JEE Main Mathematics · Chapter 2

Complex Numbers

Algebra of complex numbers, modulus and argument, Argand plane, polar and Euler form, cube roots of unity, locus problems.

NCERT Class 11JEE Main: 1–2 Qs per shift 9 min read

1. Chapter at a Glance

A complex number is a number of the form z = x + iy, where x and y are real and i = √(−1), so i² = −1. Complex numbers let us solve every polynomial equation and also describe points and rotations in a plane. JEE Main asks one or two questions every shift, mostly on modulus and argument, locus of z, powers of i and ω, and rotation.

What JEE asks most
  • Simplifying expressions using powers of i and cube roots of unity (ω).
  • Finding the locus of z from conditions like |z − a| = k|z − b|.
  • Modulus and principal argument; maximum and minimum of |z| using the triangle inequality.
  • Solving equations such as z² + |z| = 0 by putting z = x + iy.

2. Basics and Algebra of Complex Numbers

For z = x + iy, Re(z) = x is the real part and Im(z) = y is the imaginary part. Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal. Complex numbers cannot be compared: statements like 2 + 3i > 1 + i have no meaning.

Powers of i

The powers of i repeat in a cycle of 4: i¹ = i, i² = −1, i³ = −i, i⁴ = 1. So in depends only on the remainder when n is divided by 4. The sum of any four consecutive powers of i is 0.

Conjugate

The conjugate of z = x + iy is z̄ = x − iy (mirror image in the real axis). Useful results:

  • z + z̄ = 2 Re(z), z − z̄ = 2i Im(z), z·z̄ = x² + y² = |z|².
  • conj(z1 ± z2) = z̄1 ± z̄2, conj(z1z2) = z̄1z̄2.
  • z is purely real ⇔ z = z̄; z is purely imaginary (or 0) ⇔ z = −z̄.

Division

To divide, multiply the numerator and the denominator by the conjugate of the denominator: (a + ib)/(c + id) = (a + ib)(c − id)/(c² + d²).

3. Modulus, Argument and the Argand Plane

We plot z = x + iy as the point (x, y) in the Argand plane. Its distance from the origin is the modulus |z| = √(x² + y²), and the angle it makes with the positive real axis is the argument θ.

ReIm z = x + iy z̄ = x − iy θ r = |z|
z and its conjugate are mirror images in the real axis; |z| = r and arg z = θ.

Principal argument

The principal argument lies in (−π, π]. With α = tan−1|y/x|:

Position of zPrincipal argument
1st quadrant (x > 0, y > 0)α
2nd quadrant (x < 0, y > 0)π − α
3rd quadrant (x < 0, y < 0)−(π − α)
4th quadrant (x > 0, y < 0)−α

Polar and Euler forms

z = r(cos θ + i sin θ) = r eiθ

In this form multiplication is easy: moduli multiply and arguments add.

  • |z1z2| = |z1||z2|, arg(z1z2) = arg z1 + arg z2 (up to 2π).
  • |z1/z2| = |z1|/|z2|, arg(z1/z2) = arg z1 − arg z2.
  • De Moivre's theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ.

Triangle inequality

| |z1| − |z2| | ≤ |z1 ± z2| ≤ |z1| + |z2|

Equality |z1 + z2| = |z1| + |z2| holds when z1 and z2 point in the same direction (equal arguments).

4. Cube Roots of Unity

The equation z³ = 1 has three roots: 1, ω and ω², where

ω = (−1 + i√3)/2 = ei2π/3,   ω² = (−1 − i√3)/2
  • 1 + ω + ω² = 0 and ω³ = 1.
  • ω̄ = ω² and |ω| = 1.
  • They are the vertices of an equilateral triangle on the unit circle.
  • Useful factorisations: x² + x + 1 = (x − ω)(x − ω²); a³ + b³ = (a + b)(a + bω)(a + bω²).
  • In general, the n-th roots of unity ei2πk/n (k = 0 to n − 1) add up to 0 and form a regular n-gon.

