De Moivre's Theorem

Here |z| = √(a²+b²) and sgn(b) = +1 if b ≥ 0, −1 if b < 0. The two square roots are always negatives of each other. Derived by setting (x+iy)² = a+ib and solving the system x²−y²=a, 2xy=b using x²+y²=|z|.

For any integer n, raising cis θ to the power n simply multiplies the angle. In Euler's form: (e^(iθ))^n = e^(inθ) — the result is immediate. Proof for positive integers uses induction; extends to negative integers via (cis θ)^(−1) = cis(−θ).

For rational exponent p/q (in lowest terms), there are exactly q distinct values, obtained by taking k = 0, 1, 2, …, q−1. Unlike the integer case, the result is not unique — rational powers of a complex number are multi-valued.

Every non-zero complex number z = r·cis θ has exactly n distinct nth roots. They lie on a circle of radius r^(1/n) centred at the origin, spaced equally at angular intervals of 2π/n — forming a regular n-gon. In Euler form: r^(1/n) · e^(i(θ+2kπ)/n).

The n solutions of zⁿ = 1. They are equally spaced on the unit circle, forming a regular n-gon with one vertex always at (1, 0). If ω = e^(2πi/n) is a primitive nth root, then the complete set is {1, ω, ω², …, ωⁿ⁻¹}.

The sum of all nth roots of unity is always zero — geometrically, the n equally spaced vectors cancel. The product is (−1)^(n+1): it equals +1 for odd n and −1 for even n. Both follow from Vieta's formulas applied to zⁿ − 1 = 0.

The three solutions of z³ = 1. In Euler form: 1 = e^(0), ω = e^(2πi/3), ω² = e^(4πi/3). They sit at the vertices of an equilateral triangle on the unit circle. ω and ω² are complex conjugates of each other.

Three identities memorised together. ω³ = 1 is the defining relation. 1+ω+ω² = 0 follows from the sum of cube roots of unity (cn32 with n=3) — it is the most-used identity in JEE. ω² = ω̄ since ω and ω² are conjugates.

The symmetric factorization of a³+b³+c³−3abc over ℂ. When a+b+c = 0, the identity gives a³+b³+c³ = 3abc directly — a classic JEE trick. Also: a²+b²+c²−ab−bc−ca = (a+ωb+ω²c)(a+ω²b+ωc).

Expand the RHS using the binomial theorem, then separate real and imaginary parts. Real part gives cos nθ as a polynomial in cos θ; imaginary part gives sin nθ. For example, cos 3θ = 4cos³θ − 3cos θ and sin 3θ = 3sin θ − 4sin³θ both fall out of (cis θ)³.