Complex Numbers and Quadratic Equations

🎯 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).

Quadratic Equations

NCERT Chapter 4 pairs complex numbers with quadratic equations: once i is available, every quadratic has roots.

🎯 Key Points

• Quadratic formula: x = (-b ± √(b²-4ac)) / 2a — works for ANY quadratic, even when factorization fails
• Discriminant D = b²-4ac: D>0 → 2 distinct real roots; D=0 → 1 repeated root; D<0 → 2 complex conjugate roots
• Vieta's formulas: sum of roots α+β = -b/a, product αβ = c/a — lets you build equations from given roots without solving
• Vertex (minimum if a>0, maximum if a<0) occurs at x = -b/2a; irrational roots always come in conjugate pairs (p+√q and p-√q)

Quadratic Equations

A quadratic equation is of the form ax² + bx + c = 0 (where a ≠ 0). The solutions are called roots.

Methods of Solving

• Factorization: Write as (x - p)(x - q) = 0 and read off p, q as roots. Works when roots are rational.
• Quadratic formula: x = (-b ± √(b²-4ac)) / 2a: works always.
• Completing the square: Rearrange to (x + b/2a)² = (b²-4ac)/4a².

Discriminant (D = b² - 4ac)

• D > 0 : Two distinct real roots
• D = 0 : Two equal real roots (repeated root)
• D < 0 : No real roots (complex roots)

Vieta's Formulas

For roots α and β: α + β = -b/a and αβ = c/a. Use these to form equations given roots.

Nature of Roots

For rational roots, D must be a perfect square. If one root is irrational (like 2+√3), the other is its conjugate (2-√3).

Key Facts

• Maximum/minimum of ax² + bx + c is at x = -b/2a
• Sum of squares of roots: α² + β² = (α+β)² - 2αβ
• Every quadratic has exactly 2 roots (may be complex)

Forming a Quadratic Equation from Its Roots

If a quadratic has roots α and β, it can be reconstructed as x² − (α + β)x + αβ = 0, i.e. x² − (sum of roots)x + (product of roots) = 0. Any non-zero multiple k[x² − (α+β)x + αβ] = 0 is the same equation.

• Roots 2 and 5 → sum = 7, product = 10 → x² − 7x + 10 = 0.
• Roots 3 + √2 and 3 − √2 → sum = 6, product = 9 − 2 = 7 → x² − 6x + 7 = 0.
• To build an equation whose roots are related to α, β (say 1/α, 1/β or α², β²), first compute the new sum and product from Vieta's formulas, then plug into x² − (sum)x + product = 0.

Transforming Roots

• Reciprocal roots (1/α, 1/β): new sum = (α+β)/αβ, new product = 1/αβ. Shortcut: reverse the coefficients, so ax²+bx+c → cx²+bx+a.
• Roots increased by h (α+h, β+h): replace x by (x−h) in the original equation.
• Roots scaled by k (kα, kβ): replace x by x/k.

Symmetric Functions of the Roots

Expressions unchanged when α and β are swapped can be written using only the sum s = α+β = −b/a and product p = αβ = c/a, so the roots need never be found individually:

• α² + β² = s² − 2p
• (α − β)² = s² − 4p = D/a²  →  |α − β| = √D / |a|
• α³ + β³ = s³ − 3ps = s(s² − 3p)
• α² β + α β² = αβ(α + β) = ps
• 1/α + 1/β = s/p,   1/α² + 1/β² = (s² − 2p)/p²

Sign of a Quadratic Expression

The sign of f(x) = ax² + bx + c across the real line depends on the discriminant D and the leading coefficient a:

• D < 0: f(x) never changes sign — it keeps the sign of a for every real x (always positive if a>0, always negative if a<0).
• D = 0: f(x) keeps the sign of a everywhere except at the single repeated root x = −b/2a, where it is 0.
• D > 0: with real roots α < β, f(x) has the sign of a outside [α, β] and the opposite sign of a between the roots (α < x < β).

This is the basis of the "same sign as a outside the roots, opposite between the roots" rule used to solve quadratic inequalities like ax² + bx + c > 0.

Maximum and Minimum Value of a Quadratic

Since f(x) = ax² + bx + c = a(x + b/2a)² + (4ac − b²)/4a, the extreme value occurs at the vertex x = −b/2a and equals (4ac − b²)/4a = −D/4a.

• If a > 0 the parabola opens upward, so this is the minimum value and the range is [−D/4a, ∞).
• If a < 0 the parabola opens downward, so this is the maximum value and the range is (−∞, −D/4a].
• Example: for f(x) = 2x² − 8x + 3, x = −(−8)/(2·2) = 2 and minimum value = f(2) = 8 − 16 + 3 = −5.

🚀 JEE Advanced Edge

Common roots condition: Two quadratics a₁x²+b₁x+c₁=0 and a₂x²+b₂x+c₂=0 share BOTH roots only if their coefficients are proportional (a₁/a₂ = b₁/b₂ = c₁/c₂). If they share exactly ONE common root α, eliminating x² between the two equations gives α = (b₁c₂-b₂c₁)/(a₂b₁-a₁b₂) — a frequently tested elimination technique that avoids solving either quadratic directly.

Location of roots relative to a number: For f(x) = ax²+bx+c with a>0, both roots exceed a value k exactly when THREE conditions hold simultaneously: D≥0, f(k)>0, and the vertex x-coordinate -b/2a > k. Checking only the discriminant (a common mistake) is insufficient — all three conditions are needed together.

Worked problem: Find the value of k for which x²-(k+1)x+k²+k-8=0 has roots that are reciprocals of each other (i.e. product of roots = 1). Approach: Product of roots = c/a = k²+k-8. Setting this equal to 1: k²+k-8=1, so k²+k-9=0, giving k = (-1±√37)/2.

Worked Example: Sum and Product of Roots

If α and β are the roots of 2x² − 5x + 3 = 0, find α² + β² and α³ + β³.

By Vieta's formulas: α + β = 5/2, αβ = 3/2.

α² + β² = (α + β)² − 2αβ = 25/4 − 3 = 13/4.

α³ + β³ = (α + β)(α² − αβ + β²) = (5/2)(13/4 − 3/2) = (5/2)(7/4) = 35/8. Vieta's formulas eliminate the need to find α and β individually — use them whenever asked for symmetric expressions.

Worked Example: Equation Reducible to Quadratic

Solve x^(2/3) − 9x^(1/3) + 8 = 0.

Substitute t = x^(1/3): t² − 9t + 8 = 0 → (t − 1)(t − 8) = 0 → t = 1 or t = 8.

Reverse: x^(1/3) = 1 → x = 1; x^(1/3) = 8 → x = 512. Recognising a quadratic substitution is the key skill for this question type.