5. Geometry with Complex Numbers

  • |z − z1| is the distance between z and z1.
  • |z − z0| = r is a circle with centre z0 and radius r.
  • |z − z1| = |z − z2| is the perpendicular bisector of the segment joining z1 and z2.
  • |z − z1| = k|z − z2| (k ≠ 1) is a circle (called the circle of Apollonius).
  • |z − z1| + |z − z2| = 2a with 2a > |z1 − z2| is an ellipse with foci z1, z2.
  • arg((z − z1)/(z − z2)) = θ is an arc of a circle through z1 and z2.
Rotation

Multiplying by eiθ rotates a point anticlockwise by θ about the origin. If segment AB is rotated about A by angle θ to lie along AC, then

(zC − zA)/(zB − zA) = (|AC|/|AB|) eiθ

Multiplying by i rotates by 90° anticlockwise.

6. Formula Master Table

ResultFormula
Modulus|x + iy| = √(x² + y²)
Product with conjugatez·z̄ = |z|²
Reciprocal1/z = z̄/|z|²
Euler formz = reiθ = r(cos θ + i sin θ)
De Moivre(cos θ + i sin θ)n = cos nθ + i sin nθ
Parallelogram law|z1+z2|² + |z1−z2|² = 2(|z1|² + |z2|²)
Cube roots of unity1 + ω + ω² = 0, ω³ = 1
Square root of a + ib (b > 0)±[√((|z|+a)/2) + i√((|z|−a)/2)]
Circle|z − z0| = r
Section formulaz = (mz2 + nz1)/(m + n)
Centroid of triangle(z1 + z2 + z3)/3
Equilateral triangle conditionz1² + z2² + z3² = z1z2 + z2z3 + z3z1

7. Mind Map

Complex Numbers
Algebra
  • z = x + iy
  • Powers of i (cycle 4)
  • Conjugate, division
Polar form
  • |z|, arg z
  • reiθ
  • De Moivre
Roots of unity
  • 1, ω, ω²
  • 1 + ω + ω² = 0
  • n-th roots → regular polygon
Geometry
  • Distance |z − a|
  • Locus: circle, bisector, ellipse
  • Rotation eiθ

8. Tips and Tricks

Shortcuts
  • (1 + i)² = 2i and (1 − i)² = −2i. So (1 + i)/(1 − i) = i and (1 − i)/(1 + i) = −i.
  • With ω: 1 + ω = −ω², 1 + ω² = −ω and ω + ω² = −1 — replace sums quickly before raising to powers.
  • For locus questions, put z = x + iy only when the geometry is not obvious. |z − a| = |z − b| is simply a perpendicular bisector — no algebra needed.
  • Maximum and minimum of |z| when |z − z0| = r: they are |z0| + r and | |z0| − r |.
  • If |z| = 1, then z̄ = 1/z. This single fact solves many "show that it is purely imaginary" questions.
Mistakes students make
  • Writing √(−4)·√(−9) = √36 = 6. Correct: 2i · 3i = −6. The rule √a·√b = √(ab) fails when both a and b are negative.
  • Using tan−1(y/x) directly as the argument without checking the quadrant.
  • Forgetting that the principal argument is in (−π, π], not [0, 2π).

9. Important One-liners

  • i4k = 1, i4k+1 = i, i4k+2 = −1, i4k+3 = −i.
  • 0 is the only complex number that is both purely real and purely imaginary.
  • arg(z̄) = −arg(z) (for z not on the negative real axis) and |z̄| = |z|.
  • arg(z) + arg(z̄) = 0 and arg(−z) = arg(z) ± π.
  • If |z| = 1 and z ≠ −1, then (z − 1)/(z + 1) is purely imaginary.
  • The n-th roots of unity sum to 0 and their product is (−1)n−1.
  • Complex roots of a polynomial with real coefficients occur in conjugate pairs.
  • |z1 + z2| = |z1 − z2| means z1 and z2 are perpendicular as vectors from the origin (arg z1 − arg z2 = ±π/2).
  • eiπ + 1 = 0 (Euler's identity).

10. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

If z = (1 + i)/(1 − i), then z2026 equals:

  1. 1
  2. i
  3. −1
  4. −i
Show solution

Multiply numerator and denominator by (1 + i): z = (1 + i)²/(1 − i²) = 2i/2 = i.

2026 = 4 × 506 + 2, so i2026 = i² = −1.

Answer: (C) −1

Q2Medium

The locus of z satisfying |z − 3| = 2|z − 1| is a circle whose radius is:

  1. 2/3
  2. 4/3
  3. 5/3
  4. 2
Show solution

Put z = x + iy: (x − 3)² + y² = 4[(x − 1)² + y²].

x² − 6x + 9 + y² = 4x² − 8x + 4 + 4y² ⇒ 3x² + 3y² − 2x − 5 = 0 ⇒ x² + y² − (2/3)x − 5/3 = 0.

Centre (1/3, 0); radius = √(1/9 + 5/3) = √(16/9) = 4/3.

Answer: (B) 4/3

Q3Medium

If ω is a non-real cube root of unity, then (1 + ω − ω²)³ equals:

  1. 8
  2. −8
  3. 0
  4. 8ω
Show solution

Since 1 + ω + ω² = 0, we have 1 + ω = −ω². So 1 + ω − ω² = −ω² − ω² = −2ω².

(−2ω²)³ = −8ω6 = −8(ω³)² = −8.

Answer: (B) −8

Q4Easy

The principal argument of z = −1 − i√3 is:

  1. π/3
  2. 2π/3
  3. −2π/3
  4. 4π/3
Show solution

z lies in the third quadrant. α = tan−1(√3/1) = π/3, so the principal argument = −(π − π/3) = −2π/3.

4π/3 describes the same direction but is not in (−π, π].

Answer: (C) −2π/3

Q5Numerical

Find the number of complex numbers z satisfying z² + |z| = 0.

Show solution

Let z = x + iy. Then x² − y² + √(x² + y²) + 2ixy = 0, so xy = 0 and x² − y² + √(x² + y²) = 0.

If y = 0: x² + |x| = 0 ⇒ x = 0. If x = 0: −y² + |y| = 0 ⇒ |y| = 0 or 1 ⇒ y = 0, 1, −1.

Solutions: 0, i, −i.

Answer: 3

Q6Numerical

Find the value of 1 + i + i² + i³ + … + i100.

Show solution

There are 101 terms. Every block of four consecutive powers adds up to 0. The first 100 terms (i⁰ to i99) form 25 such blocks, leaving only i100 = (i⁴)25 = 1.

Answer: 1

Q7Hard

If |z − 3 − 4i| = 2, then the maximum value of |z| is:

  1. 5
  2. 7
  3. 3
  4. 9
Show solution

z lies on a circle with centre 3 + 4i and radius 2. The distance of the centre from the origin is |3 + 4i| = 5.

The farthest point of the circle from the origin is at distance 5 + 2 = 7 (and the nearest at 5 − 2 = 3).

Answer: (B) 7

Q8Hard

If |z| = 1 and z ≠ −1, then w = (z − 1)/(z + 1) is always:

  1. Purely real
  2. Purely imaginary
  3. Of modulus 1
  4. Zero
Show solution

Since |z| = 1, z̄ = 1/z. Then w̄ = (z̄ − 1)/(z̄ + 1) = (1/z − 1)/(1/z + 1) = (1 − z)/(1 + z) = −w.

w̄ = −w means w is purely imaginary (it is 0 when z = 1, and 0 is also counted as purely imaginary here).

Answer: (B)

11. Quick Revision Summary

Remember before the exam
  • in repeats every 4; (1 + i)/(1 − i) = i.
  • |z|² = z z̄; 1/z = z̄/|z|²; if |z| = 1 then z̄ = 1/z.
  • Principal argument lies in (−π, π] — always check the quadrant.
  • Product: multiply moduli, add arguments; De Moivre for powers.
  • 1 + ω + ω² = 0, ω³ = 1, ω̄ = ω².
  • |z − a| = r is a circle; |z − a| = |z − b| is a perpendicular bisector; max/min of |z| on a circle = |centre| ± r.
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