Complex Numbers

🎯 Key Points

• Powers of i cycle every 4: i¹=i, i²=-1, i³=-i, i⁴=1 — to simplify iⁿ, just use the remainder of n÷4
• z·z̄ = |z|² (multiplying by conjugate eliminates i, the basis of complex division)
• De Moivre's theorem (cosθ+isinθ)ⁿ = cos(nθ)+isin(nθ) makes high powers/roots of complex numbers easy via polar form, instead of repeated multiplication
• Cube roots of unity: 1, ω, ω² with 1+ω+ω²=0 and ω³=1 — these two identities solve almost every cube-root-of-unity simplification problem

Complex Numbers

A complex number is of the form z = a + ib where a, b ∈ R and i = √(−1). Here a is the real part and b is the imaginary part.

Basic Definitions

• i = √(−1), i² = −1, i³ = −i, i⁴ = 1 (cycle of period 4)
• Conjugate: z̄ = a − ib; z × z̄ = a² + b² = |z|²
• Modulus: |z| = √(a² + b²) ≥ 0
• Argument (arg z): θ = tan⁻¹(b/a), the angle z makes with positive x-axis

Algebraic Operations

• Addition: (a+ib) + (c+id) = (a+c) + i(b+d)
• Multiplication: (a+ib)(c+id) = (ac−bd) + i(ad+bc)
• Division: z₁/z₂ = (z₁ × z̄₂) / |z₂|²
• |z₁ z₂| = |z₁||z₂|; arg(z₁z₂) = arg z₁ + arg z₂

Polar and Exponential Form

• Polar: z = r(cos θ + i sin θ), where r = |z|, θ = arg z
• Euler form: z = re^(iθ)

De Moivre Theorem

(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. Used to find powers and roots of complex numbers.

Cube Roots of Unity

• The three cube roots of 1 are: 1, ω, ω² where ω = (−1 + i√3)/2
• Key identities: 1 + ω + ω² = 0; ω³ = 1

Argand Plane (Locus)

Complex numbers are plotted as points (a, b) in the Argand plane. |z − z₀| = r represents a circle; Re(z) = k is a vertical line; Im(z) = k is a horizontal line.

Worked Example: Modulus, Argument, and Polar Form

Let z = 1 + i. Then |z| = sqrt(1^2 + 1^2) = sqrt(2), and arg(z) = tan^-1(1/1) = 45° = π/4 (since z lies in the first quadrant).

Polar form: z = sqrt(2)(cos45° + i sin45°), and the Euler form is z = sqrt(2) e^(iπ/4).

Worked Example: De Moivre Theorem Application

Find (1 + i)^8 using De Moivre theorem.

Using the polar form above: (1+i)^8 = [sqrt(2)]^8 [cos(8 x 45°) + i sin(8 x 45°)] = 16 [cos360° + i sin360°] = 16(1 + 0i) = 16.

This can be checked directly: (1+i)^2 = 2i, so (1+i)^4 = (2i)^2 = -4, and (1+i)^8 = (-4)^2 = 16, confirming the result. De Moivre theorem is especially powerful for high powers where direct multiplication would be tedious.

Cube Roots of Unity: Properties

• The cube roots of unity are 1, ω, ω^2, where ω = (-1 + i√3)/2 and ω^2 = (-1 - i√3)/2 = ω̄ (the conjugate of ω).
• Key identities: 1 + ω + ω^2 = 0 and ω^3 = 1, used repeatedly to simplify higher powers (e.g. ω^7 = ω^(6+1) = (ω^3)^2 x ω = ω).
• ω and ω^2 are also the roots of x^2 + x + 1 = 0, the factor obtained after dividing x^3 - 1 by (x - 1).
• Worked example: Simplify (1 + ω - ω^2)(1 - ω + ω^2). Since 1 + ω + ω^2 = 0, we get 1 + ω = -ω^2 and 1 + ω^2 = -ω. So the product becomes (-ω^2 - ω^2)(-ω - ω) = (-2ω^2)(-2ω) = 4ω^3 = 4 x 1 = 4.

