Complex Numbers: De Moivre's Theorem & JEE Main Guide

Complex numbers is a chapter with two distinct halves in JEE Main. The first half is algebraic: operations with complex numbers, conjugates, modulus, argument, polar form. The second half is geometric: Argand plane, loci, De Moivre's theorem, and roots of unity. Both halves are tested in every session, typically in 2-3 questions, and the chapter has the rare property that a deep understanding of one half directly enhances the other. This guide builds both.

Fundamentals: Modulus, Argument, and Polar Form

For z = a + bi, modulus |z| = √(a² + b²), argument arg(z) = arctan(b/a) adjusted to the correct quadrant (principal argument ∈ (−π, π]). Polar form: z = r(cosθ + i sinθ) = re^{iθ} (Euler's form). Key properties: |z₁z₂| = |z₁||z₂|, arg(z₁z₂) = arg(z₁) + arg(z₂); |z₁/z₂| = |z₁|/|z₂|, arg(z₁/z₂) = arg(z₁) − arg(z₂). These multiplicative properties of modulus and argument are why polar/Euler form is superior to Cartesian form for products and quotients.

Conjugate: z̄ = a − bi. Properties: z·z̄ = |z|², z + z̄ = 2Re(z), z − z̄ = 2i·Im(z). The identity 1/z = z̄/|z|² lets you rationalise complex fractions by multiplying numerator and denominator by the conjugate of the denominator. JEE uses these properties to simplify expressions like (a+bi)/(c+di) and to prove locus conditions by manipulating |z − z₁| and |z − z₂|. Take a free complex numbers mock to practise modulus-argument calculations.

De Moivre's Theorem

De Moivre's theorem: (cosθ + i sinθ)ⁿ = cos(nθ) + i sin(nθ) for all rational n. In Euler notation: (e^{iθ})ⁿ = e^{inθ}. Applications: (1) Computing high powers of complex numbers — write in polar form and multiply the argument by the power; (2) Finding multiple-angle trigonometric identities — expand (cosθ + i sinθ)ⁿ by the binomial theorem, then equate real and imaginary parts; (3) Finding nth roots — z^(1/n) has n distinct values, equally spaced in argument on a circle of radius |z|^(1/n).

The most tested application: find all values of (a+bi)^(1/n). Step 1: convert a+bi to polar form r·e^{iθ}. Step 2: the nth roots are r^(1/n) · e^{i(θ + 2kπ)/n} for k = 0, 1, ..., n−1. These n roots are equally spaced on a circle of radius r^(1/n), separated by argument 2π/n. The geometric picture — equally spaced points on a circle — is the key insight for roots-of-unity problems.

Roots of Unity and Their Properties

The nth roots of unity satisfy zⁿ = 1. In polar form, z = e^{2πik/n} for k = 0, 1, ..., n−1. The cube roots of unity are particularly important: 1, ω = e^{2πi/3}, ω² = e^{4πi/3}, where ω satisfies 1 + ω + ω² = 0 and ω³ = 1. The key algebraic properties: if ω is a primitive nth root, then 1 + ω + ω² + ... + ω^{n−1} = 0 (sum of all nth roots = 0) and the product of all nth roots = (−1)^{n+1} (last coefficient of zⁿ − 1 = 0).

JEE uses cube roots most frequently: (1) simplifying expressions like 1/(1+ω) + 1/(1+ω²); (2) proving divisibility conditions (a sum of the form aⁿ + bⁿ + cⁿ is divisible by a+b+c if 1, ω, ω² are roots); (3) finding the value of complex geometric sums. Locus problems on the Argand plane: |z − z₁| = k|z − z₂| is a circle (Apollonius circle) when k ≠ 1, or a perpendicular bisector when k = 1. These locus results connect directly to the coordinate geometry in our circle guide. For the algebraic connections, see our complex numbers and geometry guide.