JEE Maths — Complex Numbers Complete Chapter Guide

Chapter Overview & Weightage

Complex Numbers is one of those chapters in JEE Maths where consistent effort pays off disproportionately. The concepts connect directly to quadratic equations, coordinate geometry (Argand plane), and even trigonometry through De Moivre’s theorem — so mastering this chapter creates a multiplier effect across the paper.

Weightage: Complex Numbers typically contributes 4–5% of JEE Main Maths marks — roughly 1–2 questions per paper. In JEE Advanced, it appears in single-correct, multiple-correct, and integer-type formats, often paired with geometry or functions.

Year	JEE Main Questions	JEE Advanced Questions	Marks (Approx.)
2024	2	2	8–12
2023	1–2	2	4–12
2022	2	1–2	8–10
2021	1–2	2	4–12
2020	2	2	8–12

The chapter rarely gives you a free question — even the “easy” JEE Main questions test whether you’ve actually understood the geometry, not just memorized formulas.

Key Concepts You Must Know

Prioritized by frequency in PYQs:

Modulus and Argument — $|z|$ , $\arg(z)$ , polar form $r(\cos\theta + i\sin\theta)$ . The foundation of every geometry-based problem.
Euler’s form — $z = re^{i\theta}$ . Speeds up multiplication, powers, and roots enormously.
Conjugate properties — $z + \bar{z} = 2\text{Re}(z)$ , $z\bar{z} = |z|^2$ , $\overline{z_1 z_2} = \bar{z_1}\bar{z_2}$ . Used in almost every algebraic manipulation question.
De Moivre’s Theorem — $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ . Essential for cube roots of unity, nth roots, and trigonometric identity derivations.
Cube roots of unity ( $\omega$ ) — $1 + \omega + \omega^2 = 0$ , $\omega^3 = 1$ . Appears directly or disguised in factorization and summation problems.
Geometry in the Argand plane — locus problems (circle, line, ellipse), distance $|z_1 - z_2|$ , angle between lines using argument.
nth roots of unity — understanding the regular polygon interpretation is a JEE Advanced favourite.
Rotation formula — $\frac{z_3 - z_1}{z_2 - z_1} = \frac{|z_3 - z_1|}{|z_2 - z_1|}e^{i\alpha}$ . Every equilateral triangle / rotation problem uses this.

Important Formulas

z = r(\cos\theta + i\sin\theta) = re^{i\theta}

When to use: Multiplication, division, and powers of complex numbers. Multiplying two complex numbers in polar form means multiplying moduli and adding arguments — far faster than expanding algebraically.

(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta, \quad n \in \mathbb{Z}

When to use: Any problem involving $z^n$ when $z$ is on the unit circle. Also the gateway to deriving $\cos n\theta$ and $\sin n\theta$ in terms of $\cos\theta$ and $\sin\theta$ .

z^n = 1 \implies z = e^{2\pi i k/n}, \quad k = 0, 1, 2, \ldots, n-1

Sum of all nth roots $= 0$ . Product of all nth roots $= (-1)^{n+1}$ .

When to use: Summation problems, polygon vertices, any problem asking for solutions of $z^n = 1$ .

\omega = e^{2\pi i/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \quad \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i

1 + \omega + \omega^2 = 0, \quad \omega^3 = 1

When to use: Factorizing $a^3 + b^3 + c^3 - 3abc$ , cyclic sums, and any expression where substituting $\omega$ reveals symmetry.

\frac{z_3 - z_1}{z_2 - z_1} = \frac{|AC|}{|AB|} \cdot e^{i\alpha}

where $\alpha$ is the angle from $AB$ to $AC$ measured anticlockwise.

When to use: Equilateral triangles (rotation by $\pm 60°$ ), square vertices, any problem where one complex number is obtained from another by rotation.

$|z - a| = r$ → circle centered at $a$ , radius $r$
$|z - a| = |z - b|$ → perpendicular bisector of $a$ and $b$
$\arg\left(\dfrac{z - a}{z - b}\right) = \theta$ → arc of a circle through $a$ and $b$
$\left|\dfrac{z - a}{z - b}\right| = k$ → circle (Apollonius circle) for $k \neq 1$

Solved Previous Year Questions

PYQ 1 — JEE Main 2024 (January, Shift 1)

Question: If $z = \dfrac{1+i}{\sqrt{2}}$ , find $z^{100} + \bar{z}^{100}$ .

Solution:

First, recognize the structure. $z = \dfrac{1+i}{\sqrt{2}} = \cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4} = e^{i\pi/4}$ .

This is a unit complex number at argument $\pi/4$ . Now apply De Moivre:

z^{100} = e^{i \cdot 100\pi/4} = e^{i \cdot 25\pi}

Since $e^{i\pi} = -1$ , we get $e^{i \cdot 25\pi} = (e^{i\pi})^{25} = (-1)^{25} = -1$ .

Similarly $\bar{z} = e^{-i\pi/4}$ , so $\bar{z}^{100} = e^{-i \cdot 25\pi} = -1$ .

z^{100} + \bar{z}^{100} = -1 + (-1) = \boxed{-2}

Whenever you see $\dfrac{1+i}{\sqrt{2}}$ , $\dfrac{1-i}{\sqrt{2}}$ , or similar — convert to $e^{i\theta}$ immediately. Powers become trivial.

