Complex Numbers and Quadratic Equations – JEE Main Maths Formula Sheet & Class 11 Notes | All Formulas, Laws & Results

Imaginary Numbers, Powers of Iota, and Complex Number Basics

Why Powers of i and Complex Number Algebra Are Direct JEE Main Formula Questions

Imaginary number i (iota) in complex numbers: i = √(–1), so i² = –1, i³ = –i, i⁴ = 1. This four-cycle repeats: i⁴ⁿ = 1, i⁴ⁿ⁺¹ = i, i⁴ⁿ⁺² = –1, i⁴ⁿ⁺³ = –i for any integer n. To find iⁿ: divide n by 4 and use the remainder r → iⁿ = iʳ. Also: i⁻¹ = –i; i⁻² = –1; i⁻³ = i; i⁻⁴ = 1. The sum of four consecutive powers of i is always zero: iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ = 0. Important: √a · √b = √(ab) only when at least one of a, b is non-negative. If a < 0 and b < 0: √a · √b = i√|a| · i√|b| = –√(|a||b|) ≠ √(ab).

Complex number z = a + ib where a, b ∈ ℝ, i = √–1. Re(z) = a (real part); Im(z) = b (imaginary part). z is purely real if Im(z) = 0; purely imaginary if Re(z) = 0. The number 0 = 0 + 0i is both purely real and purely imaginary. Two complex numbers are equal iff their real parts are equal AND their imaginary parts are equal: a + ib = c + id ⟺ a = c AND b = d. Complex numbers cannot be compared by > or < (ordering is not defined for non-real complex numbers).

Algebra of complex numbers: Addition: (a+ib) + (c+id) = (a+c) + i(b+d). Subtraction: (a+ib) – (c+id) = (a–c) + i(b–d). Multiplication: (a+ib)(c+id) = (ac–bd) + i(ad+bc). Division: (a+ib)/(c+id) = [(ac+bd) + i(bc–ad)] / (c²+d²) — multiply numerator and denominator by the conjugate (c–id). Download the Free PDF for Complex Numbers and Quadratic Equations for all algebra worked examples.

Conjugate, Modulus, and Their Properties — Complex Numbers Formulas

Why Conjugate and Modulus Properties Are the Core Complex Numbers Formulas for JEE Main

Conjugate z̄: for z = a + ib, z̄ = a – ib (change sign of imaginary part, reflect about real axis). Properties of conjugate in complex numbers: (z̄)̄ = z; z = z̄ iff z is purely real; z = –z̄ iff z is purely imaginary; z + z̄ = 2Re(z); z – z̄ = 2i·Im(z); Re(z) = (z + z̄)/2; Im(z) = (z – z̄)/(2i); z₁ + z₂ overline = z̄₁ + z̄₂; z₁ · z₂ overline = z̄₁ · z̄₂; (z₁/z₂) overline = z̄₁/z̄₂; (zⁿ) overline = (z̄)ⁿ; z · z̄ = |z|²; if f(z) is a polynomial with real coefficients then overline of f(z) = f(z̄).

Modulus |z| of complex number z = x + iy: |z| = √(x² + y²) = √(Re(z)² + Im(z)²) = distance from origin in Argand plane. Properties of modulus of complex numbers: |z| ≥ 0; |z| = 0 iff z = 0; |z₁z₂| = |z₁||z₂|; |z₁/z₂| = |z₁|/|z₂| (z₂ ≠ 0); |zⁿ| = |z|ⁿ; |z̄| = |z|; z·z̄ = |z|²; |Re(z)| ≤ |z|; |Im(z)| ≤ |z|; |z₁ ± z₂|² = |z₁|² + |z₂|² ± 2Re(z₁z̄₂). Key result: |az₁ – bz₂|² + |bz₁ + az₂|² = (a² + b²)(|z₁|² + |z₂|²) for a, b ∈ ℝ.

Triangle inequality for complex numbers (JEE Main formula): |z₁ + z₂| ≤ |z₁| + |z₂| (equality iff arg(z₁/z₂) = 0, i.e., z₁, z₂ have the same argument — parallel vectors). |z₁ – z₂| ≥ ||z₁| – |z₂|| (equality iff arg(z₁/z₂) = 0). Combined: ||z₁| – |z₂|| ≤ |z₁ ± z₂| ≤ |z₁| + |z₂|. Parallelogram law: |z₁ + z₂|² + |z₁ – z₂|² = 2(|z₁|² + |z₂|²). If |z| = a then: |z₁| has greatest value = (a + √(a²+4))/2 and least value = (–a + √(a²+4))/2 for the condition |z – 1/z| = a. Download the Free PDF for all Complex Numbers modulus formulas and triangle inequality proofs.

