Chapter 13: Complex Numbers – Exercise 13.1

Complex Numbers – Exercise 13.1 – Q.1(i)

We know that 

i² = – 1

i³ = – i

i⁴  = 1

In order to find iⁿ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

n = 4p + q, o ≤ q < 4

Then in = i^4o+q

= i^4o ×i^q

= (i⁴)^q × i^q

= i^p × i^q

= i^q             [∵ 1^p-1]

Hence iⁿ = i^q, where o ≤ q < 4

∴  i⁴⁵⁷ = i^4×114 × i¹

= i¹

= i

Complex Numbers – Exercise 13.1 – Q.1(ii)

We know that

i² =  – 1

i³ =  – i

i⁴ = 1

In order to find iⁿ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

n = 4p + q, o ≤ q < 4

Then iⁿ = i^4o+q

= i^4o × i^q

= (i⁴)^p ×  i^q

= 1^p × i^q

= i^q          [∵ 1^p-1]

Hence iⁿ = i^q, where o ≤ q < 4

∴ i⁵²⁸ = i^4×132

= (i⁴)¹³²

= 1¹³²

= 1

∴  (i⁵²⁸) = 1

Complex Numbers – Exercise 13.1 – Q.1(iii)

We know that

i² = – 1

i³ = – i

i⁴ = 1

In order to find iⁿ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

n = 4p + q, o ≤ q < 4

Then iⁿ = i^4o+q

= i^4o × i^q

= (i⁴)^p ×  i^q

= 1^p × i^q

= i^q            [∵  1^p-1]

Hence iⁿ = i^q, where o ≤ q < 4

Complex Numbers – Exercise 13.1 – Q.1(iv)

We know that

i² = – 1

i³ = – i

i⁴ = 1

In order to find iⁿ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

n = 4p + q, o ≤ q < 4

Then iⁿ = i^4o+q

= i^4o × i^q

= (i⁴)^p ×  i^q

= 1^p × i^q

= i^q            [∵ 1^p-1]

Hence iⁿ = i^q, where o ≤ q < 4

Complex Numbers – Exercise 13.1 – Q.1(v)

We know that

i² = – 1

i³ = – i

i⁴ = 1

In order to find iⁿ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

n = 4p + q, o ≤ q < 4

Then iⁿ = i^4o+q

= i^4o × i^q

= (i⁴)^p ×  i^q

= 1^p × i^q

= i^q                   [∵ 1^p-1]

Hence iⁿ = i^q, where o ≤ q < 4

Complex Numbers – Exercise 13.1 – Q.1(vi)

We know that

i² = – 1

i³ = – i

i⁴ = 1

In order to find iⁿ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

n = 4p + q, o ≤ q < 4

Then iⁿ = i^4o+q

= i^4o × i^q

= (i⁴)^p ×  i^q

= 1^p × i^q

= i^q                   [∵ 1^p-1]

Hence iⁿ = i^q, where o ≤ q < 4

(i⁷⁷ + i⁷⁰ + i⁸⁷ + i⁴¹⁴)³ = (i^4×19 × i¹ + i^4×17 × i² + i^4×21 × i³ + i^4×103 × i²)³

= (1 × i + 1 × i² + 1 × i³ + 1 × i²)³

= (i – 1 –  i –1)³

= ( – 2)³

= – 8

∴  (i⁷⁷ + i⁷⁰ + i⁸⁷ + i⁴¹⁴)³ = – 8

Complex Numbers – Exercise 13.1 – Q.1(vii)

i² =  – 1

i³ = – i

i⁴ = 1

In order to find iⁿ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

n = 4p + q, o ≤ q < 4

Then iⁿ = i^4o+q

= i^4o × i^q

= (i⁴)^p ×  i^q

= 1^p × i^q

= i^q                   [∵ 1^p-1]

Hence iⁿ = i^q, where o ≤ q < 4

∴  i³⁰ + i⁴⁰ + i⁶⁰ = i^4×7 ×i² + i^4×10 + i^4×15

= 1 × i² + 1 + 1

= – 1 + 1 + 1

= 1

∴  i³⁰ + i⁴⁰ + i⁶⁰ = 1

Complex Numbers – Exercise 13.1 – Q.1(viii)

We know that

i² = – 1

i³ = – i

i⁴ = 1

In order to find iⁿ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

n = 4p + q, o ≤ q < 4

Then iⁿ = i^4o+q

= i^4o × i^q

= (i⁴)^p ×  i^q

= 1^p × i^q

= i^q          [∵ 1^p-1]

Hence iⁿ = i^q, where o ≤ q < 4

i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰ = i^4×12 × i¹ + i^4×17 + i^4×22 × i¹ + i^4×27 × i²

= 1 × i + 1 + 1 × i + 1 × i²

= i + 1 + i - 1

= 2i

∴ i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰ = 2i

Complex Numbers – Exercise 13.1 – Q.2

1 + i¹⁰  +  i²⁰ + i³⁰ = 1 + i^4×2 × i² + i^4×5 + i^4×7 × i²

= 1 + 1 × i² + 1 + 1 × i²

= 1 - 1 + 1 - 1

=  0, which is real number