Complex Numbers and Quadratic Equations: JEE Main Maths formulas and notes

Q: How do I solve “Simplify a complex expression or find its modulus and argument” questions in JEE Main?

A fraction or power of complex numbers is given. Write it as x + iy, or find |z| and arg z. 1. For a quotient multiply top and bottom by the conjugate of the bottom. 2. Use z·z̄ = |z|² and the rule |z₁z₂| = |z₁||z₂|. 3. For large powers convert to polar form and use De Moivre. Example: (1 + i)⁸ = 16.

Q: How do I solve “Locus of a point z” questions in JEE Main?

A condition on |z − a| or arg is given. Identify the curve. 1. Read |z − a| as a distance from the point a. 2. Equal distances to two points give a line; a fixed distance gives a circle. 3. Check by writing z = x + iy when in doubt. Example: |z − 1| = |z + 1| is the imaginary axis.

Q: How do I solve “Roots of unity and powers of ω” questions in JEE Main?

An expression in ω, ω² or in nth roots of unity is given. 1. Reduce each power of ω with n mod 3. 2. Replace 1 + ω + ω² by 0 wherever it appears. 3. For nth roots of unity use that their sum is 0. Example: (1 + ω)⁷ = −ω².

Q: How do I solve “Nature, sum and product of roots” questions in JEE Main?

A quadratic with a parameter, or a relation between the roots, is given. 1. Write α + β = −b/a and αβ = c/a. 2. Express the asked quantity through them. 3. Use D = b² − 4ac for the nature of the roots or the range of a parameter. Example: x² − 6x + 8 = 0 gives α² + β² = 20.

Q: How do I solve “Common roots and location of roots” questions in JEE Main?

Two equations share a root, or the roots must lie in a given interval. 1. Eliminate the squared term to get the common root and substitute it. 2. For location use D, the vertex −b/2a and the sign of f at the given point. 3. Sketch the parabola to avoid missing a case. Example: x² − 4x + 5 > 0 for all x (D 0).

Complex Numbers and Quadratic Equations give 2 to 3 questions in JEE Mains in most years: a simplification or a locus, a power of ω, and a quadratic with a condition on its roots. About 8 hours in 5 days is enough. The chapter is the base for sequences, conics and calculus, so the effort pays again.

Practise this chapter free in Rank mode

Open Physica, choose Rank mode, JEE Mains, then Maths and this chapter for the guide, formulas, most-asked questions, past-year questions and a mock test.

Quick revision

i⁴ = 1; reduce iⁿ by n mod 4; z·z̄ = |z|²; |z₁z₂| = |z₁||z₂|
1 + ω + ω² = 0, ω³ = 1; nth roots of unity sum to 0
|z − z₀| = r circle; |z − z₁| = |z − z₂| perpendicular bisector
x² − (sum)x + product = 0; D = b² − 4ac decides the nature of roots
a > 0 and D < 0 ⇒ f(x) > 0 for all x

Formulas with solved examples

Complex numbers: algebra

Powers of i and the basic operations

i² = −1
iⁿ depends on n mod 4: i⁰ = 1, i¹ = i, i² = −1, i³ = −i
(a + ib) + (c + id) = (a + c) + i(b + d)
(a + ib)(c + id) = (ac − bd) + i(ad + bc)
a + ib = c + id ⇔ a = c and b = d

How to apply. Divide n by 4 and keep the remainder. For a quotient multiply the top and bottom by the conjugate of the bottom.

Example. i²⁰²³: 2023 leaves remainder 3 on division by 4, so i²⁰²³ = i³ = −i. And (3 + 4i)/(1 − 2i) = −1 + 2i.

Square root of a complex number

√(a + ib) = ±( √((|z| + a)/2) + i·sign(b)·√((|z| − a)/2) ) where |z| = √(a² + b²)
or put √(a + ib) = x + iy and compare real and imaginary parts

How to apply. Square x + iy, equate the real and imaginary parts, and use x² + y² = |z| as a helper.

Example. √(3 + 4i) = ±(2 + i), because (2 + i)² = 3 + 4i.