Argument of Complex Number — Principal Argument Formula by Quadrant

Why Argument Formulas for All Quadrants Are the Most-Tested Complex Numbers JEE Main Topic

Argument (amplitude) of z = x + iy: the angle θ that the line OP makes with the positive real axis (measured anticlockwise). arg(z) = θ where tan θ = y/x. Argument is not unique — if θ is one value, then θ + 2kπ (k ∈ ℤ) are all valid. The principal argument Arg(z) lies in the interval (–π, π]. To find principal argument, let α = tan⁻¹|y/x| (always positive, acute angle), then:

Principal argument by quadrant (Complex Numbers formula — JEE Main):

1st Quadrant (x > 0, y > 0): Arg(z) = α

2nd Quadrant (x < 0, y > 0): Arg(z) = π – α

3rd Quadrant (x < 0, y < 0): Arg(z) = –(π – α) = α – π

4th Quadrant (x > 0, y < 0): Arg(z) = –α

Positive real axis (x > 0, y = 0): Arg(z) = 0

Negative real axis (x < 0, y = 0): Arg(z) = π

Positive imaginary axis (x = 0, y > 0): Arg(z) = π/2

Negative imaginary axis (x = 0, y < 0): Arg(z) = –π/2

Properties of argument of complex numbers (JEE Main formulas): arg(z₁z₂) = arg(z₁) + arg(z₂) + 2kπ (k ∈ ℤ). arg(z₁/z₂) = arg(z₁) – arg(z₂) + 2kπ. arg(zⁿ) = n·arg(z) + 2kπ. arg(z̄) = –arg(z). arg(z/z̄) = 2·arg(z) + 2kπ (or simply 2·Arg(z) when z is in upper half plane). arg(z) = 0 or π iff z is real. arg(z) = π/2 or –π/2 iff z is purely imaginary. If arg(z₁/z₂) = π/2 then Re(z₁z̄₂) = 0 (z₁ and z₂ are perpendicular). If arg(z₁/z₂) = 0 then Im(z₁z̄₂) = 0 (z₁ and z₂ are parallel). Download the Free PDF for Complex Numbers quadrant argument table and full argument formula list.

Polar Form, Euler's Form, Rotation Formula, and De Moivre's Theorem

Why These Complex Numbers Formulas Are Critical for JEE Main Geometry and nth Powers

Polar (Trigonometric) form of complex number: z = r(cosθ + i sinθ) where r = |z| (modulus) and θ = arg(z) (argument). Short notation: cis(θ) = cosθ + i sinθ. Any complex number can be written as z = |z| · cis(Arg z). For two complex numbers in polar form: z₁z₂ = r₁r₂ · cis(θ₁+θ₂) (multiply moduli, add arguments). z₁/z₂ = (r₁/r₂) · cis(θ₁–θ₂) (divide moduli, subtract arguments).

Euler's form (exponential form) of complex number: z = re^(iθ) = r(cosθ + i sinθ). Euler's formula: e^(iθ) = cosθ + i sinθ; e^(–iθ) = cosθ – i sinθ. Hence: cosθ = (e^(iθ) + e^(–iθ))/2; sinθ = (e^(iθ) – e^(–iθ))/(2i). Special values: e^(iπ) + 1 = 0 (Euler's identity). e^(iπ/2) = i; e^(iπ) = –1; e^(i·2π) = 1.

Rotation formula for complex numbers (JEE Main geometry formula): If z₁, z₂, z₃ are vertices of a triangle, and the angle at vertex A(z₁) from AB to AC is θ (measured anticlockwise), then: (z₃ – z₁)/(z₂ – z₁) = |AC|/|AB| · e^(iθ). The rotation formula rotates vector z₂–z₁ by angle θ to give the direction of z₃–z₁. If multiplication by e^(iπ/2) = i rotates anticlockwise by 90°; multiplication by –i rotates clockwise by 90°.

