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State Board · Class 11 · Mathematics Part 2

Lesson 1: Complex Numbers

5 concepts · 40 practice questions with answers · Free to read

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Complex numbers and their algebra

From daily life

In an electric circuit we must state a current and its angle together; one number is not enough, we need a pair. Such a pair is a complex number. Building on classes 9–10 : Since classes 9–10 we solved quadratic equations; x² = −1 had no real answer. Now we take a new number i for it.

Simple definition

i is a number with i² = −1. A number a + bi (a, b real) is a complex number; a is the real part and b the imaginary part. The number a − bi is its conjugate.

Rule 1: The cycle of powers of i — i¹ = i, i² = −1, i³ = −i, i⁴ = 1, then the same again. So for iⁿ divide n by 4 and look at the remainder: 0 → 1, 1 → i, 2 → −1, 3 → −i.

ReImi−1−i1each multiplication by i turns the point 90° anticlockwise

In the figure : The four powers of i are four points; each multiplication by i turns the point through 90°.

Rule 2: The four operations — Addition: real with real, imaginary with imaginary. Multiplication: the usual product of two brackets with i² = −1. Division: multiply numerator and denominator by the conjugate of the denominator. Two complex numbers are equal only when their real parts and their imaginary parts are equal.

How is the formula obtained? : Since i² = −1 we get i⁴ = (i²)² = 1; so after i⁴ the powers return to the same four values. In division (a + bi)(a − bi) = a² + b² is real, so we multiply by the conjugate to make the denominator real.

  • Easy : (3 + 2i) + (1 − 5i) = 4 − 3i.
  • Medium : (2 + 3i)(4 − i) = 8 − 2i + 12i − 3i² = 8 + 10i + 3 = 11 + 10i.
  • Exam-type : (5 + i) ÷ (2 − 3i) = (5 + i)(2 + 3i) ÷ 13 = (10 + 15i + 2i − 3) ÷ 13 = (7 + 17i)/13. Also i⁴⁷: the remainder of 47 ÷ 4 is 3, so i⁴⁷ = −i.
Remember

The imaginary part is b, not bi; for 4 − 7i it is −7.

Common mistake

Writing √(−4) × √(−9) = √36 = 6. The correct value is 2i × 3i = 6i² = −6.

Shortcut

Any four consecutive powers add up to 0: i + i² + i³ + i⁴ = 0. Memory clue : "i, −1, −i, 1: back at the fourth".

CET trap

i⁴ⁿ⁺² is −1, not 1; when the remainder is 2 take i².

CET-style example : i⁹⁹ + i¹⁰⁰ + i¹⁰¹ + i¹⁰² = ? (A) 0 (B) 1 (C) i (D) −1

Answer : (A). −i + 1 + i − 1 = 0. Time-saving method : When you see four consecutive powers the answer is 0 at once.

Spot the difference

Pointz + conjugatez × conjugate
Result2a (real)a² + b² (real)
When zeroa = 0never, when z ≠ 0
Useto get the real partto make the denominator real

Practice questions

Easy
  1. 1Find (2 + 5i) + (3 − 2i).
  2. 2State the value of i¹³.
  3. 3Write the conjugate of 4 − 7i.
Medium
  1. 4Find (3 − i)(2 + 4i).
  2. 5Find (1 + i)².
  3. 6Simplify 1 ÷ (3 + 4i).
Challenging
  1. 7Simplify (4 + 3i) ÷ (1 − 2i).
  2. 8Find the value of i + i² + … + i⁵⁰.
Show answers (try them yourself first)
  1. 15 + 3i.
  2. 2i¹³: the remainder of 13 ÷ 4 is 1; so i.
  3. 34 + 7i.
  4. 46 + 12i − 2i − 4i² = 10 + 10i.
  5. 51 + 2i + i² = 2i.
  6. 6(3 − 4i)/25.
  7. 7(4 + 3i)(1 + 2i) ÷ 5 = (4 + 8i + 3i − 6) ÷ 5 = (−2 + 11i)/5.
  8. 848 terms make 12 groups, each adding to 0; the rest is i⁴⁹ + i⁵⁰ = i − 1.

