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Chapter 3 – Further Complex Numbers: Write down a complex number, z, in modulus-argument form as either z=r(cos + isin ) or z = re i. apply de Moivre’s theorem to find trigonometric identities To find the nth roots of a complex number. Edexcel Further Pure 2

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In FP1, we have calculated the modulus and argument, now we are going to use this in the form z=r(cos + i sin ). Chapter 3 – Further Complex Numbers: FP1 Recap

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

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

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

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

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

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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 3 θ (i sin θ ) + 6 cos 2 θ (i sin θ )² + 4 cos θ (i sin θ ) 3 + (i sin θ ) 4 = cos 4 θ + i 4 cos 3 θ sin θ – 6 cos 2 θ sin² θ – i 4 cos θ sin 3 θ + sin 4 θ = cos 4 θ – 6 cos 2 θ sin² θ + sin 4 θ + i (4 cos 3 θ sin θ – 4 cos θ sin 3 θ ) Comparing real parts cos 4 θ = cos 4 θ – 6 cos 2 θ sin² θ + sin 4 θ Comparing imaginary parts sin 4 θ = 4 cos 3 θ sin θ – 4 cos θ sin 3 θ

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cos θ = 1 / 2 (e iθ + e -iθ ) sin θ = 1 / 2i (e iθ – e -iθ ) Now 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 θand also (z + 1/z) n = 2 cos nθ z - 1/z = 2i sin θ and also (z - 1/z) n = 2i sin nθ Chapter 3 – Further Complex Numbers

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

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Chapter 3 – Further Complex Numbers: Nth roots Solve the equation z 3 = 1

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Chapter 3 – Further Complex Numbers: Nth roots

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Exam Questions 1. (a) Express 32 cos 6 θ in the form p cos 6 θ + q cos 4 θ + r cos 2 θ + 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)

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

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Exam Answers (b) = z 5 – 5z 3 + 10z – M1, A1 = (2i sin ) 5 = 32i sin 5 = 2i sin 5 – 10i sin 3 + 20i sin M1, A1 sin 5 = (sin 5 – 5sin 3 + 10 sin ) A1 (Total 10 marks)

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Section 7.7 Previously, when we encountered square roots of negative numbers in solving equations, we would say “no real solution” or “not a real number”.

Section 7.7 Previously, when we encountered square roots of negative numbers in solving equations, we would say “no real solution” or “not a real number”.

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