify (2 + 5i) − (3 − 4i)
= (2 − 3) + i(5 + 4) = −1 + 9i

Example 2: Multiply (1 + 2i)(1 − 2i)
= 1 − (2i)² = 1 − (−4) = 5

Example 3: Divide (4 + 3i)/(2 − i)
= (4 + 3i)(2 + i)/(2² + (−1)²)
= (8 + 4i + 6i + 3i²)/5
= (5 + 10i)/5 = 1 + 2i

Common Mistakes Students Make

1. Forgetting i² = −1
   Students often keep i² in final answers – this is wrong. Always replace i² with −1.
2. Leaving denominator in complex form
   Division answers must be simplified with conjugates.
3. Mixing up real and imaginary parts
   Always combine like terms carefully.
4. Sign errors in multiplication
   Remember: (a + ib)(a − ib) = a² + b² (positive!).

JEE, KCET, and COMEDK Strategy

• JEE Main: Expect simplification, division, and expansion-based questions. Speed comes from recognizing patterns.
• KCET: One guaranteed 5-mark problem, often direct computation.
• COMEDK: Tests conceptual understanding, especially conjugates.

Tip: In competitive exams, always write answers in a + ib form.

Quick Formula Table

JEE & KCET Style Solved Problems

Problem 1 (JEE Main, Simplification)

Simplify: (2 + i)(2 − i)(3 + 2i).

Solution:
(2 + i)(2 − i) = 4 − i² = 5
So = 5(3 + 2i) = 15 + 10i

Problem 2 (KCET)

Divide: (1 + 3i)/(2 − i).

Solution:
= (1 + 3i)(2 + i)/(2² + (−1)²)
= (2 + i + 6i + 3i²)/5
= (−1 + 7i)/5

Problem 3 (JEE Main, Conceptual)

If z₁ = 3 + 2i and z₂ = 3 − 2i, find z₁z₂.

Solution:
(3 + 2i)(3 − 2i) = 9 − (2i)² = 13

Problem 4 (COMEDK)

Prove: (a + ib)(a − ib) = a² + b².

Solution:
= a² − iab + iab − (i²)b² = a² + b²

Problem 5 (JEE Main Trick)

Simplify: (1 + i)⁴.

Solution:
(1 + i)² = 1 + 2i + i² = 2i
So (1 + i)⁴ = (2i)² = −4

Problem 6 (KCET)

If z = (3 + 4i)/(1 − 2i), find Re(z) and Im(z).

Solution:
Multiply numerator & denominator by (1 + 2i):
= (3 + 4i)(1 + 2i)/(1 + 4)
= (3 + 6i + 4i + 8i²)/5
= (−5 + 10i)/5 = −1 + 2i

Problem 7 (Advanced JEE)

Find the multiplicative inverse of 2 + 3i.

Solution:
1/(2 + 3i) = (2 − 3i)/(2² + 3²) = (2 − 3i)/13

Practice Exercises (Try Yourself)

1. Simplify (5 + 2i) − (3 − 6i).
2. Compute (2 + 3i)(2 − 3i).
3. Simplify (7 − 4i)/(3 + i).
4. If z = 4 − 3i, find z̅ and z × z̅.
5. Simplify (1 + i)⁶.
6. Show that (x + iy)(x − iy) = x² + y².
7. If (2 + i)/(1 − i) = a + ib, find a and b.
8. Compute ((1 + i)(1 − i))/(1 + 2i).

FAQs

Q1. What is the simplest way to divide two complex numbers?
Multiply numerator and denominator by the conjugate of the denominator.

Q2. Why does z × z̅ = |z|² always give a real number?
Because (a + ib)(a − ib) = a² + b², which is purely real.

Q3. How important is this topic for JEE?
Very important. Many JEE algebra simplification and quadratic equation questions are based on this.

Q4. Is algebra of complex numbers needed for calculus?
Yes, especially in integration problems involving trigonometric substitutions.

Q5. Can KCET ask theory-based questions here?
Mostly no – they stick to direct computation, but sometimes definitions of conjugate/modulus appear.

Summary & Key Takeaways

• Complex numbers follow all algebraic rules.
• Addition/Subtraction = combine real and imaginary parts.
• Multiplication = expand and replace i² with −1.
• Division = multiply by conjugate.
• Conjugates are powerful shortcuts in exam problems.
• Expect 8 marks in JEE Main and 5 marks in KCET/COMEDK.