Chapter 1: Complex Numbers

Complete Notes & Exercise 1.1 Solutions

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Introduction to Complex Numbers

In real numbers, the square root of a negative number is undefined. To handle such situations, imaginary numbers and complex numbers were introduced.

Imaginary Number

i = √−1

Powers of i:

i¹ = i
i² = -1
i³ = -i
i⁴ = 1 (repeats every 4 powers)

Complex Number

A complex number is a number in the form a + bi, where a and b are real numbers, and i = √−1.

Pure Imaginary Number

A complex number of the form 0 + bi, such as 2i or -5i.

Conjugate of Complex Number

The conjugate of a + bi is a - bi. It is used in division of complex numbers.

Operations on Complex Numbers

Addition/Subtraction: (3 + 2i) + (1 - 4i) = 4 - 2i

Multiplication: (2 + 3i)(1 - i) = 5 + i

Division: (1 + i)/(1 - i) = i

Exercise 1.1 – Solutions

Question 1: Simplify and express in terms of i

i. √−3

= √3 × √−1

= √3i Answer

ii. 6√−4

= 6 × √4 × √−1

= 6 × 2i

= 12i

iii. √(−4/9)

= √−4 / √9

= 2i / 3

iv. −√−20

= −√(4×5) × √−1

= −2√5i

v. 4 − √−60

= 4 − √(4×15) × √−1

= 4 − 2√15i

vi. √−8 × √−2

= √8i × √2i

= √16 × i²

= 4 × (−1)

= −4

Question 2: Simplify

i. (4 − i) + (5 + 5i)

= (4+5) + (-1+5)i

= 9 + 4i

ii. (7 − 6i) − (5 − 6i)

= (7-5) + (-6+6)i

= 2

iii. (−2 + 8i) − (7 + 3i)

= (-2-7) + (8-3)i

= −9 + 5i

iv. (4 − 2i) − (5 − 2i)

= (4-5) + (-2+2)i

= −1

v. (2 + 4i)(1 + 2i)

= 2*1 + 2*2i + 4i*1 + 4i*2i

= 2 + 4i + 4i + 8i²

= 2 + 8i + 8(-1)

= −6 + 8i

vi. (1 − 4i)(2 − 3i)

= 1×2 + 1×(−3i) + (−4i)×2 + (−4i)×(−3i)

= 2 − 3i − 8i + 12i²

= 2 − 11i + 12(−1)

= 2 − 11i − 12

= −10 − 11i

vii. −8i(2 − 2i)

= (−8i)×2 + (−8i)×(−2i)

= −16i + 16i²

= −16i + 16(−1)

= −16i − 16

= −16 − 16i

viii. (3 + 2i)²

= (3 + 2i)(3 + 2i)

= 3×3 + 3×2i + 2i×3 + 2i×2i

= 9 + 6i + 6i + 4i²

= 9 + 12i + 4(−1)

= 9 + 12i − 4

= 5 + 12i

ix. (3 − 6i)(3 + 6i)

= a² − b² where a = 3, b = 6i

= 3² − (6i)²

= 9 − (36)(−1)

= 9 + 36

= 45

x. (−5 − 3i)²

= (−5)² + 2(−5)(−3i) + (−3i)²

= 25 + 30i + 9(−1)

= 25 + 30i − 9

= 16 + 30i

xi. (1 + √2i)(1 − √3i)

= 1×1 + 1×(−√3i) + √2i×1 + √2i×(−√3i)

= 1 − √3i + √2i − √6(i²)

= 1 − √3i + √2i + √6

= (1 + √6) + (√2 − √3)i

xii. (√2 + i)(√2 − i)

= a² − b² where a = √2, b = i

= (√2)² − (i)²

= 2 − (−1)

= 2 + 1

= 3

Question 3: Simplify

i. i⁰ = 1

· Any number raised to the power 0 is 1.

· So, i⁰ = 1

ii. i¹¹ = i³ = −i

· The powers of i repeat every 4 terms:

· i¹ = i, i² = −1, i³ = −i, i⁴ = 1, ...

