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Mathematics Grade XI

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Chapter 4: Complex Numbers and Quadratic Equations

Class 11 • Mathematics • Chapter 4

Complex Numbers and Quadratic Equations

Extending the number line so that every quadratic equation has a solution.

CBSE / NCERT Aligned   School Revise

Chapter Roadmap

The Imaginary Unit i • Complex Numbers • Algebra of Complex Numbers • Powers of i • Modulus & Conjugate • Argand Plane • Polar Form • Quadratic Equations with Complex Roots
1

Why We Needed a New Kind of Number

For centuries, mathematicians were stuck on one stubborn question: what number, when multiplied by itself, gives a negative result? No real number works, because a positive times a positive is positive and a negative times a negative is also positive. Equations as simple as x² + 1 = 0 had no solution at all. Rather than give up, mathematicians invented a brand-new number to fill the gap.

That number is the imaginary unit i, defined so that i² = −1. Combining it with ordinary real numbers produces complex numbers of the form a + bi, which extend the one-dimensional number line into a two-dimensional plane. Far from being a curiosity, complex numbers are the working language of electrical engineering, signal processing, quantum mechanics and computer graphics. This chapter teaches you to build, combine, measure and picture them, and to solve quadratic equations that real numbers alone cannot.

Foundation
The imaginary unit i satisfies i² = −1. A complex number is written z = a + bi, where a is the real part Re(z) and b is the imaginary part Im(z). The set of complex numbers is written C and contains every real number (taking b = 0).
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Key Terms You Must Know
Term Meaning Example
Imaginary unit i The number defined by i² = −1; it is not on the real number line. i = √(−1)
Complex number z A number of the form a + bi with a, b real. 3 + 4i
Real part Re(z) The a in a + bi. Re(3 + 4i) = 3
Imaginary part Im(z) The b in a + bi (the coefficient of i, without i). Im(3 + 4i) = 4
Conjugate z̅ The reflection a − bi, formed by flipping the sign of the imaginary part. Conjugate of 3 + 4i is 3 − 4i
Modulus |z| The distance of z from the origin, √(a² + b²). |3 + 4i| = 5
Argand plane The plane that pictures a + bi as the point (a, b). 3 + 4i plots at (3, 4)
3

Core Concepts, Step by Step

1. The Imaginary Unit and Powers of i

Everything starts from i² = −1. From this, higher powers cycle through just four values: i¹ = i, i² = −1, i³ = −i, and i⁴ = 1, after which the pattern repeats. To find any power, divide the exponent by 4 and use the remainder, so for instance i¹⁰ has remainder 2, so i¹⁰ = i² = −1. Square roots of negatives are written using i too: √(−9) = 3i, and in general √(−a) = i√a for positive a.

2. Complex Numbers and Their Equality

A complex number z = a + bi packs a real part and an imaginary part into one object. Two complex numbers are equal only when both parts match: a + bi = c + di means a = c and b = d. A complex number is purely real when b = 0 and purely imaginary when a = 0 (but b ≠ 0). This single rule for equality is what lets us solve equations by comparing real and imaginary parts separately.

3. Algebra of Complex Numbers

Complex numbers add and subtract part by part: (a + bi) + (c + di) = (a + c) + (b + d)i. They multiply using the distributive rule together with i² = −1: (a + bi)(c + di) = (ac − bd) + (ad + bc)i. To divide, multiply top and bottom by the conjugate of the denominator, which clears i from below: this turns the denominator into the real number c² + d².

4. Modulus and Conjugate

The conjugate of z = a + bi is z̅ = a − bi. Multiplying a complex number by its conjugate always gives a non-negative real number: z · z̅ = a² + b². The modulus |z| = √(a² + b²) measures how far z sits from the origin, so |z|² = z · z̅. Useful properties include |z₁z₂| = |z₁| |z₂|, the conjugate of a sum equals the sum of conjugates, and the conjugate of a product equals the product of conjugates.

