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De Moivre’s theorem, n-th roots and square roots of complex numbers

الأسبوع 6 · نظرية ديموافر والجذور النونية للعدد المركب والجذران التربيعيان والمعادلات التربيعية ذات المعاملات المركبة · 3 دروس
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De Moivre’s theorem: multiple angles in terms of powers of sine and cosine

Pure Mathematics Sec 3 Term 1, Unit 2, Lesson 2
GoalUse De Moivre’s theorem to find powers of complex numbers and to write cos nθ and sin nθ with powers of cos θ and sin θ.
New words
De Moivre’s theorem · (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ for n = 1, 2, 3, …multiple angle · an angle such as 2θ, 3θ or 4θ, a whole-number multiple of θequal complex numbers · two complex numbers are equal when their real parts are equal and their imaginary parts are equal
Explanation

Last week you saw that multiplying two complex numbers in trigonometric form multiplies the moduli and adds the amplitudes. Multiply a number by itself n times and the modulus is raised to the power n, while the amplitude is added n times. For a number on the unit circle this is De Moivre’s theorem (in Arabic نظرية ديموافر): (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ for n = 1, 2, 3, … It gives powers at once, with no repeated multiplication. If the number is not yet in trigonometric form, find its modulus r and amplitude θ first; then the power is rⁿ(cos nθ + i sin nθ), and any amplitude beyond 360° can be reduced by subtracting full turns. The theorem has a second use. The left side can also be expanded with the binomial theorem of Unit 1, remembering i² = −1, i³ = −i and i⁴ = 1. Both expansions are the same complex number, so their real parts agree and their imaginary parts agree. The real parts give cos nθ and the imaginary parts give sin nθ, written with powers of cos θ and sin θ. To keep lines short, write c for cos θ and s for sin θ. For n = 2: (c + is)² = c² − s² + 2cs·i, so cos 2θ = c² − s² and sin 2θ = 2sc, formulas you already know. At the end use s² + c² = 1 to remove one of the two letters if you want only one.

xy−11−110z (30°)z² (60°)z³ (90°)
Worked examples
Find (−1 + √3 i)⁵. The modulus is √(1 + 3) = 2. The point lies in the second quadrant and its amplitude is 120°. So the power is 2⁵(cos 600° + i sin 600°). Since 600° = 360° + 240°, this is 32(cos 240° + i sin 240°) = 32(−1/2 − (√3/2)i) = −16 − 16√3 i.
Express sin 4θ and cos 4θ with c and s. Expand: (c + is)⁴ = c⁴ + 4c³(is) + 6c²(is)² + 4c(is)³ + (is)⁴ = (c⁴ − 6c²s² + s⁴) + i(4c³s − 4cs³). By De Moivre this equals cos 4θ + i sin 4θ. Imaginary parts: sin 4θ = 4c³s − 4cs³. Real parts: cos 4θ = c⁴ − 6c²s² + s⁴. Check with θ = 30°, where sin 120° = √3/2: the right side is 4(√3/2)³(1/2) − 4(√3/2)(1/2)³ = 3√3/4 − √3/4 = √3/2. It agrees.
Find the real part of ((1 + i)/√2)³⁰. Since (1 + i)/√2 = cos 45° + i sin 45°, the power is cos 1350° + i sin 1350°. Now 1350° = 3 × 360° + 270°, so it equals cos 270° + i sin 270° = −i. The real part is 0.
Class activity

Check by hand. Let z = cos 60° + i sin 60° = 1/2 + (√3/2)i. Multiply z by itself with ordinary algebra to get z² and then z³, and compare z³ with cos 180° + i sin 180° = −1.

Practice
1
Write (cos 10° + i sin 10°)¹² in the form a + bi.
2
Find (1 − i)¹⁰. First write 1 − i with its modulus and amplitude.
3
Expand (c + is)³ and read off cos 3θ and sin 3θ in terms of c and s. Then use s² = 1 − c² and c² = 1 − s² to get one-letter forms.
4
From Example 2, cos 4θ = c⁴ − 6c²s² + s⁴. Write cos 4θ with c only.
5
Let z = cos 36° + i sin 36°. Find z⁵ and z¹⁰.