· 11 mod 4 = 3 → i¹¹ = i³ = −i

iii. i²⁸ = i⁰ = 1

· 28 mod 4 = 0, so i²⁸ = i⁰ = 1

iv. (−i)²¹ = −i

· (−i)²¹ = (−1)²¹ × i²¹ = (−1) × i¹

· = −i

v. (3i)³ = −27i

· (3i)³ = 3³ × i³ = 27 × (−i) = −27i

vi. (−2i)⁴ = 16

· (−2i)⁴ = (−2)⁴ × i⁴ = 16 × 1 = 16

Question 4: Simplify in the form a + bi

i. 9 + i⁶

Step 1: Recall that powers of i repeat every 4 steps:
i¹ = i, i² = -1, i³ = -i, i⁴ = 1, then repeats...
i⁶ = i² = -1
Step 2: Substitute in the expression:
9 + i⁶ = 9 + (-1)

= 8

ii. −17 + i⁸ = −16

Step 1: i⁸ = (i⁴)² = 1² = 1
Step 2: −17 + i⁸ = −17 + 1

= −16

iii. i⁴ − 13i = 1 − 13i

Step 1: i⁴ = 1
Step 2: i⁴ − 13i

= 1 − 13i

iv. i³ + 2i = i

Step 1: i³ = -i
Step 2: -i + 2i

= i

v. i⁵ + i⁷ = 0

Step 1: i⁵ = i (since i⁴ = 1, i⁵ = i⁴·i = 1·i = i)
Step 2: i⁷ = i³ = -i
Step 3: i⁵ + i⁷ = i + (-i)

= 0

vi. i⁷⁰ − i¹⁰⁰ = 0

Step 1: Find i⁷⁰ and i¹⁰⁰
→ i⁷⁰ = i² = -1
→ i¹⁰⁰ = i⁰ = 1
Step 2: i⁷⁰ − i¹⁰⁰ = -1 − 1

= -2

Question 5: Divide and simplify in the form a + bi

i. (3 / (4 − i)) × (4 + i)/(4 + i)

· = (3 × (4 + i)) / ((4 − i)(4 + i))

· = (12 + 3i) / (4² + 1²)

· = (12 + 3i) / 17

ii. (3i / (6 + 5i)) × (6 − 5i)/(6 − 5i)

· = (3i × (6 − 5i)) / ((6 + 5i)(6 − 5i))

· = (18i − 15i²) / (36 + 25)

· = (18i + 15) / 61

· = (15 + 18i) / 61

iii. (3 − i√5)/(3 + i√5)

· = [(3 − i√5)(3 − i√5)] / [(3 + i√5)(3 − i√5)]

· = (9 − 6i√5 + (i√5)²) / (9 + 5)

· = (9 − 6i√5 − 5) / 14

· = (4 − 6i√5) / 14

· = (2/7) − (3/7)i√5

iv. (2 + 7i)/5i

· = [(2 + 7i) × i] / (5i × i)

· = (2i + 7i²) / (5i²)

· = (2i − 7) / (−5)

· = (−7/−5) + (2i/−5)

· = 7/5 − 2/5i

v. (4 + 5i)/(4 − 5i)

· = [(4 + 5i)(4 + 5i)] / [(4 − 5i)(4 + 5i)]

· = (16 + 40i + 25i²) / (16 + 25)

· = (16 + 40i − 25) / 41

· = (−9 + 40i) / 41

· = −9/41 + 40/41i

vi. (3 + 2i)/(2 + i)

· = [(3 + 2i)(2 − i)] / [(2 + i)(2 − i)]

· = (6 − 3i + 4i − 2i²) / (4 + 1)

· = (6 + i + 2) / 5

· = (8 + i) / 5

· = 8/5 + 1/5i

vii. (5 + i)/(1 + 2i)

· = [(5 + i)(1 − 2i)] / [(1 + 2i)(1 − 2i)]

· = (5 − 10i + 1i − 2i²) / (1 + 4)

· = (5 − 9i + 2) / 5

· = (7 − 9i) / 5

· = 7/5 − 9/5i

viii. (a + ib)/(a − ib)

· = [(a + ib)(a + ib)] / [(a − ib)(a + ib)]

· = (a + ib)² / (a² + b²)

ix. (1 + i)/(1 − i)²

· (1 − i)² = 1 − 2i + i² = 1 − 2i − 1 = −2i

· Then (1 + i)/(−2i) = [(1 + i) × i] / (−2i × i)

· = (i + i²) / (−2 × −1) = (i − 1)/2

· = −0.5 + 0.5i

x. ((2 + 2i)²)/((1 + i)²)

· Numerator: (2 + 2i)² = 4 + 8i + 4i² = 4 + 8i − 4 = 8i

· Denominator: (1 + i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i

· Then (8i)/(2i) = 4

Developing Skilled Knowledge – Conceptual Answers

1. An imaginary number is a number of the form bi, where b is a real number and i = √−1.

2. Imaginary numbers are not real numbers because they include √−1.

3. Examples of pure imaginary numbers: 2i, −5i, (1/3)i.

4. Powers of i form a repeating cycle: i, −1, −i, 1.

5. Product of (2 − 3i)(2 − 3i) = −5 − 12i (not purely imaginary).

6. Yes, zz̅ = (a + bi)(a − bi) = a² + b² ∈ ℝ (real number).

Grade 10 - Exercise 1.1 Complete Solution