What is a complex number?

A complex number is a number of the form z = a + ib, where a is the real part, b is the imaginary part, and i = √(-1) is the imaginary unit. The set of complex numbers is denoted by ℂ. Complex numbers extend the real number system to solve equations like x² + 1 = 0.

What is the Argand plane?

The Argand plane (or complex plane) is a two-dimensional plane used to represent complex numbers geometrically. The horizontal axis represents the real part and the vertical axis represents the imaginary part. A complex number z = a + ib is represented as the point (a, b) in this plane.

What is the polar form of a complex number?

The polar form of a complex number z = a + ib is z = r(cos θ + i sin θ) or z = r·e^(iθ), where r = |z| = √(a² + b²) is the modulus and θ = arg(z) = tan⁻¹(b/a) is the argument. This form is especially useful for multiplication, division, and finding powers of complex numbers.

What is De Moivre's Theorem?

De Moivre's Theorem states that for any complex number z = r(cos θ + i sin θ) and integer n: zⁿ = rⁿ(cos nθ + i sin nθ). This theorem is essential for finding powers and roots of complex numbers, and is frequently tested in JEE.

What are the cube roots of unity?

The cube roots of unity are the three solutions of x³ = 1. They are: 1, ω = (-1 + i√3)/2, and ω² = (-1 - i√3)/2. Important properties: 1 + ω + ω² = 0, ω³ = 1, and ω̄ = ω². These appear frequently in JEE questions.

How to find argument of a complex number?

The argument (arg z) of a complex number z = a + ib is the angle θ that the line joining origin to z makes with positive real axis. Principal argument lies in (-π, π]. For a > 0: θ = tan⁻¹(b/a). For a 0, -π/2 if b < 0.

Complex Numbers for JEE 2025-26 | Complete Notes with Formulas & Solved Examples

Introduction to Complex Numbers

Complex numbers extend the real number system to include solutions of equations like x² + 1 = 0. They are one of the most beautiful and powerful tools in mathematics, with applications in physics, engineering, and pure mathematics.

1.1 The Imaginary Unit

Definition of i (iota)

The imaginary unit i is defined as the square root of -1.

i = \sqrt{-1} \quad \text{or equivalently} \quad i^2 = -1

1.2 Powers of i

Cyclic Nature of Powers of i

\[i^1 = i\]

\[i^2 = -1\]

\[i^3 = -i\]

\[i^4 = 1\]

General Formula:

i^n = i^{n \mod 4}

Powers of i repeat with period 4.

💡 Quick Memory Trick

Remember "I Minus, Minus I, One"

i¹ = I
i² = Minus 1
i³ = Minus I
i⁴ = One

📝 Solved Example 1

Question: Find the value of i¹⁰⁷ + i¹⁰⁸ + i¹⁰⁹ + i¹¹⁰

Solution:

Step 1: Find remainder when divided by 4

107 ÷ 4 → remainder = 3 → i¹⁰⁷ = i³ = -i

108 ÷ 4 → remainder = 0 → i¹⁰⁸ = i⁴ = 1

109 ÷ 4 → remainder = 1 → i¹⁰⁹ = i¹ = i

110 ÷ 4 → remainder = 2 → i¹¹⁰ = i² = -1

Step 2: Add all values

\[= -i + 1 + i + (-1)\]

\[= 0\]

1.3 Definition of Complex Number

Standard Form

A complex number z is a number of the form:

\[z = a + ib\]

where:

a = Real part = Re(z)
b = Imaginary part = Im(z)
a, b ∈ ℝ (real numbers)

Set Notation:

\mathbb{C} = \{a + ib : a, b \in \mathbb{R}, i = \sqrt{-1}\}

1.4 Equality of Complex Numbers

Two complex numbers z₁ = a₁ + ib₁ and z₂ = a₂ + ib₂ are equal if and only if:

a_1 = a_2 \quad \text{and} \quad b_1 = b_2

In other words: Real parts are equal AND imaginary parts are equal.

📝 Solved Example 2

Question: Find real numbers x and y if (3x - 2) + (2y + 1)i = 7 + 5i

Solution:

Comparing real and imaginary parts:

Real parts: 3x - 2 = 7

3x = 9 \implies x = 3

Imaginary parts: 2y + 1 = 5

2y = 4 \implies y = 2

x = 3, \quad y = 2

Algebra of Complex Numbers

Complex numbers follow algebraic operations similar to real numbers, with the additional rule that i² = -1. Mastering these operations is essential for JEE.

