Edexcel Further Pure 2

1. 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 = reiϴ. apply de Moivre’s theorem to find trigonometric identities To find the nth roots of a complex number.

2. 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(cosϴ + i sin ϴ).

3. Chapter 3 – Further Complex Numbers Consider the function eiθ Remember Maclaurin/Taylor’s Comparing with the other series expansions – we see that we have cos and sin This is called Euler’s relation

4. Chapter 3 – Further Complex Numbers m Complex numbers written in the modulus argument form R(cos + i sin ) can now be rewritten using Euler’s relation as Rei 5i 4i 3i 2i i Z Therefore z = 5eiπ/3 represents the complex number with modulus 5 and argument = π/3 -5 -4 -3 -2 -1 0 1 2 3 4 5 e -i -2i -3i -4i -5i

5. Chapter 3 – Further Complex Numbers

6. Chapter 3 – Further Complex Numbers: Proof  when n = 0  (cos θ + i sin θ)0 = cos 0 + i sin 0 (cosθ + i sin θ)n = cos nθ + i sin nθ true for when n < 0  (cos θ + i sin θ)n = (cos θ + i sin θ)-m where n = – m, m > 0 1 1 = = = cos mθ + i sin mθ (cos θ + i sin θ)m (cos mθ + i sin mθ)(cos mθ – i sin mθ) using De Moivre’s theorem Rationalising the denominator cos mθ – i sin mθ = cos mθ – i sin mθ = cos (– mθ) + i sin (– mθ) = cos (nθ) + i sin (nθ)

7. Exercise 3C, Page 31 • Use de Moivre’s theorem to simplify question 1.

8. Chapter 3 – Further Complex Numbers: Binomial You need to be able to apply the following binomial expansion found in C2:

9. Chapter 3 – Further Complex Numbers: Binomial Use De Moivre’s theorem to show that cos 4θ = cos4θ – 6cos2θ sin²θ + sin4θ and sin 4θ = 4cos3θ sin θ – 4cosθ sin3θ cos 4θ + i sin 4θ = (cos θ + i sin θ)4 = cos4θ + 4 cos3θ (i sin θ) + 6 cos2θ (i sin θ)² + 4 cosθ (i sin θ)3 + (i sin θ)4 = cos4θ + i 4 cos3θ sin θ – 6 cos2θ sin²θ – i 4 cosθ sin3θ + sin4θ = cos4θ – 6 cos2θ sin²θ + sin4θ + i (4 cos3θ sin θ – 4 cosθ sin3θ) Comparing real parts cos 4θ = cos4θ – 6 cos2θ sin²θ + sin4θ Comparing imaginary parts sin 4θ = 4 cos3θ sin θ – 4 cosθ sin3θ

10. Chapter 3 – Further Complex Numbers Now z = cos θ + i sin θ = eiθ and so z-1 = cos θ – i sin θ = e-iθ By alternately adding and subtracting these equation we obtain cos θ = 1/2(eiθ+ e-iθ) sin θ = 1/2i (eiθ– e-iθ) z + 1/z = 2 cos θ and also (z + 1/z)n = 2 cos nθ z - 1/z = 2i sin θ and also (z - 1/z)n = 2i sin nθ

11. Chapter 3 – Further Complex Numbers

12. Exercise 3D, Page 36 • Answer the following questions: Questions 1, 2, 3, 4 and 5. • Extension Task: Question 7.

13. Chapter 3 – Further Complex Numbers: Nth roots Solve the equation z3 = 1

14. Chapter 3 – Further Complex Numbers: Nth roots

15. Chapter 3 – Further Complex Numbers: Nth roots

16. Exam Questions 1. (a) Express 32 cos 6θ in the form p cos 6θ + q cos 4θ + r cos 2θ + s, where p, q, r and s are integers. (5) (b) Show that sin 5q = (sin 5q – 5 sin 3q + 10 sin q ). (5) (Total 10 marks)

17. Exam Answers 1. (a) M1 = z6 + z–6 + 6(z4 + z–4) + 15(z2 + z–2) + 20 M1 64cos6 ϴ = 2 cos 6ϴ + 12 cos 4ϴ + 30 cos 2ϴ + 20 M1 32 cos6ϴ = cos 6ϴ + 6 cos 4ϴ + 15 cos 2ϴ + 10 A1, A1 (p = 1, q = 6, r = 15, s = 10) A1 any two correct (5)

18. Exam Answers (b) = z5 – 5z3 + 10z – M1, A1 = (2i sin ϴ)5 = 32i sin5ϴ = 2i sin 5ϴ– 10i sin 3ϴ+ 20i sin ϴM1, A1 sin5ϴ = (sin 5ϴ– 5sin 3ϴ+ 10 sin ϴ) A1 (Total 10 marks)