Complex Numbers and Quadratic Equations – Class 11 Maths Chapter 4 Notes & NCERT PDF

Class 11MathematicsChapter 4NCERT book: Mathematics - Class 11

Notes and a simple summary of Chapter 4 of the NCERT Class 11 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 4: Complex Numbers and Quadratic Equations

Introduction

The equation x² + 1 = 0 has no real solution, because the square of a real number is never negative. To solve such equations we extend the real numbers by introducing the symbol i, with i² = –1. This chapter defines complex numbers, their arithmetic, modulus and conjugate, and how to picture them on the Argand plane. The Historical Note records that the Indian mathematician Mahavira (850) stated clearly that a negative quantity has no square root, and Bhaskara, in his Bijaganita (1150), wrote the same.

Key Concepts

4.2 Complex Numbers

A number of the form z = a + ib, where a, b are real and i = √–1, is a complex number. a = Re z is the real part and b = Im z the imaginary part.

Two complex numbers a + ib and c + id are equal if and only if a = c and b = d.

Every real number a is the complex number a + i0.

4.3 Algebra of Complex Numbers

For z₁ = a + ib and z₂ = c + id:

4.3.1 Addition: z₁ + z₂ = (a + c) + i(b + d). It is closed, commutative and associative; 0 + i0 is the additive identity and –z is the additive inverse.

4.3.2 Difference: z₁ – z₂ = (a – c) + i(b – d).

4.3.3 Multiplication: z₁ z₂ = (ac – bd) + i(ad + bc). It is closed, commutative, associative, distributive over addition; 1 + i0 is the identity.

Multiplicative inverse of z = a + ib ≠ 0: 1/z = a/(a² + b²) + i(–b)/(a² + b²).

4.3.4 Division: z₁/z₂ = z₁ × (1/z₂), for z₂ ≠ 0.

4.3.5 Powers of i: i² = –1, i³ = –i, i⁴ = 1; in general i^(4k) = 1, i^(4k+1) = i, i^(4k+2) = –1, i^(4k+3) = –i.

4.3.6 Square roots of negative numbers: the square roots of –a (a > 0) are ±i√a. Note: √a × √b = √(ab) only when a, b are not both negative.

4.3.7 Identities: (z₁ + z₂)² = z₁² + 2z₁z₂ + z₂², (z₁ – z₂)³ = z₁³ – 3z₁²z₂ + 3z₁z₂² – z₂³, z₁² – z₂² = (z₁ + z₂)(z₁ – z₂), etc. — the same as for real numbers.

4.4 Modulus and Conjugate

For z = a + ib: |z| = √(a² + b²) and the conjugate is z̄ = a – ib.

z z̄ = |z|², so 1/z = z̄/|z|².

|z₁z₂| = |z₁||z₂|; |z₁/z₂| = |z₁|/|z₂| (z₂ ≠ 0)

conj(z₁ + z₂) = z̄₁ + z̄₂; conj(z₁z₂) = z̄₁ z̄₂

4.5 Argand Plane and Polar Representation

The complex number x + iy is shown as the point P(x, y) in the plane; the x-axis is the real axis and the y-axis the imaginary axis. This is the Argand plane.

|z| = OP is the distance of P from the origin; z̄ is the reflection of z in the real axis.

Polar form: writing x = r cos θ, y = r sin θ gives z = r(cos θ + i sin θ), where r = |z| is the modulus and θ is the argument. The value of θ with –π < θ ≤ π is the principal argument.

(In the rationalised textbook, this chapter covers complex numbers, their algebra, modulus, conjugate and polar form.)

Worked Examples

Example 1 (powers of i). i⁻³⁹ = i⁻⁴⁰ × i = i. And i⁹ + i¹⁹ = i + (–i) = 0.

Example 2 (inverse). 1/(2 – 3i) = (2 + 3i)/(2² + 3²) = 2/13 + (3/13)i.

Example 3 (product of conjugates). (3 + i√5)(3 – i√5) = 9 – 5i² = 9 + 5 = 14, a real number.

Example 4 (cube). (5 – 3i)³ = 125 – 3(25)(3i) + 3(5)(3i)² – (3i)³ = 125 – 225i – 135 + 27i = –10 – 198i.