De Moivre's Theorem for complex numbers (JEE Main formula): For integer n: (cosθ + i sinθ)ⁿ = cos(nθ) + i sin(nθ). Equivalently: (e^(iθ))ⁿ = e^(inθ). For rational p/q: (cosθ + i sinθ)^(p/q) = cos((2kπ+θ)p/q) + i sin((2kπ+θ)p/q) for k = 0, 1, 2, …, q–1 (gives q distinct values). Applications: (1) Finding nth powers of complex numbers. (2) Proving trigonometric identities (cos nθ and sin nθ in terms of cos θ and sin θ). (3) Finding nth roots of complex numbers. Download the Free PDF for Complex Numbers De Moivre applications and worked examples for JEE Main.

Cube Roots of Unity ω — Properties and Factorisation Formulas

Why Cube Roots of Unity ω Properties Are the Most Formula-Rich JEE Main Complex Numbers Section

Cube roots of unity are the three solutions of z³ = 1. They are: z = 1, ω = (–1 + i√3)/2, ω² = (–1 – i√3)/2. Here ω and ω² are the two imaginary cube roots of unity (complex conjugates of each other).

Key properties of cube roots of unity ω (Complex Numbers JEE Main formulas):

1 + ω + ω² = 0 (sum of all cube roots of unity = 0)

ω³ = 1 (defining property)

ω̄ = ω² (conjugate of ω is ω²)

1/ω = ω²; 1/ω² = ω

ω^(3k) = 1; ω^(3k+1) = ω; ω^(3k+2) = ω² for integer k

1 + ωʳ + ω²ʳ = 3 if r is a multiple of 3; = 0 if r is not a multiple of 3

Polar form of ω: ω = cis(2π/3) = e^(i·2π/3); ω² = cis(4π/3) = e^(i·4π/3)

The three cube roots of unity form vertices of an equilateral triangle on the unit circle in the Argand plane, with vertices at (1,0), (–1/2, √3/2), (–1/2, –√3/2).

Factorisation formulas using cube roots of unity ω (JEE Main):

a³ – b³ = (a–b)(a–ωb)(a–ω²b)

a³ + b³ = (a+b)(a+ωb)(a+ω²b)

a² + ab + b² = (a–ωb)(a–ω²b)

x² + x + 1 = (x–ω)(x–ω²)

a³ + b³ + c³ – 3abc = (a+b+c)(a+ωb+ω²c)(a+ω²b+ωc)

If a + b + c = 0 then a³ + b³ + c³ = 3abc (special case)

1 + ω + ω² = 0 is used extensively in simplification: whenever x³ = 1 and x ≠ 1, replace x = ω or x = ω² and use 1 + ω + ω² = 0. Download the Free PDF for all cube roots of unity formulas and factorisation identities in Complex Numbers JEE Main.

nth Roots of Unity — Formula, Sum, and Product

Why nth Roots of Unity Formulas Appear in JEE Main as Product and Sum Questions

nth roots of unity are the n solutions of zⁿ = 1. Let ε = e^(i·2π/n) = cis(2π/n). Then the n roots are: 1, ε, ε², ε³, …, εⁿ⁻¹. These are n equally spaced points on the unit circle (|z| = 1), forming the vertices of a regular n-gon with one vertex at (1, 0) in the Argand plane. Common ratio of the GP: e^(i·2π/n).

Properties of nth roots of unity (Complex Numbers JEE Main formulas):

Sum: 1 + ε + ε² + … + εⁿ⁻¹ = 0 (sum of all nth roots of unity = 0)

Product: 1 · ε · ε² · … · εⁿ⁻¹ = εⁿ⁻¹⁺ⁿ⁻²⁺…⁺¹⁺⁰ = ε^(n(n–1)/2) = (–1)ⁿ⁻¹

Sum of pth powers: 1^p + ε^p + ε^(2p) + … + ε^((n–1)p) = n if p is a multiple of n; = 0 otherwise

The nth roots of unity are the roots of the polynomial xⁿ – 1 = 0, which factors as (x–1)(x–ε)(x–ε²)…(x–εⁿ⁻¹)

Sum of all nth roots of unity = coefficient of xⁿ⁻¹ in xⁿ–1 = 0

Product of all nth roots of unity = constant term of xⁿ–1 (with sign (–1)ⁿ) = (–1)ⁿ · (–1) = (–1)ⁿ⁺¹. For even n: product = –1; for odd n: product = –1 also. Using: 1·ε·ε²·…·εⁿ⁻¹ = (–1)ⁿ⁻¹. Download the Free PDF for nth roots of unity formula derivations and JEE Main application examples.