5. The Argand Plane and Polar Form

We picture z = a + bi as the point (a, b) on a plane called the Argand plane, with the real part along the horizontal axis and the imaginary part along the vertical axis. The figure below shows how the sign of each part decides the quadrant. The distance from the origin is the modulus r = |z|, and the angle the line Oz makes with the positive real axis is the argument θ. This gives the polar form z = r(cos θ + i sin θ).

The Argand plane: plotting z = a + bi

Argand plane: z = a + bi with modulus r and argument theta

6. Quadratic Equations with Complex Roots

A quadratic ax² + bx + c = 0 (with real a, b, c and a ≠ 0) is solved by x = (−b ± √(b² − 4ac)) / 2a. When the discriminant b² − 4ac is negative, the square root is imaginary, and the two roots become complex conjugates of each other, a + bi and a − bi. Every quadratic now has exactly two roots in C, which is the payoff of inventing complex numbers, and it foreshadows the Fundamental Theorem of Algebra.
4

Key Results & Proofs

Complex numbers rest on a handful of properties that are properly proved from the single fact i² = −1. Here are the key ones, derived line by line.

Result 1: Powers of i Repeat with Period 4

Statement. For any integer n, iⁿ takes only four values: i⁰ = 1, i¹ = i, i² = −1, i³ = −i.

Proof

Start from i² = −1 and keep multiplying by i.

i² = −1 definition of i
i³ = i² · i = −i multiply by i
i⁴ = i² · i² = (−1)(−1) = 1 multiply
i⁵ = i⁴ · i = i the pattern starts again

Because i⁴ = 1, multiplying by i⁴ changes nothing, so iⁿ equals i raised to the remainder of n divided by 4.
Result 2: z times its Conjugate equals |z|²

Statement. For z = a + bi, z · z̅ = a² + b² = |z|².

Proof

Multiply z by its conjugate z̅ = a − bi.

z · z̅ = (a + bi)(a − bi) write out the conjugate
  = a² − (bi)² difference of two squares
  = a² − b²i² expand (bi)²
  = a² + b² i² = −1
|z|² = a² + b² definition of modulus

So z · z̅ is always a non-negative real number, exactly equal to |z|². This is the trick behind dividing complex numbers.
Result 3: The Modulus is Multiplicative

Statement. For any complex numbers z₁ and z₂, |z₁ z₂| = |z₁| · |z₂|.

Proof

Square the left side and use |z|² = z · z̅ together with (z₁z₂)̅ = z₁̅ z₂̅.

|z₁ z₂|² = (z₁ z₂)(z₁ z₂)̅ since |z|² = z · z̅ (Result 2)
  = (z₁ z₂)(z₁̅ z₂̅) conjugate of a product = product of conjugates
  = (z₁ z₁̅)(z₂ z₂̅) rearrange the factors
  = |z₁|² · |z₂|² each pair = its modulus squared
|z₁ z₂| = |z₁| · |z₂| take the non-negative square root

The square-root step is valid because a modulus is never negative.
Result 4: Complex Roots Come in Conjugate Pairs

Statement. If ax² + bx + c = 0 has real coefficients (a ≠ 0) and b² − 4ac < 0, then its two roots are complex conjugates of each other.

Proof

Apply the quadratic formula and pull the i out of the negative square root.

x = (−b ± √(b² − 4ac)) / 2a quadratic formula
Since b² − 4ac < 0, the quantity 4ac − b² is positive, and √(b² − 4ac) = i√(4ac − b²).
x = (−b ± i√(4ac − b²)) / 2a √(−1) = i
x₁ = −b/2a + i·√(4ac − b²)/2a the + root
x₂ = −b/2a − i·√(4ac − b²)/2a the − root

The two roots share the same real part (−b/2a) and have opposite imaginary parts, so x₂ is the conjugate of x₁. Complex roots of a real quadratic always occur in conjugate pairs.
5

Worked Examples
Example 1

Question: Simplify i³⁹.

▶ Show full working

Powers of i repeat every 4 (i, −1, −i, 1), so use the remainder when the power is divided by 4.

