De Moivre's Theorem and Roots of Unity
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Direct answer
For an integer n, (cos θ + i sin θ)^n = cos nθ + i sin nθ — De Moivre's theorem, the engine behind powers and roots of complex numbers. The n nth-roots of r(cos θ + i sin θ) are r^(1/n)[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)] for k = 0, 1, ..., n − 1, sitting at equal angular spacing 2π/n on a circle of radius r^(1/n). For n = 3 and modulus 1 the roots are 1, ω, ω^2 with ω = −1/2 + i√3/2, obeying the twin identities ω^3 = 1 and 1 + ω + ω^2 = 0 on which JEE algebra leans constantly.
What you must remember
- Cube roots of unity: ω = −1/2 + i√3/2 and ω^2 = −1/2 − i√3/2 are conjugates; ω^3 = 1 and 1 + ω + ω^2 = 0, so (1 + ω)(1 + ω^2) = 1.
- nth roots of unity: n points on the unit circle at arguments 2kπ/n; their sum is 0 and their product is (−1)^(n − 1).
- Root geometry: the n roots form the vertices of a regular n-gon centred at the origin, each a rotation of the next by 2π/n.
- Fractional caution: for non-integer exponents the theorem becomes multi-valued — an integer power has one value, a qth root has q.
- Polar arithmetic: multiplication adds arguments and multiplies moduli; division subtracts arguments — De Moivre is the iterated form of this.
- Factorisation hooks: x^3 − 1 = (x − 1)(x^2 + x + 1), and x^2 + x + 1 = 0 has exactly the roots ω and ω^2.
- Principal argument: θ taken in (−π, π]; answers asking for roots "in order" must respect this branch.
Reducing an omega expression
Evaluate (1 + ω − ω^2)(1 − ω + ω^2). Since 1 + ω = −ω^2, the first bracket is −ω^2 − ω^2 = −2ω^2; since 1 + ω^2 = −ω, the second bracket is −ω − ω = −2ω. The product is 4ω^3 = 4. Converting ω into Cartesian form a + ib would work eventually and waste three lines; the identities do the whole job, which is the point of the chapter — treat ω as a symbol with two reduction rules, not as a decimal.
The same economy scales up. Any power a^9 of a cube root of unity collapses as a^9 = (a^3)^3 = 1; any expression like ω^50 + ω^100 reduces by dividing exponents by 3 and keeping remainders (here ω^2 + ω = −1). And equations of the type x^3 = 8 own the three roots 2, 2ω, 2ω^2 — write all three, because the examiner counts the missed complex pair as the wrong answer.
Where marks leak
JEE Main evaluates directly: a power of a complex number in polar form, a root listing, an omega identity. JEE Advanced prefers sums over roots — α + α^2 + ... + α^n for α a root of unity, or factorisations of x^9 − 1 into cyclotomic pieces. The three recurring losses: quoting (cos θ + i sin θ)^(1/3) = cos(θ/3) + i sin(θ/3) as the only cube root instead of one of three; forgetting k runs 0 to n − 1, which truncates or inflates the root list; and ignoring the principal-argument branch when the angle is given beyond (−π, π]. The modulus-argument to Cartesian conversion errors finish the list — carry √3/2, not 0.866.
Frequently asked questions
What is the value of 1 + ω + ω^2?
Zero — this identity, with ω^3 = 1, reduces every polynomial expression in ω.
What is the product of all nth roots of unity?
(−1)^(n − 1): the product of roots of x^n − 1 = 0 read off Vieta's formulas.
How many distinct cube roots does 8 have?
Three: 2, 2ω and 2ω^2, spaced 120 degrees apart on the circle of radius 2.
Is De Moivre's theorem valid for fractional powers?
Only with care — fractional exponents make it multi-valued, producing several answers that must all be listed.
What is the modulus of ω?
1, like every root of unity; all of them lie on the unit circle.