Square root of a complex number and quadratic equations

From daily life

If the graph of a quadratic never touches the x-axis it has no real answer, yet we still want an exact answer. Complex numbers give it. Building on classes 9–10 : Since classes 9–10 we know D = b² − 4ac; here we see what happens when D is negative.

Simple definition

A square root of z = a + bi is a number x + yi with (x + yi)² = a + bi. Every non-zero number has two square roots, one the negative of the other.

Rule 1: Finding the square root — Write x² − y² = a and 2xy = b. Take r = √(a² + b²). Then x = ±√((r + a)/2), y = ±√((r − a)/2), with signs chosen so that xy has the sign of b.

ReIm123452 + 3i2 − 3ix² − 4x + 13 = 0b² − 4ac = 16 − 52 = −36no real rootstwo conjugate rootssum = 4product = 13the two roots are mirror images in the real axis

In the figure : The two roots of x² − 4x + 13 = 0 are 2 + 3i and 2 − 3i; mirror images in the real axis.

Rule 2: Quadratic with negative D — In ax² + bx + c = 0 (a, b, c real) with D < 0 the roots are x = (−b ± i√(−D)) ÷ 2a, a conjugate pair. The sum = −b/a and the product = c/a stay the same.

How is the formula obtained? : (x + yi)² = x² − y² + 2xyi. Matching the real part with a and the imaginary part with b gives two equations. Adding x² + y² = r (because |z| = r) separates x² and y².

  • Easy : √(−9) = ±3i.
  • Medium : √(8 − 6i): r = 10, x = √((10 + 8)/2) = 3, y = √((10 − 8)/2) = 1; b is negative, so ±(3 − i).
  • Exam-type : x² − 4x + 13 = 0: D = 16 − 52 = −36; x = (4 ± 6i) ÷ 2 = 2 ± 3i.
Remember

A complex root of an equation with real coefficients always comes in a conjugate pair.

Common mistake

Giving only one sign to the square root. √(8 − 6i) = ±(3 − i), both roots.

Shortcut

Given one root, the other is the conjugate; p = −(sum), q = product. Memory clue : "The pair is conjugate: sum and product are real".

CET trap

When D is negative the roots are not real, but "no roots" is a wrong option; the roots are complex.

CET-style example : One root of x² + px + q = 0 (p, q real) is 3 − 2i. Find p, q. (A) −6, 13 (B) 6, 13 (C) −6, 5 (D) 6, 5

Answer : (A). The other root is 3 + 2i; sum 6 so p = −6; product 9 + 4 = 13 so q = 13. Time-saving method : Do not solve the equation; take the sum and product directly.

Spot the difference

DRootsGraph
D > 0two distinct realcuts the x-axis twice
D = 0two equal realtouches the x-axis
D < 0conjugate complex pairdoes not meet the x-axis

Practice questions

Easy
  1. 1Find √(−25).
  2. 2Solve x² + 4 = 0.
  3. 3Find D of x² − 2x + 5 = 0.
Medium
  1. 4Find √(5 + 12i).
  2. 5Solve x² − 6x + 25 = 0.
  3. 6Solve x² + x + 1 = 0.
Challenging
  1. 7Write the real-coefficient quadratic that has 1 − 2i as a root.
  2. 8Find √(−7 + 24i).
Show answers (try them yourself first)
  1. 1±5i.
  2. 2x² = −4, so x = ±2i.
  3. 34 − 20 = −16.
  4. 4r = 13, x = √9 = 3, y = √4 = 2; ±(3 + 2i).
  5. 5D = 36 − 100 = −64; x = (6 ± 8i) ÷ 2 = 3 ± 4i.
  6. 6x = (−1 ± i√3) ÷ 2.
  7. 7The other root is 1 + 2i; sum 2, product 5; x² − 2x + 5 = 0.
  8. 8r = 25, x = √((25 − 7)/2) = 3, y = √((25 + 7)/2) = 4; ±(3 + 4i).

Argand plane, modulus, argument and polar form

From daily life

To give a place on a map we state "distance and direction". The polar form of a complex number does the same: the distance from the origin and the angle from the axis. Building on classes 9–10 : Since classes 9–10 we know coordinates and sin, cos; we treat a + bi as the point (a, b) and use them.