Modulus, conjugate and argument

Conjugate and modulus

z̄ = x − iy
z + z̄ = 2x
z − z̄ = 2iy
z·z̄ = |z|²
|z₁z₂| = |z₁||z₂|
|z₁/z₂| = |z₁|/|z₂|
|z̄| = |z|
|zⁿ| = |z|ⁿ
conj of a product = product of conjugates
z is real ⇔ z = z̄
z is purely imaginary ⇔ z + z̄ = 0 (and z ≠ 0)

How to apply. Use z·z̄ = |z|² to remove a modulus or to divide. Use z = z̄ to test whether a number is real.

Example. |3 − 4i| = 5 and (3 − 4i)(3 + 4i) = 25.

Triangle inequalities

|z₁ + z₂| ≤ |z₁| + |z₂|
|z₁ − z₂| ≥ ||z₁| − |z₂||
|z₁ + z₂|² + |z₁ − z₂|² = 2(|z₁|² + |z₂|²)
equality when z₁ and z₂ have the same direction (first one)

How to apply. Use them to find the greatest and least value of |z₁ + z₂| when |z₁| and |z₂| are given.

Example. If |z₁| = 3 and |z₂| = 4 then |z₁ + z₂| lies between 1 and 7.

Argument

arg z = θ where tan θ = y/x, with the quadrant of (x, y)
principal value −π < θ ≤ π
arg(z₁z₂) = arg z₁ + arg z₂ (mod 2π)
arg(z₁/z₂) = arg z₁ − arg z₂
arg z̄ = − arg z
arg(−z) = arg z ± π

How to apply. Draw the point in the plane first to pick the quadrant, then find the angle from tan⁻¹|y/x|.

Example. arg(−1 + i) = 3π/4, since the point is in the second quadrant with a 45° slope.

Polar form, De Moivre and roots of unity

Polar form and De Moivre

z = r(cos θ + i sin θ) = r e^{iθ}
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
to multiply, multiply moduli and add arguments
to find a power, raise the modulus and multiply the argument

How to apply. Convert to polar form for large powers: find r and θ first.

Example. (1 + i)⁸: modulus √2, argument π/4, so the result has modulus 16 and argument 2π, i.e. 16.

Cube roots of unity

ω = (−1 + i√3)/2, ω² = (−1 − i√3)/2
ω³ = 1
1 + ω + ω² = 0
ω̄ = ω²
ωⁿ depends on n mod 3
1 + ω = −ω² and 1 + ω² = −ω

How to apply. Reduce the power of ω with n mod 3. Replace 1 + ω + ω² by 0 whenever it appears.

Example. 1 + ω + ω² = 0, so (1 + ω)⁷ = (−ω²)⁷ = −ω¹⁴ = −ω².

nth roots of unity

the roots of zⁿ = 1 are cos(2kπ/n) + i sin(2kπ/n), k = 0, 1, …, n − 1
their sum is 0
their product is (−1)ⁿ⁺¹
they lie on the unit circle and form a regular n-gon

How to apply. Use the sum rule for sums like 1 + ω + ω² + … and the polygon picture for geometry questions.

Example. The 4th roots of unity are 1, i, −1, −i and their sum is 0.

Locus and geometry

Distance and standard loci

|z − z₁| = distance between z and z₁
|z − z₀| = r is a circle (centre z₀, radius r)
|z − z₁| = |z − z₂| is the perpendicular bisector
|z − z₁| + |z − z₂| = k (k > |z₁ − z₂|) is an ellipse
arg((z − z₁)/(z − z₂)) = constant is an arc of a circle
three points are collinear when (z₃ − z₁)/(z₂ − z₁) is real

How to apply. Write z = x + iy and simplify, or read the geometric meaning of the modulus as a distance.

Example. |z − 1| = |z + 1| is the imaginary axis (the perpendicular bisector of 1 and −1).

Quadratic equations: roots and nature

Roots, sum and product

ax² + bx + c = 0
x = (−b ± √D)/2a with D = b² − 4ac
α + β = −b/a
αβ = c/a
equation with roots α, β: x² − (α + β)x + αβ = 0
α² + β² = (α + β)² − 2αβ
α³ + β³ = (α + β)³ − 3αβ(α + β)
|α − β| = √D/|a|

How to apply. Express the quantity asked in terms of α + β and αβ first, then substitute.