Complex Numbers in Geometry — Distance, Section Formula, Area, Rotation

Why Geometric Applications of Complex Numbers Are a JEE Main Multi-Mark Section

The Argand plane (also called complex plane or Gaussian plane) represents the complex number z = x + iy as the point P(x, y). Every geometric formula in coordinate geometry has a complex number counterpart — and these complex number geometry formulas appear regularly in JEE Main maths.

Distance formula for complex numbers: Distance between points z₁ and z₂ = |z₁ – z₂|. This is the most used complex number geometry formula in JEE Main.

Section formula for complex numbers: Point dividing z₁ and z₂ internally in ratio m:n = (mz₂ + nz₁)/(m+n). Externally in ratio m:n = (mz₂ – nz₁)/(m–n). Midpoint = (z₁+z₂)/2.

Area of triangle for complex numbers (JEE Main formula): Area of triangle with vertices z₁, z₂, z₃ = (1/4i) |Δ| where Δ = determinant of the 3×3 matrix with rows (z₁, z̄₁, 1), (z₂, z̄₂, 1), (z₃, z̄₃, 1). Alternative: Area = (i/4)[(z₁(z̄₂–z̄₃) + z₂(z̄₃–z̄₁) + z₃(z̄₁–z̄₂))].

Complex slope of line segment for complex numbers: Complex slope of line joining z₁ and z₂ = (z₁–z₂)/(z̄₁–z̄₂). Two lines are parallel if their complex slopes are equal. Two lines are perpendicular if the sum of their complex slopes is zero.

Collinearity of three complex numbers: z₁, z₂, z₃ are collinear iff (z₃–z₁)/(z₂–z₁) is real (argument is 0 or π). Equivalently: Im[(z₂–z₁)/(z₃–z₁)] = 0. Determinant condition: |z₁ z̄₁ 1; z₂ z̄₂ 1; z₃ z̄₃ 1| = 0.

Rotation formula for complex numbers (key JEE Main geometry tool): (z₃–z₁)/(z₂–z₁) = |AC|/|AB| · e^(iθ) where θ = angle at A from AB to AC (anticlockwise positive). If three complex numbers form an equilateral triangle with centroid at origin: z₁² + z₂² + z₃² = z₁z₂ + z₂z₃ + z₃z₁. Download the Free PDF for Complex Numbers geometry formulas and JEE Main application examples.

Quadratic Equations — Discriminant, Nature of Roots, and Vieta's Formulas

Why Quadratic Equations Discriminant and Vieta's Formulas Are Core JEE Main Topics

For the quadratic equation ax² + bx + c = 0 (a ≠ 0, a, b, c ∈ ℝ): the two roots are given by x = (–b ± √D)/(2a) where D = b² – 4ac is the discriminant.

Nature of roots — discriminant formula for quadratic equations (JEE Main):

D > 0: two distinct real roots. D = 0: two equal (repeated) real roots, each = –b/(2a). D < 0: two complex conjugate roots. D > 0 and D is a perfect square (with rational a,b,c): rational unequal roots. D > 0 and D not a perfect square (with rational a,b,c): irrational unequal roots (occur in conjugate pairs).

Imaginary roots always occur in conjugate pairs (if a,b,c ∈ ℝ): if α + iβ is a root, then α – iβ is also a root. Irrational roots always occur in conjugate pairs (if a,b,c ∈ ℚ): if α + √β is a root, then α – √β is also a root.

Vieta's formulas for quadratic equations (sum and product of roots — JEE Main): If α, β are roots of ax² + bx + c = 0, then: α + β = –b/a (sum of roots); αβ = c/a (product of roots); |α – β| = √D/|a| (difference of roots). The quadratic equation with roots α and β is: x² – (α+β)x + αβ = 0, i.e., x² – (sum)x + (product) = 0.

Special root conditions for quadratic equations (JEE Main short cuts): a + b + c = 0 → x = 1 is a root, other root = c/a. a – b + c = 0 → x = –1 is a root, other root = –c/a. Roots equal in magnitude, opposite sign → b = 0 and ac < 0. Roots are reciprocals of each other → c = a. Roots reciprocal of roots of cx² + bx + a = 0. Both roots positive → b/a < 0 and c/a > 0 and D ≥ 0. Both roots negative → b/a > 0 and c/a > 0 and D ≥ 0. Roots opposite in sign → c/a < 0. Download the Free PDF for all quadratic equation root shortcut formulas for JEE Main.