Simple definition

The plane where a + bi is the point (a, b) is the Argand plane. The distance r = √(a² + b²) from the origin is the modulus. The angle θ measured anticlockwise from the positive x-axis is the argument; its value in (−π, π] is the principal argument.

Rule 1: Properties of the modulus — ∣z₁z₂∣ = ∣z₁∣∣z₂∣; ∣z₁/z₂∣ = ∣z₁∣/∣z₂∣; the distance between z₁ and z₂ = ∣z₁ − z₂∣; ∣z∣² = z × conjugate.

ReImθ34z = 3 + 4ir = 5|z| = r = 5tan θ = 4/3z = r(cos θ + i sin θ)

In the figure : z = 3 + 4i is the point (3, 4); r = 5 and tan θ = 4/3.

Rule 2: Argument and polar form — First take the acute angle α = tan⁻¹ ∣b/a∣; then by quadrant θ: 1st α, 2nd π − α, 3rd −(π − α), 4th −α. z = r(cos θ + i sin θ).

How is the formula obtained? : The point (a, b) gives a right triangle: a = r cos θ, b = r sin θ. So a + bi = r cos θ + i r sin θ = r(cos θ + i sin θ). Pythagoras gives r² = a² + b².

  • Easy : ∣3 + 4i∣ = √(9 + 16) = 5.
  • Medium : −1 + i: 2nd quadrant, α = tan⁻¹ 1 = π/4; θ = π − π/4 = 3π/4.
  • Exam-type : 1 − i√3: r = 2, α = π/3, 4th quadrant so θ = −π/3; polar form 2(cos(−π/3) + i sin(−π/3)).
Remember

Before finding the argument see which quadrant the point is in; do not stop at tan⁻¹.

Common mistake

Taking θ = tan⁻¹(−1) = −π/4 for −1 + i. The point is in the 2nd quadrant, so it is 3π/4.

Shortcut

To find the modulus of a quotient do not divide: modulus of the numerator ÷ modulus of the denominator. Memory clue : "In a product or quotient the modulus multiplies or divides".

CET trap

The modulus of z₁ − z₂ gives the distance; it is not ∣z₁∣ − ∣z₂∣.

CET-style example : ∣(3 + 4i) ÷ (5 − 12i)∣ = ? (A) 5/13 (B) 13/5 (C) 1/13 (D) 7/17

Answer : (A). ∣3 + 4i∣ = 5 and ∣5 − 12i∣ = 13. Time-saving method : Find the two moduli separately without dividing.

Spot the difference

QuadrantSigns of a, bθ
1st+, +α
2nd−, +π − α
3rd−, −−(π − α)
4th+, −−α

Practice questions

Easy
  1. 1Find ∣−6 + 8i∣.
  2. 2In which quadrant is the point (2, −3)?
  3. 3Find ∣5 − 12i∣.
Medium
  1. 4Find the principal argument of 1 + i.
  2. 5Find the principal argument of −1 − i.
  3. 6Find the distance between 2 + i and 5 + 5i.
Challenging
  1. 7Write −2 + 2i in polar form.
  2. 8Find ∣(6 + 8i)(5 − 12i)∣.
Show answers (try them yourself first)
  1. 1√(36 + 64) = 10.
  2. 24th quadrant (x positive, y negative).
  3. 3√(25 + 144) = 13.
  4. 4α = π/4, first quadrant; θ = π/4.
  5. 5α = π/4, third quadrant; θ = −3π/4.
  6. 6z₁ − z₂ = 3 + 4i; modulus 5.
  7. 7r = 2√2, 2nd quadrant, θ = 3π/4; 2√2(cos 3π/4 + i sin 3π/4).
  8. 810 × 13 = 130.

Exponential form and De Moivre's theorem

From daily life

A clock hand moves round the same circle every day; after one full turn it is back at the start. Powers of a complex number work the same way on turns of the angle. Building on classes 9–10 : Since classes 9–10 we know angles and the sum formulas of sin and cos; De Moivre’s theorem is a result of them.

Simple definition

e^(iθ) = cos θ + i sin θ (θ in radians). So z = r(cos θ + i sin θ) can be written in exponential form as z = re^(iθ). Special case: e^(iπ) = −1.