39 ÷ 4 = 9 remainder 3 find the remainder
i³⁹ = i³ a remainder of 3 behaves like i³
  = −i since i³ = −i

Answer: i³⁹ = −i.
Example 2

Question: Write √(−25) in terms of i.

▶ Show full working

Separate the −1, since √(−1) = i.

√(−25) = √25 × √(−1) split off the −1
  = 5 × i √25 = 5 and √(−1) = i
  = 5i  

Answer: √(−25) = 5i.
Example 3

Question: Add (3 + 2i) and (4 − 5i).

▶ Show full working

Add the real parts together and the imaginary parts together.

real part = 3 + 4 = 7 add the plain numbers
imaginary part = 2 + (−5) = −3 add the i-parts
sum = 7 − 3i combine

Answer: 7 − 3i.
Example 4

Question: Multiply (2 + 3i)(1 − 4i).

▶ Show full working

Expand the brackets, then use i² = −1.

(2 + 3i)(1 − 4i) = 2 − 8i + 3i − 12i² expand every pair of terms
  = 2 − 5i − 12(−1) combine −8i + 3i; use i² = −1
  = 2 − 5i + 12 simplify
  = 14 − 5i group real and imaginary

Answer: 14 − 5i.
Example 5

Question: Express (1 + i)/(1 − i) in the form a + bi.

▶ Show full working

Multiply top and bottom by the conjugate of the denominator, 1 + i, to clear the i.

(1 + i)/(1 − i) = [(1 + i)(1 + i)] / [(1 − i)(1 + i)] multiply by the conjugate
top = (1 + i)² = 1 + 2i + i² = 2i expand the top
bottom = 1 − i² = 1 + 1 = 2 expand the bottom (now real)
  = 2i / 2 put together
  = i simplify

Answer: (1 + i)/(1 − i) = i, i.e. 0 + 1i.
Example 6

Question: Find the conjugate and modulus of z = 5 − 12i.

▶ Show full working

The conjugate flips the imaginary sign; the modulus is √(real² + imaginary²).

conjugate = 5 + 12i flip the sign of −12i
|z| = √(5² + (−12)²) distance from the origin
  = √(25 + 144) square each part
  = √169 = 13 add and take the root

Answer: Conjugate 5 + 12i; modulus 13.
Example 7

Question: Find the real numbers x and y if (x + 2) + (y − 3)i = 5 + 4i.

▶ Show full working

Equal complex numbers have equal real parts and equal imaginary parts.

x + 2 = 5 match the real parts
x = 3 subtract 2
y − 3 = 4 match the imaginary parts
y = 7 add 3

Answer: x = 3, y = 7.
Example 8

Question: Verify that z · z̅ = |z|² for z = 3 + 4i.

▶ Show full working

Work out z · z̅ and |z|² separately, then compare.

z · z̅ = (3 + 4i)(3 − 4i) multiply by the conjugate
  = 9 − 16i² = 9 + 16 = 25 i² = −1
|z|² = (√(9 + 16))² = 25 modulus squared

Answer: z · z̅ = |z|² = 25, verified.
Example 9

Question: Plot z = −2 + 3i on the Argand plane and state its quadrant.

▶ Show full working

Plot at (real, imaginary) = (−2, 3).

Real part −2 → 2 units left; imaginary part 3 → 3 units up.
Left and up → the point lies in the second quadrant.

Answer: Point (−2, 3), second quadrant.
Example 10

Question: Express z = 1 + i in polar form.

▶ Show full working

Find r = |z| and the angle θ, then write z = r(cos θ + i sin θ).

r = √(1² + 1²) = √2 modulus
θ = 45° = π/4 the point (1, 1) is on the 45° line
z = √2 (cos π/4 + i sin π/4) polar form

Answer: z = √2 (cos π/4 + i sin π/4).
Example 11

Question: Solve x² + 4 = 0.

▶ Show full working

Isolate x², then take the square root using i.

x² + 4 = 0 given
x² = −4 subtract 4
x = ±√(−4) = ±2i √(−1) = i

Answer: x = 2i or x = −2i.
Example 12

Question: Solve x² + x + 1 = 0.

