De Moivre's theorem and roots of unity (IB Math AA HL)
Updated October 2026
Quick answer: De Moivre's theorem: (r cis θ)ⁿ = rⁿ cis nθ. To solve zⁿ = w, write w = r cis(θ + 2πk); the n roots are r1/n cis((θ + 2πk)/n) for k = 0, 1, …, n − 1.
(cosθ + i·sinθ)n = cos(nθ) + i·sin(nθ). Use to find nth roots of unity, expand cos(3θ), or prove trig identities.
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(r cis θ)ⁿ = rⁿ cis nθ: raise the modulus to the power n and multiply the argument by n.
nth roots: write the right-hand side in modulus-argument form with + 2πk, take the nth root of the modulus and divide the argument by n, for k = 0 to n − 1. Then adjust each argument into the range the question asks for.
The n roots lie on a circle of radius r1/n, equally spaced 2π/n apart, and add up to zero (for n ≥ 2).
For multiple-angle identities, expand (cos θ + i sin θ)ⁿ with the binomial theorem and compare real and imaginary parts with cos nθ + i sin nθ.
If |z| = 1, then z + 1/z = 2 cos θ and zⁿ + z⁻ⁿ = 2 cos nθ.
Worked example
Solve z³ = 8i, giving your answers in the form r cis θ with −π < θ ≤ π.
8i = 8 cis(π/2 + 2πk).
z = 81/3 cis((π/2 + 2πk)/3) = 2 cis(π/6 + 2πk/3).
k = 0: 2 cis(π/6). k = 1: 2 cis(5π/6). k = 2: 2 cis(3π/2) = 2 cis(−π/2).
Check the last root: 2 cis(−π/2) = −2i, and (−2i)³ = −8i³ = 8i.
Answer: z = 2 cis(π/6), 2 cis(5π/6) or 2 cis(−π/2).
How do you find the nth roots of a complex number?
Write the number as r cis(θ + 2πk). The roots are r^(1/n) cis((θ + 2πk)/n) for k = 0, 1, …, n − 1. There are always n of them, equally spaced round a circle.
Where can I practise de Moivre's theorem and roots of unity questions?
On the Complex Numbers questions page: 46 exam-style HL AA questions with full mark schemes, plus free step-by-step worked examples.
Can I use my calculator for de Moivre's theorem and roots of unity?
In HL AA, Paper 1 is non-calculator and the other papers allow a GDC, so learn the exact method as well as the calculator one.