Rule 1: De Moivre's theorem — (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. So zⁿ = rⁿ(cos nθ + i sin nθ). In a product the moduli multiply and the angles add.

zz²z³z at 30°z² at 60°z³ at 90°angle × nlength stays 1(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ

In the figure : If z has angle 30° then z² has 60° and z³ has 90°; the modulus 1 stays the same.

Rule 2: The nth roots — The n roots of z = r(cos θ + i sin θ): r^(1/n) × (cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)), k = 0, 1, …, n − 1. They lie on one circle at equal gaps.

How is the formula obtained? : (cos a + i sin a)(cos b + i sin b) = cos(a + b) + i sin(a + b) (by the sum formulas of sin and cos). Repeating it n times multiplies the angle by n and gives the formula.

  • Easy : e^(iπ/2) = cos 90° + i sin 90° = i.
  • Medium : (1 + i)⁸: r = √2, θ = π/4; (√2)⁸ = 16 and 8θ = 2π; so 16.
  • Exam-type : (√3 + i)⁵: r = 2, θ = π/6; 2⁵ = 32, 5θ = 5π/6; 32(−√3/2 + i/2) = −16√3 + 16i.
Remember

Keep the angle in radians; using degrees in e^(iθ) is wrong.

Common mistake

Writing (cos θ + i sin θ)ⁿ = cosⁿ θ + i sinⁿ θ. The correct form is cos nθ + i sin nθ.

Shortcut

For a large power first try a small power: (1 − i)² = −2i. Memory clue : "(1 ± i)² = ±2i".

CET trap

For nth roots use all n values of k; with only k = 0 the roots are incomplete.

CET-style example : (1 − i)¹² = ? (A) −64 (B) 64 (C) 64i (D) −64i

Answer : (A). (1 − i)² = −2i, so (−2i)⁶ = 64 × i⁶ = 64 × (−1) = −64. Time-saving method : Use the power of (1 − i)² without finding the angle.

Spot the difference

FormHow it is writtenMost convenient for
Algebraica + biaddition, subtraction
Polarr(cos θ + i sin θ)seeing angle and modulus
Exponentialre^(iθ)products, powers, roots

Practice questions

Easy
  1. 1State the value of e^(iπ).
  2. 2Find the value of (cos 20° + i sin 20°)⁹.
  3. 3Write 2(cos π/6 + i sin π/6) in exponential form.
Medium
  1. 4Find (1 + i)⁴.
  2. 5Find (√3 − i)⁶.
  3. 6Simplify (cos θ + i sin θ)³ (cos θ − i sin θ)².
Challenging
  1. 7Find the square root of i.
  2. 8Find (1 + i√3)⁹.
Show answers (try them yourself first)
  1. 1cos π + i sin π = −1.
  2. 2cos 180° + i sin 180° = −1.
  3. 32e^(iπ/6).
  4. 4r = √2, θ = π/4; (√2)⁴ = 4 and 4θ = π; so −4.
  5. 5r = 2, θ = −π/6; 2⁶ = 64 and 6θ = −π; so −64.
  6. 6(cos θ + i sin θ)³ (cos θ + i sin θ)⁻² = cos θ + i sin θ.
  7. 7i = cos 90° + i sin 90°; the roots are ±(cos 45° + i sin 45°) = ±(1 + i)/√2.
  8. 8r = 2, θ = π/3; 2⁹ = 512 and 9θ = 3π; so −512.

Cube roots of unity

From daily life

On a circular staircase three equal steps make one full turn; after three steps we are back at the start. The three cube roots of 1 lie on a circle 120° apart in the same way. Building on classes 9–10 : Since classes 9–10 we factorised expressions like a³ − b³; here we find all roots of x³ = 1.

Simple definition

The three roots of x³ = 1 are 1, ω = (−1 + i√3)/2 and ω² = (−1 − i√3)/2. ω and ω² are conjugates of each other.

Rule 1: Main properties — ω³ = 1 and 1 + ω + ω² = 0. For ωⁿ divide n by 3: remainder 0 → 1, 1 → ω, 2 → ω². Also 1 + ω = −ω², 1 + ω² = −ω, 1/ω = ω².