Symmetric Functions of Roots, Common Roots, and Formation of Equations

Why Symmetric Functions and Common Root Conditions Are JEE Main Quadratic Equations Questions

Symmetric functions of roots in quadratic equations (JEE Main formulas): α² + β² = (α+β)² – 2αβ. α³ + β³ = (α+β)³ – 3αβ(α+β). α² – β² = (α+β)(α–β). α⁴ + β⁴ = (α²+β²)² – 2α²β² = [(α+β)²–2αβ]² – 2(αβ)². 1/α + 1/β = (α+β)/(αβ) = –b/c. α²β + αβ² = αβ(α+β) = (–b/a)(c/a) = –bc/a². (α–β)² = (α+β)² – 4αβ = D/a². α/β + β/α = (α²+β²)/(αβ) = [(α+β)²–2αβ]/(αβ).

Common roots condition for quadratic equations: If α is a common root of a₁x²+b₁x+c₁=0 and a₂x²+b₂x+c₂=0, then the condition for one common root is: (a₁b₂–a₂b₁)(b₁c₂–b₂c₁) = (c₁a₂–c₂a₁)². The common root value = (c₁a₂–c₂a₁)/(a₁b₂–a₂b₁) = (b₁c₂–b₂c₁)/(c₁a₂–c₂a₁). For both roots common: a₁/a₂ = b₁/b₂ = c₁/c₂ (equations are proportional/identical). If two quadratic equations with real coefficients share a common imaginary or irrational root, then both roots are common (both equations are identical).

Formation of quadratic equation with given roots: x² – (sum of roots)x + (product of roots) = 0. If roots are α and β: x² – (α+β)x + αβ = 0. To form equation with roots that are functions of original roots (1/α, 1/β): replace x by 1/x and simplify → cx² + bx + a = 0. Roots –α, –β: replace x by –x → ax² – bx + c = 0. Roots α+k, β+k: replace x by x–k. Roots kα, kβ: replace x by x/k. Download the Free PDF for all symmetric function formulas and common root conditions for Quadratic Equations JEE Main.

Location of Roots and Sign Analysis of Quadratic Expression

Why Location of Roots Conditions Are the Highest-Concept JEE Main Quadratic Equations Section

Location of roots problems in quadratic equations ask: given f(x) = ax² + bx + c with roots α ≤ β, determine conditions on a, b, c so that roots satisfy specific positioning relative to real numbers k, k₁, k₂. These are among the most analytically rich JEE Main maths questions in this chapter.

Location of roots conditions for quadratic equations f(x) = ax² + bx + c:

(1) k lies between the roots (α < k < β): Conditions: (i) D > 0; (ii) a·f(k) < 0. (a·f(k) < 0 means k is between the roots regardless of sign of a.)

(2) Both roots greater than k (k < α ≤ β): Conditions: (i) D ≥ 0; (ii) a·f(k) > 0; (iii) –b/(2a) > k (vertex is to the right of k).

(3) Both roots less than k (α ≤ β < k): Conditions: (i) D ≥ 0; (ii) a·f(k) > 0; (iii) –b/(2a) < k (vertex is to the left of k).

(4) Exactly one root lies in (k₁, k₂): Conditions: (i) D > 0; (ii) f(k₁)·f(k₂) < 0.

(5) Both roots lie in (k₁, k₂): Conditions: (i) D ≥ 0; (ii) a·f(k₁) > 0; (iii) a·f(k₂) > 0; (iv) k₁ < –b/(2a) < k₂.

(6) Both k₁ and k₂ lie between roots (k₁, k₂ are between α and β): Conditions: (i) D > 0; (ii) a·f(k₁) < 0; (iii) a·f(k₂) < 0.

Sign analysis of quadratic expression f(x) = ax²+bx+c: If a > 0 and D < 0: f(x) > 0 for all x ∈ ℝ (positive definite). If a < 0 and D < 0: f(x) < 0 for all x ∈ ℝ (negative definite). If a > 0 and D > 0 with roots α < β: f(x) > 0 for x ∈ (–∞,α) ∪ (β,∞); f(x) < 0 for x ∈ (α,β). Range of f: if a > 0, minimum value = –D/(4a) at x = –b/(2a); if a < 0, maximum value = –D/(4a). Download the Free PDF for all location of roots conditions for Quadratic Equations JEE Main with diagrams.