Conjugate Properties and Equations

• (z1 + z2)bar = z1bar + z2bar; (z1 z2)bar = z1bar x z2bar; (z1/z2)bar = z1bar/z2bar (z2 ≠ 0)
• z + zbar = 2 Re(z) (always real); z - zbar = 2i Im(z) (always purely imaginary)
• z is purely real if z = zbar; z is purely imaginary if z + zbar = 0
• For a quadratic ax^2 + bx + c = 0 with real coefficients and negative discriminant, the two complex roots are always conjugates of each other.

Modulus Properties

• |z| ≥ 0, and |z| = 0 ⇔ z = 0; |z| = |z̄| = |−z|.
• |z₁ z₂| = |z₁||z₂|; |z₁/z₂| = |z₁|/|z₂| (z₂ ≠ 0); |zⁿ| = |z|ⁿ.
• z·z̄ = |z|² and |z|² = (Re z)² + (Im z)².
• Triangle inequality: |z₁ + z₂| ≤ |z₁| + |z₂|, with equality when z₁, z₂ have the same argument.
• ||z₁| − |z₂|| ≤ |z₁ − z₂| (reverse triangle inequality).

Argument Properties

• arg(z₁ z₂) = arg z₁ + arg z₂; arg(z₁/z₂) = arg z₁ − arg z₂ (added/subtracted modulo 2π).
• arg(z̄) = −arg(z); the principal argument lies in (−π, π].
• Quadrant matters when using θ = tan⁻¹(b/a): add or subtract π when a < 0 to land in the correct quadrant.
• z is purely real ⇔ arg z = 0 or π; z is purely imaginary ⇔ arg z = ±π/2.

Euler Form and Multiplication as Rotation

• Euler form: z = r·e^(iθ) where r = |z|, θ = arg z; this makes products and powers trivial: (r₁e^(iθ₁))(r₂e^(iθ₂)) = r₁r₂ e^(i(θ₁+θ₂)).
• Multiplying z by e^(iα) rotates the vector z by angle α about the origin without changing its length.
• Multiplying by i (= e^(iπ/2)) rotates z by 90° anticlockwise; multiplying by −1 rotates by 180°.
• e^(iπ) + 1 = 0 (Euler's identity), obtained by setting θ = π.

Square Root of a Complex Number

• To find √(a + ib), set √(a + ib) = x + iy, square both sides: x² − y² = a and 2xy = b.
• Solve together with x² + y² = √(a² + b²) = |a + ib| to get x² = (|z| + a)/2 and y² = (|z| − a)/2.
• The sign of y is chosen so that 2xy has the same sign as b; there are two square roots, ±(x + iy).
• Worked example: √(3 + 4i): |z| = 5, so x² = (5+3)/2 = 4 → x = ±2, y² = (5−3)/2 = 1 → y = ±1. Since b = 4 > 0, xy > 0, giving √(3 + 4i) = ±(2 + i).

🚀 JEE Advanced Edge

nth roots of a complex number via polar form: To find all n distinct nth roots of z = r(cosθ+isinθ), use z^(1/n) = r^(1/n)[cos((θ+2kπ)/n) + isin((θ+2kπ)/n)] for k = 0,1,...,n-1 — the "+2kπ" term is essential because complex numbers have infinitely many equivalent angle representations, and stepping k through n values generates all n roots, evenly spaced around a circle of radius r^(1/n).

|z₁+z₂|² identity for geometry problems: |z₁+z₂|² = |z₁|²+|z₂|²+2Re(z₁z̄₂) — this expands the modulus of a sum without needing to convert to a+ib form first, and is the standard tool for proving locus/triangle-inequality-style complex number problems directly in modulus-argument language.

Worked problem: Find all cube roots of z = 8 (i.e. solve x³ = 8). Approach: Write 8 = 8(cos0° + isin0°). Cube roots: x = 8^(1/3)[cos(0+2kπ)/3 + isin(0+2kπ)/3] for k=0,1,2. k=0: x=2(cos0+isin0)=2. k=1: x=2(cos120°+isin120°)=2(-1/2+i√3/2)=-1+i√3. k=2: x=2(cos240°+isin240°)=-1-i√3. The three roots are 2, -1+i√3, -1-i√3 (i.e. 2, 2ω, 2ω²).