Example 5 (polar form). z = 1 + i√3: r = √(1 + 3) = 2, cos θ = 1/2, sin θ = √3/2, so θ = π/3 and z = 2(cos π/3 + i sin π/3).

Example 6 (second quadrant). z = –1 + i: r = √2, and the point (–1, 1) lies in quadrant II, so θ = 3π/4 and z = √2(cos 3π/4 + i sin 3π/4).

Example 7 (division). Express (5 + √2 i)/(1 – √2 i) in the form a + ib. Multiply numerator and denominator by the conjugate 1 + √2 i:

denominator: (1 – √2 i)(1 + √2 i) = 1 + 2 = 3

numerator: (5 + √2 i)(1 + √2 i) = 5 + 5√2 i + √2 i + 2i² = 3 + 6√2 i

result: (3 + 6√2 i)/3 = 1 + 2√2 i

Why √a × √b fails for two negatives. √(–3) × √(–3) = (i√3)(i√3) = 3i² = –3, whereas √((–3)(–3)) = √9 = 3. So the rule √a √b = √(ab) must not be applied when both a and b are negative — always rewrite √(–a) as i√a first.

Reading the Argand plane. Points on the real axis are real numbers (b = 0); points on the imaginary axis, such as 0 + 2i, are purely imaginary. Moving from z to z̄ flips the point across the real axis, and |z| stays the same because the distance from the origin is unchanged.

Formula Sheet

i² = –1; i^(4k) = 1, i^(4k+1) = i, i^(4k+2) = –1, i^(4k+3) = –i

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

1/(a + ib) = (a – ib)/(a² + b²)

|z| = √(a² + b²); z z̄ = |z|²

z = r(cos θ + i sin θ), r = |z|, –π < θ ≤ π

Exam Focus

Express in the form a + ib: multiply numerator and denominator by the conjugate of the denominator.

Polar form questions: find r first, then find θ from cos θ and sin θ, choosing the quadrant from the signs of x and y.

Proofs with conjugates: use z z̄ = |z|² and the rules for conjugates of sums and products.

Key Terms

Complex number (सम्मिश्र संख्या), Imaginary unit i (काल्पनिक इकाई)

Real part (वास्तविक भाग), Imaginary part (काल्पनिक भाग)

Conjugate (संयुग्मी), Modulus (मापांक)

Multiplicative inverse (गुणात्मक प्रतिलोम)

Argand plane (आर्गंड तल), Real axis (वास्तविक अक्ष), Imaginary axis (काल्पनिक अक्ष)

Polar form (ध्रुवीय रूप), Argument (कोणांक), Principal argument (मुख्य कोणांक)

Common Mistakes

Using √(–a) × √(–b) = √(ab); in fact i√a × i√b = –√(ab).

Forgetting that i² = –1 while expanding products.

Choosing the wrong argument by ignoring the quadrant of the point.

Calling the imaginary part ib instead of the real number b.

Leaving i in the denominator instead of rationalising with the conjugate.

💡 Key Learning Points

✓Understand why i = √–1 is needed and write complex numbers as a + ib
✓Add, subtract, multiply and divide complex numbers and find multiplicative inverses
✓Evaluate powers of i using the cycle i, –1, –i, 1
✓Find the modulus and conjugate of a complex number and use z z̄ = |z|²
✓Represent complex numbers on the Argand plane and convert to polar form

👨‍🏫 Teaching Tips

→Open with x² + 1 = 0 to show the need for new numbers, then tell the Mahavira and Bhaskara story
→Draw the cycle of powers of i as a four-step circle to make i^n quick to evaluate
→Treat multiplication like expanding brackets, stressing i² = –1 at the last step
→Show division as 'rationalising the denominator' with the conjugate
→Plot several numbers and their conjugates on the Argand plane
→Have students find the polar form of numbers in all four quadrants

📋 Assessment Questions

Can the student express products and quotients of complex numbers in the form a + ib?
Can the student evaluate i^n for positive and negative n?
Can the student find the modulus, conjugate and inverse of a complex number?
Can the student write a complex number in polar form with the correct argument?

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