Topic 7 of 8 · Form 6
Complex numbers and de Moivre's theorem
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ, for any integer n (proved by induction for positive n).
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de Moivre's theorem
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ, for any integer n (proved by induction for positive n).
Exponential form: z = re^(iθ); eⁱᶿ = cos θ + i sin θ.
Multiple angles
Worked example: Express cos 3θ in terms of cos θ.
(c + is)³ = c³ + 3ic²s − 3cs² − is³; real part: cos 3θ = c³ − 3cs² = c³ − 3c(1 − c²) = 4cos³θ − 3cos θ
Powers of sine and cosine
Using z + 1/z = 2cos θ and z − 1/z = 2i sin θ: e.g. cos³θ = ¼(cos 3θ + 3cos θ), useful in integration.
nth roots
zⁿ = re^(iθ) has n roots: r^(1/n) e^(i(θ + 2kπ)/n), k = 0, 1, …, n − 1, equally spaced on a circle in the Argand diagram.
Worked example: Solve z³ = 8.
z = 2e^(2kπi/3): z = 2, 2e^(2πi/3) = −1 + i√3, 2e^(−2πi/3) = −1 − i√3
Roots of unity
The n nth roots of unity sum to zero; the cube roots are 1, ω, ω² with 1 + ω + ω² = 0.
Summing series
Use C + iS with geometric series of complex exponentials to sum series such as Σ cos rθ.
Check yourself
- Simplify (cos π/6 + i sin π/6)⁶.
- What is the sum of the fifth roots of unity?
- Express sin 2θ using de Moivre's theorem.
Show answers
- cos π + i sin π = −1
- 0
- Imaginary part of (c + is)² = 2 sin θ cos θ
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