1ωω²ω = (−1 + i√3)/2ω² = (−1 − i√3)/21 + ω + ω² = 0ω³ = 1angles 0°, 120°, 240°: an equilateral triangle

In the figure : The three roots at 0°, 120°, 240°; joined they form an equilateral triangle.

Rule 2: Factors — x³ − 1 = (x − 1)(x − ω)(x − ω²) and a³ + b³ = (a + b)(a + ωb)(a + ω²b).

How is the formula obtained? : x³ − 1 = (x − 1)(x² + x + 1). The roots of x² + x + 1 = 0 are (−1 ± i√3)/2, which are ω and ω². Since ω is a root of x³ = 1 we have ω³ = 1; and from x² + x + 1 = 0 we get 1 + ω + ω² = 0.

  • Easy : ω⁷: the remainder of 7 ÷ 3 is 1, so ω; ω³⁰ = 1.
  • Medium : (1 + ω)³ = (−ω²)³ = −ω⁶ = −1.
  • Exam-type : (1 + ω − ω²)³: 1 + ω = −ω², so −ω² − ω² = −2ω²; the cube is −8ω⁶ = −8.
Remember

Learn 1 + ω + ω² = 0 by heart; most questions are solved with it.

Common mistake

Writing ω² = −ω. The correct form is ω² = −1 − ω.

Shortcut

Replace 1 + ω by −ω² and 1 + ω² by −ω; and 1/ω² = ω. Memory clue : "One plus ω is minus ω squared".

CET trap

In ωⁿ divide n by 3, not by 4; dividing by 4 is for i.

CET-style example : 1/(1 + ω) + 1/(1 + ω²) = ? (A) 1 (B) −1 (C) 0 (D) ω

Answer : (A). 1/(−ω²) + 1/(−ω) = −ω − ω² = 1. Time-saving method : Make the denominators −ω² and −ω and use 1/ω² = ω.

Spot the difference

Remainder when n is divided by 3ωⁿExample
01ω⁹ = 1
1ωω¹⁰⁰ = ω (100 = 99 + 1)
2ω²ω⁸ = ω²

Practice questions

Easy
  1. 1State the value of ω⁹.
  2. 2State the value of ω¹⁰.
  3. 3State the value of ω + ω².
Medium
  1. 4Find the value of ω¹⁰⁰.
  2. 5Write 1 + ω² in terms of ω.
  3. 6Find (ω + ω²)³.
Challenging
  1. 7Find (1 − ω + ω²)(1 + ω − ω²).
  2. 8Find (2 + ω)(2 + ω²).
Show answers (try them yourself first)
  1. 1The remainder of 9 ÷ 3 is 0; 1.
  2. 2Remainder 1; ω.
  3. 3Since 1 + ω + ω² = 0 it is −1.
  4. 4The remainder of 100 ÷ 3 is 1; ω.
  5. 51 + ω² = −ω.
  6. 6(−1)³ = −1.
  7. 7(−2ω)(−2ω²) = 4ω³ = 4.
  8. 84 + 2(ω + ω²) + ω³ = 4 − 2 + 1 = 3.

Lesson 1: Complex numbers — quick revision

TopicFormula / point
Powers of ii¹ = i, i² = −1, i³ = −i, i⁴ = 1; remainder by 4
Operationsin division multiply by the conjugate of the denominator; z × conjugate = a² + b²
Square rootx = ±√((r + a)/2), y = ±√((r − a)/2); xy has the sign of b
QuadraticD < 0: x = (−b ± i√(−D)) ÷ 2a; roots are conjugate
Modulus / argumentr = √(a² + b²); α = tan⁻¹ ∣b/a∣, θ by quadrant
De Moivre(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ; e^(iθ) = cos θ + i sin θ
Cube rootsω³ = 1; 1 + ω + ω² = 0; for ωⁿ remainder by 3

🎯 Most important points for the exam

  • In iⁿ divide by 4, in ωⁿ by 3.
  • Square roots and roots of a quadratic come two at a time.
  • Check the quadrant when finding the argument.
  • The modulus of a quotient is the quotient of moduli.
  • In ω questions use 1 + ω + ω² = 0.
Study mantra

"Four for i, three for ω; a quadrant for the angle".