Example. For x² − 6x + 8 = 0: α + β = 6, αβ = 8, so α² + β² = 36 − 16 = 20 and the roots are 2 and 4.

Nature of roots

D > 0: two real distinct roots
D = 0: equal real roots
D < 0: two complex conjugate roots (for real a, b, c)
D a perfect square with rational a, b, c: rational roots
irrational and complex roots come in conjugate pairs for real or rational coefficients

How to apply. Find D first. For a parameter k, form D as an expression in k and solve the inequality.

Example. x² − 6x + 13 = 0 has D = 36 − 52 = −16 < 0, so the roots are 3 ± 2i.

Location of roots and common roots

Common roots and conditions

one common root: eliminate x² between the two equations to get the common root, then substitute
both roots common: a₁/a₂ = b₁/b₂ = c₁/c₂
roots in a ratio m : n: (m + n)²·ac = mn·b²
one root double the other: 2b² = 9ac
roots reciprocal: a = c
roots equal in size and opposite in sign: b = 0

How to apply. Write the condition in terms of α and β and then convert to a, b, c using the sum and product.

Example. Roots in ratio 1 : 2 give 2b² = 9ac, e.g. x² − 3x + 2 = 0 (roots 1 and 2): 2·9 = 18 = 9·1·2.

Location of roots and sign of a quadratic

for f(x) = ax² + bx + c with a > 0: both roots greater than k needs D ≥ 0, −b/2a > k and f(k) > 0
k between the roots needs f(k) < 0
exactly one root in (p, q) needs f(p)·f(q) < 0
f(x) > 0 for all real x needs a > 0 and D < 0

How to apply. Sketch the parabola. Pick out the three things to check: D, the position of the vertex −b/2a, and the sign of f at the given point.

Example. x² − 4x + 5 > 0 for all x because a = 1 > 0 and D = 16 − 20 = −4 < 0.

Most asked question types

How do I solve “Simplify a complex expression or find its modulus and argument” questions in JEE Main?

A fraction or power of complex numbers is given. Write it as x + iy, or find |z| and arg z. 1. For a quotient multiply top and bottom by the conjugate of the bottom. 2. Use z·z̄ = |z|² and the rule |z₁z₂| = |z₁||z₂|. 3. For large powers convert to polar form and use De Moivre. Example: (1 + i)⁸ = 16.

How do I solve “Locus of a point z” questions in JEE Main?

A condition on |z − a| or arg is given. Identify the curve. 1. Read |z − a| as a distance from the point a. 2. Equal distances to two points give a line; a fixed distance gives a circle. 3. Check by writing z = x + iy when in doubt. Example: |z − 1| = |z + 1| is the imaginary axis.

How do I solve “Roots of unity and powers of ω” questions in JEE Main?

An expression in ω, ω² or in nth roots of unity is given. 1. Reduce each power of ω with n mod 3. 2. Replace 1 + ω + ω² by 0 wherever it appears. 3. For nth roots of unity use that their sum is 0. Example: (1 + ω)⁷ = −ω².

How do I solve “Nature, sum and product of roots” questions in JEE Main?

A quadratic with a parameter, or a relation between the roots, is given. 1. Write α + β = −b/a and αβ = c/a. 2. Express the asked quantity through them. 3. Use D = b² − 4ac for the nature of the roots or the range of a parameter. Example: x² − 6x + 8 = 0 gives α² + β² = 20.

How do I solve “Common roots and location of roots” questions in JEE Main?

Two equations share a root, or the roots must lie in a given interval. 1. Eliminate the squared term to get the common root and substitute it. 2. For location use D, the vertex −b/2a and the sign of f at the given point. 3. Sketch the parabola to avoid missing a case. Example: x² − 4x + 5 > 0 for all x (D < 0, a > 0).

Common myths and the truth

√(−a)√(−b) = √(ab). Not for negative numbers: √(−1)√(−1) = i·i = −1, not 1.
Every quadratic has two real roots. Real roots need D ≥ 0. D < 0 gives a conjugate pair of complex roots.

Mistakes to avoid

Forgetting the quadrant when finding arg z.
Treating |z|² as z².
Using D = b² − 4ac when the coefficients are not real.
Missing the case D = 0 in a condition on the roots.

More: all JEE Main chapters.