▶ Show full working

Read off a, b, c; a negative discriminant gives a conjugate pair of complex roots.

a = 1, b = 1, c = 1; discriminant b² − 4ac = 1 − 4 = −3 (negative).
x = (−b ± √(b² − 4ac)) / 2a quadratic formula
  = (−1 ± √(−3)) / 2 substitute
  = (−1 ± i√3) / 2 √(−3) = i√3

Answer: x = (−1 ± i√3)/2.
6

Where You Meet This in Real Life

Electrical engineering

Alternating-current circuits are analysed with complex numbers, where the modulus gives the size of a voltage or current and the argument gives its phase.

Signal processing

The Fourier transform behind audio compression, image formats and mobile networks is built entirely on complex numbers.

Quantum mechanics

The wave function describing a particle is complex-valued; physics at the smallest scale simply cannot be written without i.

Computer graphics and fractals

The famous Mandelbrot set is generated by repeatedly squaring a complex number, producing the intricate patterns you see in fractal art.

Control systems

Engineers test whether aircraft, robots and power grids stay stable by examining complex roots of their characteristic equations.
7

Practice Sets A-D

Practice Set A – Basics

A1. Simplify i¹².

▶ Reveal full working

Use the cycle of 4: divide the power by 4.

12 ÷ 4 = 3 remainder 0 remainder 0
i¹² = i⁴ = 1 remainder 0 means i⁴

Answer: i¹² = 1.

A2. Write √(−49) using i.

▶ Reveal full working

Split off √(−1) = i.

√(−49) = √49 × √(−1) separate
  = 7i √49 = 7

Answer: 7i.

A3. State the real and imaginary parts of 6 − 9i.

▶ Reveal full working

Real part = plain number; imaginary part = coefficient of i, with its sign.

Re(z) = 6 the plain number
Im(z) = −9 the coefficient of i

Answer: Re = 6, Im = −9.

A4. Write the conjugate of −3 + 7i.

▶ Reveal full working

Flip only the imaginary sign.

conjugate = −3 − 7i change +7i to −7i

Answer: −3 − 7i.

Practice Set B – Conceptual

B1. Why does i² = −1 give every quadratic two roots in C?

▶ Reveal full working

A negative discriminant used to block real solutions.

With i² = −1, √(negative) now exists as a multiple of i.
So the quadratic formula always returns two values in C.

Answer: Because square roots of negatives now exist.

B2. When is a complex number purely imaginary?

▶ Reveal full working

Write z = a + bi.

‘Purely imaginary’ means no real part but a non-zero i-part.
So a = 0 and b ≠ 0.

Answer: When its real part is 0 and imaginary part is non-zero.

B3. Show that the conjugate of a real number is itself.

▶ Reveal full working

A real number is a + 0i.

conjugate of a = a − 0i = a flipping 0i changes nothing

Answer: A real number equals its own conjugate.

B4. If |z| = 0, what is z?

▶ Reveal full working

Use |z| = √(a² + b²).

a² + b² = 0 square both sides of |z| = 0
Two squares add to 0 only if both are 0, so a = 0 and b = 0.

Answer: z = 0.

Practice Set C – Application / Numerical

C1. Express (3 + 2i)/(2 − i) as a + bi.

▶ Reveal full working

Multiply top and bottom by the conjugate 2 + i.

(3 + 2i)/(2 − i) = [(3 + 2i)(2 + i)] / [(2 − i)(2 + i)] multiply by conjugate
top = 6 + 3i + 4i + 2i² = 4 + 7i expand; i² = −1
bottom = 4 − i² = 5 expand
  = (4 + 7i)/5 put together
  = 4/5 + (7/5)i split over 5

Answer: 4/5 + (7/5)i.

C2. Find the modulus of z = −6 + 8i.

▶ Reveal full working

Use |z| = √(real² + imaginary²).

|z| = √((−6)² + 8²) distance formula
  = √(36 + 64) square
  = √100 = 10 add and root

Answer: |z| = 10.

C3. Solve 2x² + 3 = 0.

▶ Reveal full working

Isolate x², then take the root with i.