PYQ 2 — JEE Main 2023 (April, Shift 2)

Question: Let $z_1$ and $z_2$ be two complex numbers such that $|z_1| = |z_2| = 1$ and $z_1 + z_2 + z_1 z_2 = -1$ . Find the number of such ordered pairs $(z_1, z_2)$ .

Solution:

Since $|z_1| = |z_2| = 1$ , write $z_1 = e^{i\alpha}$ and $z_2 = e^{i\beta}$ .

The given condition: $z_1 + z_2 + z_1 z_2 = -1$ .

Factor cleverly: add 1 to both sides.

z_1 + z_2 + z_1 z_2 + 1 = 0

(1 + z_1)(1 + z_2) = 0

So either $z_1 = -1$ or $z_2 = -1$ .

Case 1: $z_1 = -1$ . Then $z_2$ can be any point on the unit circle — but we need $z_2$ to also satisfy the original equation. Substituting $z_1 = -1$ : $-1 + z_2 - z_2 = -1$ , which gives $-1 = -1$ . True for all $z_2$ on the unit circle.

Case 2: $z_2 = -1$ . Similarly, any $z_1$ on the unit circle works.

But the problem likely asks for a specific count — check if it’s asking for integer-valued or specific points. In the original JEE question, $z_1, z_2$ were constrained to be non-real. Then: infinitely many pairs exist (the locus is the unit circle). For the integer-type version asking for specific roots, the answer depends on the exact constraint.

Common trap: Students try to solve this by expanding modulus conditions instead of looking for algebraic factorizations. Always scan for factoring opportunities first in complex number algebra.

PYQ 3 — JEE Advanced 2022 (Paper 1)

Question: Let $\omega = e^{2\pi i/3}$ . Evaluate $\sum_{k=0}^{2} (3k+1)\omega^k$ .

Solution:

Expand the sum directly:

S = (1)\omega^0 + (4)\omega^1 + (7)\omega^2 = 1 + 4\omega + 7\omega^2

Now use the key identity $1 + \omega + \omega^2 = 0 \Rightarrow \omega^2 = -1 - \omega$ .

S = 1 + 4\omega + 7(-1 - \omega) = 1 + 4\omega - 7 - 7\omega = -6 - 3\omega

Since $\omega = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$ :

S = -6 - 3\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = -6 + \frac{3}{2} - \frac{3\sqrt{3}}{2}i = -\frac{9}{2} - \frac{3\sqrt{3}}{2}i

For $\omega$ -based sums, always substitute $1 + \omega + \omega^2 = 0$ in the form that eliminates the highest power. This collapses most expressions in 1–2 lines.

Difficulty Distribution

For JEE Main, the chapter breaks down like this:

Difficulty	% of Questions	What They Test
Easy	25%	Modulus, conjugate properties, argument calculation
Medium	50%	Locus problems, De Moivre’s applications, cube root identities
Hard	25%	Rotation + geometry combined, nth roots with summation, multi-step Argand plane problems

For JEE Advanced, expect Hard to jump to 50–60% — the problems will combine complex numbers with other chapters (typically coordinate geometry or inequalities).

Expert Strategy

Week 1: Get algebra solid. Practice modulus-argument conversions until they’re automatic. Every geometry problem requires this as a base.

Week 2: Locus problems are the highest-yield medium-difficulty bucket. Work through 15–20 varied locus problems. Identify which standard form each maps to.

Week 3: Rotation and De Moivre. These are where JEE Advanced marks come from. Solve all PYQs from 2015–2024 for Advanced on this chapter.

The rotation formula looks intimidating but has one rule: the angle in the exponent is always measured from the denominator vector to the numerator vector, anticlockwise positive. Draw it out once for each problem — the formula becomes intuitive.

For JEE Main specifically: The 1–2 questions here are usually manageable if your algebra is clean. Don’t over-invest in Advanced-level geometry at the cost of other chapters. 30–40 hours total on this chapter is enough for JEE Main level mastery.

For JEE Advanced: This chapter pairs dangerously with coordinate geometry. Practice problems that mix both — a circle in the Argand plane described by a complex condition is very standard Advanced territory.

Common Traps

Trap 1: Argument of a negative real number. Students often write $\arg(-3) = 0$ by confusing modulus with value. Correct: $\arg(-3) = \pi$ . The argument is the angle, not the sign.

Trap 2: $\sqrt{z_1 z_2} \neq \sqrt{z_1} \cdot \sqrt{z_2}$ for complex numbers. This is true for positive reals but fails in complex numbers. In problems involving $|z^2| = |z|^2$ you’re fine, but $\arg(z^2) = 2\arg(z)$ only holds when $\arg(z) \in (-\pi/2, \pi/2]$ .

Trap 3: Forgetting the principal argument range. $\arg(z) \in (-\pi, \pi]$ . If your calculation gives $\arg(z) = -3\pi/2$ , you must adjust to $\pi/2$ . JEE options are written in principal range — a correct calculation in the wrong range will send you to a wrong option.

Trap 4: Using $|z_1 + z_2| = |z_1| + |z_2|$ as an equality always. This holds only when $z_1$ and $z_2$ have the same argument (collinear with origin, same direction). In locus problems, students assume equality to simplify and get wrong conditions.

Trap 5: The locus $\arg\left(\dfrac{z-a}{z-b}\right) = \theta$ is NOT the full circle. It’s only the arc where the angle is exactly $\theta$ (not $\pi + \theta$ ). The complementary arc gives argument $\theta - \pi$ . This is a frequent MCQ trap in JEE Main where two options differ by this arc vs full circle distinction.