Given w = 5 cos π 6 + i sin π 6 . Write w in exponential form.

Given w = 5 cos ⁡ π 6 + i sin ⁡ π 6 Now, w = 5 = r and θ = π 6 The exponential form is given by w = r e i θ i.e. w = 5 e i π 6

Given w = 5 cos π 6 + sin π 6 and z = 3 - 7 i 5 - 2 i . Write z in exponential form.

We have, z = 3 − 7 i 5 − 2 i Now, z = 3 − 7 i 5 − 2 i ⇒ z = 3 − 7 i 5 − 2 i ∵ z 1 z 2 = z 1 z 2 ⇒ z = 3 2 + ( − 7 ) 2 5 2 + ( − 2 ) 2 [From the definition of modulus.] ⇒ | z | = 58 29 ⇒ | z | = 2 Now, arg ⁡ z = arg 3 − 7 i 5 − 2 i ⇒ arg ⁡ z = arg 3 - 7 i - arg 5 - 2 i ∵ arg ⁡ z 1 z 2 = arg ⁡ z 1 − arg ⁡ z 2 ⇒ arg ⁡ z = tan − 1 ⁡ − 7 3 − tan − 1 ⁡ − 2 5 ⇒ argz = - π 4 Hence, the exponential form is given by z = 2 e − 1 4 i π

Given, w = 5 cos π 6 + i sin π 6 and z = 3 - 7 i 5 - 2 i . Find and simplify an expression for z w .

We have, w = 5 cos π 6 + i sin π 6 and z = 3 - 7 i 5 - 2 i Now, z = 3 − 7 i 5 − 2 i Now, z = 3 − 7 i 5 − 2 i ⇒ z = 3 − 7 i 5 − 2 i ∵ z 1 z 2 = z 1 z 2 ⇒ z = 3 2 + ( − 7 ) 2 5 2 + ( − 2 ) 2 [From the definition of modulus.] ⇒ | z | = 58 29 ⇒ | z | = 2 Now, arg ⁡ z = arg 3 − 7 i 5 − 2 i ⇒ arg ⁡ z = arg 3 - 7 i - arg 5 - 2 i ∵ arg ⁡ z 1 z 2 = arg ⁡ z 1 − arg ⁡ z 2 ⇒ arg ⁡ z = tan − 1 ⁡ − 7 3 − tan − 1 ⁡ − 2 5 ⇒ argz = - π 4 Hence, the exponential form is given by z = 2 e − 1 4 i π Now, w = 5 cos ⁡ π 6 + i sin ⁡ π 6 Now, w = 5 = r and θ = π 6 The exponential form is given by w = r e i θ i.e. w = 5 e i π 6 So, z w = 2 e − 1 4 i π 5 e π i 6 = 2 5 e i π − 1 4 − 1 6 = 2 5 e − 5 12 π i

The complex number a + b i is denoted by z . Given that | z | = 5 and arg z = π 6 , find the value of a and the value of b .

Given, z = a + b i Now, z = a 2 + b 2 [From the definition of modulus.] ⇒ 5 = a 2 + b 2 ∵ z = 5 Squaring both sides, we get: a 2 + b 2 = 5 2 = 25 . . . 1 And, tan ⁡ π 6 = b a ∵ tan α = y x , for z = x + i y and α = acute angle ⇒ 3 3 = b a ⇒ a 3 = 3 b ⇒ a = 3 b . . . 2 Putting equation 2 in equation 1 , we get: ( 3 b ) 2 + b 2 = 25 ⇒ 4 b 2 = 25 ⇒ b 2 = 25 4 ⇒ b = ± 25 4 = ± 5 2 But arg ⁡ z > 0 ⇒ b > 0 Thus, b = 5 2 So, a = 3 b = 5 3 2 .

If z = r ( cos θ + i sin θ ) . Show that z - 1 z = i ( 2 r sin θ ) .

We have, z = r ( cos ⁡ θ + i sin ⁡ θ ) = r e i θ Then, 1 z = 1 × z ¯ z × z ¯ = z ¯ z 2 = r e - i θ cos 2 θ + sin 2 θ 2 = r e - i θ ∵ cos 2 A + sin 2 A = 1 Now Consider, z - 1 z = r e i θ - r e - i θ = r e i θ - e - i θ = r cos θ + i sin θ - cos θ - i sin θ = r 2 i sin θ ∴ z - 1 z = i 2 r sin θ Hence proved

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 11 - Complex Numbers>EXERCISE 11C

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Complex Numbers, Exercise 5: EXERCISE 11C

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Complex Numbers, Exercise 5: EXERCISE 11C

Attempt the free practice questions on Chapter 11: Complex Numbers, Exercise 5: EXERCISE 11C with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.