2x² + 3 = 0 given
x² = −3/2 isolate
x = ±i√(3/2) = ±(i√6)/2 √(−1) = i; tidy the surd

Answer: x = ±(i√6)/2.

C4. Express z = −1 + i√3 in polar form.

▶ Reveal full working

Find r and the argument, then write r(cos θ + i sin θ).

r = √((−1)² + (√3)²) = √4 = 2 modulus
θ = 2π/3 point (−1, √3) is in Quadrant II
z = 2(cos 2π/3 + i sin 2π/3) polar form

Answer: z = 2(cos 2π/3 + i sin 2π/3).

Practice Set D – HOTS / Multi-step

D1. If z = 2 + 3i, find z + z̅ and z − z̅, and say what kind of number each is.

▶ Reveal full working

Write the conjugate, then add and subtract.

z̅ = 2 − 3i conjugate
z + z̅ = (2 + 3i) + (2 − 3i) = 4 the i-parts cancel → real
z − z̅ = (2 + 3i) − (2 − 3i) = 6i the real parts cancel → imaginary

Answer: z + z̅ = 4 (real); z − z̅ = 6i (purely imaginary).

D2. Find the smallest positive integer n for which (1 + i)ⁿ is purely real.

▶ Reveal full working

Raise 1 + i to powers until the result is real.

(1 + i)² = 1 + 2i + i² = 2i still imaginary, so n = 2 fails
(1 + i)⁴ = (2i)² = 4i² = −4 this is real
n = 1 and n = 2 are not real, so the smallest is n = 4.

Answer: n = 4, giving (1 + i)⁴ = −4.

D3. Solve x² − 2x + 5 = 0 and verify the roots are conjugates.

▶ Reveal full working

Use the discriminant and the quadratic formula.

a = 1, b = −2, c = 5; discriminant = 4 − 20 = −16 (negative).
x = (2 ± √(−16)) / 2 quadratic formula
  = (2 ± 4i) / 2 √(−16) = 4i
  = 1 ± 2i simplify
1 + 2i and 1 − 2i differ only in the sign of the i-part → conjugates.

Answer: x = 1 ± 2i, a conjugate pair.

D4. Show that for any complex z, |z|² = z · z̅, and use it to find |(3 + 4i)(1 − 2i)|.

▶ Reveal full working

Prove the rule, then use |z₁z₂| = |z₁| |z₂|.

z · z̅ = (a + bi)(a − bi) = a² + b² this equals |z|²
|3 + 4i| = √(9 + 16) = 5 first modulus
|1 − 2i| = √(1 + 4) = √5 second modulus
|product| = 5 × √5 = 5√5 multiply the moduli

Answer: |z|² = z · z̅; the product modulus is 5√5.

Chapter Summary

Everything in One Glance

Imaginary Unit

i² = −1; powers of i cycle through i, −1, −i, 1.

  

Complex Number

z = a + bi with real part a and imaginary part b; equal only if both parts match.
 

Algebra

Add and subtract part by part; divide by multiplying with the conjugate.

  

Modulus & Conjugate

z̅ = a − bi; |z| = √(a² + b²); z · z̅ = |z|².
 

Argand & Polar

Plot (a, b); write z = r(cos θ + i sin θ) with r = |z|.

  

Quadratics

Negative discriminant gives a conjugate pair of complex roots.
 
8

Are You Exam-Ready?

8-Point Exam Quick-Check

1 Evaluate i²³.
 
2 Write √(−81) using i.
 
3 State the real and imaginary parts of −7 + 2i.
 
4 Multiply (1 + i)(1 − i).
 
5 Find the modulus and conjugate of 8 + 6i.
 
6 In which quadrant of the Argand plane does −3 − 4i lie?
 
7 Express 1 − i in polar form.
 
8 Solve x² + 9 = 0.
 

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Class 11 Mathematics Chapter 4: Complex Numbers and Quadratic Equations, Complete Notes and Practice

These free Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations notes are aligned with the latest NCERT 2026-27 syllabus and deliver complete, exam-ready coverage for board exams and competitive foundations. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.
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