Edexcel Further Pure 2

Chapter 3 – Further Complex Numbers:

• Write down a complex number, z, in modulus-argument form as either z=r(cos ϴ + isin ϴ ) or z = re i ϴ . apply de Moivre’s theorem • to find trigonometric identities • To find the nth roots of a complex number.

Chapter 3 – Further Complex Numbers: FP1 Recap

In FP1, we have calculated the modulus and argument, now we are going to use this in the form z=r(cos ϴ + i sin ϴ ).

Chapter 3 – Further Complex Numbers

Consider the function e iθ Remember

Maclaurin/Taylor’s

Comparing with the other series expansions – we see that we have cos and sin This is called Euler’s relation

Chapter 3 – Further Complex Numbers

Complex numbers written in the modulus argument form R(cos  + i sin  ) can now be rewritten using Euler’s relation as

Re

i   m 5i Z Therefore

z = 5e

iπ/3 represents the complex number with modulus 5 and argument = π / 3 4i 3i 2i -5 -4 -3 -2 -1 0 1 2 3 4 5 -i i  e -2i -3i -4i -5i

Chapter 3 – Further Complex Numbers

Chapter 3 – Further Complex Numbers: Proof

when n = 0  (cos θ + i sin θ) 0 = cos 0 + i sin 0 when n < 0  = = (cos θ + i sin θ) 1 (cos θ + i sin θ) m 1 cos mθ + i sin mθ n = (cos θ + i sin θ) -m where n = – m, m > 0 using De Moivre’s theorem  = cos mθ – i sin mθ (cos mθ + i sin mθ)(cos mθ – i sin mθ) Rationalising the denominator = cos mθ – i sin mθ = cos (– mθ) + i sin (– mθ) = cos (nθ) + i sin (nθ) (cos θ + i sin θ) n = cos nθ + i sin nθ true for

Exercise 3C, Page 31



Use de Moivre’s theorem to simplify question 1.

Chapter 3 – Further Complex Numbers: Binomial

You need to be able to apply the following binomial expansion found in C2:

Chapter 3 – Further Complex Numbers: Binomial

Use De Moivre’s theorem to show that cos 4 θ = cos 4 θ – 6cos 2 θ sin² θ + sin 4 θ and sin 4 θ = 4cos 3 θ sin θ – 4cos θ sin 3 θ cos 4 θ + i sin 4 θ = (cos θ + i sin θ ) 4 = cos 4 θ + 4 cos 4 cos θ 3 θ (i sin θ ) + 6 cos 2 θ (i sin θ ) 3 + (i sin θ ) 4 (i sin θ )² + = cos 4 θ + i 4 cos 3 θ i 4 cos θ sin 3 sin θ θ – 6 cos + sin 4 θ 2 θ sin² θ – = cos 4 θ – 6 cos 2 θ i (4 cos 3 θ sin² θ sin θ + sin 4 θ + – 4 cos θ sin 3 θ ) Comparing real parts cos 4 θ = cos 4 θ – 6 cos 2 θ Comparing imaginary parts sin 4 θ = 4 cos 3 θ sin θ sin² – 4 cos θ θ + sin 4 θ sin 3 θ

Now

Chapter 3 – Further Complex Numbers

z = cos θ + i sin θ = e iθ and so z -1 = cos θ – i sin θ = e -iθ By alternately adding and subtracting these equation we obtain z + 1/z = 2 cos θ z - 1/z = 2i sin θ and also (z + 1/z) n = 2 cos nθ and also (z - 1/z) n = 2i sin nθ cos θ = 1 / 2 (e iθ + e -iθ ) sin θ = 1 / 2i (e iθ – e -iθ )

Chapter 3 – Further Complex Numbers

Exercise 3D, Page 36



Answer the following questions:

Questions 1, 2, 3, 4 and 5.



Extension Task:

Question 7.

Chapter 3 – Further Complex Numbers: Nth roots

Solve the equation z

3

= 1

Chapter 3 – Further Complex Numbers: Nth roots

Chapter 3 – Further Complex Numbers: Nth roots

Exam Questions

1.

(a) Express 32 cos 6

θ

r and s are integers.

in the form p cos 6

θ

+ q cos 4

θ

+ r cos 2

θ

+ s, where p, q,

(5)

(b) Show that sin 5q = (sin 5q – 5 sin 3q + 10 sin q ). (5)

(Total 10 marks)

Exam Answers

1.

(a) 1

z

6 

z

6  6

z

4  15

z

2  20  15

z

 2  6

z

 4 

z

 6 = z 6 + z –6 + 6(z 4 + z –4 ) + 15(z 2 + z –2 ) + 20 64cos 6

ϴ

= 2 cos 6

ϴ

+ 12 cos 4

ϴ

+ 30 cos 2

ϴ

+ 20 32 cos 6

ϴ

= cos 6

ϴ

+ 6 cos 4

ϴ

(p = 1, q = 6, r = 15, s = 10) + 15 cos 2

ϴ

+ 10 A1 any two correct

M1 M1 M1 A1, A1 (5)

(b)

Exam Answers

    1

z

   5 = z 5 – 5z 3 + 10z – 10 

z

5

z

3  1

z

5

M1, A1

=

z

5  1

z

5  5

z

3  1

z

3  10  1

z

(2i sin

ϴ

) 5 = 32i sin 5

ϴ

= 2i sin 5

ϴ

– 10i sin 3

ϴ

+ 20i sin

ϴ

sin 5

ϴ

= (sin 5

ϴ

– 5sin 3

ϴ

+ 10 sin

ϴ

)

M1, A1 A1 (Total 10 marks)