2.1 Addition and Subtraction

Let z₁ = a₁ + ib₁ and z₂ = a₂ + ib₂

Addition:

\[z_1 + z_2 = (a_1 + a_2) + i(b_1 + b_2)\]

Subtraction:

\[z_1 - z_2 = (a_1 - a_2) + i(b_1 - b_2)\]

Rule: Add/subtract real parts and imaginary parts separately.

2.2 Multiplication

Multiply using distributive property and use i² = -1:

z_1 \cdot z_2 = (a_1 + ib_1)(a_2 + ib_2)

\[= a_1a_2 + ia_1b_2 + ib_1a_2 + i^2b_1b_2\]

\[= (a_1a_2 - b_1b_2) + i(a_1b_2 + a_2b_1)\]

📝 Solved Example 3

Question: Find (3 + 2i)(1 - 4i)

Solution:

\[(3 + 2i)(1 - 4i)\]

\[= 3(1) + 3(-4i) + 2i(1) + 2i(-4i)\]

\[= 3 - 12i + 2i - 8i^2\]

\[= 3 - 10i - 8(-1)\]

\[= 3 - 10i + 8\]

\[= 11 - 10i\]

2.3 Division

Method: Multiply by Conjugate

To divide, multiply numerator and denominator by the conjugate of the denominator:

\frac{z_1}{z_2} = \frac{a_1 + ib_1}{a_2 + ib_2} = \frac{(a_1 + ib_1)(a_2 - ib_2)}{(a_2 + ib_2)(a_2 - ib_2)}

= \frac{(a_1a_2 + b_1b_2) + i(a_2b_1 - a_1b_2)}{a_2^2 + b_2^2}

📝 Solved Example 4

Question: Express \(\frac{2 + 3i}{1 - 2i}\) in the form a + ib

Solution:

Multiply by conjugate of denominator (1 + 2i):

\frac{2 + 3i}{1 - 2i} \times \frac{1 + 2i}{1 + 2i}

Numerator:

\[(2 + 3i)(1 + 2i) = 2 + 4i + 3i + 6i^2 = 2 + 7i - 6 = -4 + 7i\]

Denominator:

\[(1 - 2i)(1 + 2i) = 1 - 4i^2 = 1 + 4 = 5\]

Result:

\frac{-4 + 7i}{5} = -\frac{4}{5} + \frac{7}{5}i

2.4 Properties of Operations

Property	Addition	Multiplication
Closure	z₁ + z₂ ∈ ℂ	z₁ · z₂ ∈ ℂ
Commutative	z₁ + z₂ = z₂ + z₁	z₁ · z₂ = z₂ · z₁
Associative	(z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)	(z₁ · z₂) · z₃ = z₁ · (z₂ · z₃)
Identity	z + 0 = z	z · 1 = z
Inverse	z + (-z) = 0	z · z⁻¹ = 1 (z ≠ 0)
Distributive	z₁(z₂ + z₃) = z₁z₂ + z₁z₃

Conjugate and Modulus

Conjugate and modulus are two fundamental concepts in complex numbers. They appear in almost every JEE problem and their properties must be memorized perfectly.

3.1 Complex Conjugate

Definition

The conjugate of z = a + ib is denoted by z̄ (z bar) and is defined as:

\bar{z} = a - ib

Geometrically: Reflection of z about the real axis.

3.2 Properties of Conjugate

Essential Properties (Must Memorize!)

1. Double Conjugate:

\overline{\bar{z}} = z

2. Sum of z and z̄:

z + \bar{z} = 2 \text{Re}(z)

3. Difference:

z - \bar{z} = 2i \cdot \text{Im}(z)

4. Product:

z \cdot \bar{z} = |z|^2

5. Addition:

\overline{z_1 + z_2} = \bar{z_1} + \bar{z_2}

6. Multiplication:

\overline{z_1 \cdot z_2} = \bar{z_1} \cdot \bar{z_2}

7. Division:

\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\bar{z_1}}{\bar{z_2}}

8. Power:

\overline{z^n} = (\bar{z})^n

⚠️ Important: When is z Real or Purely Imaginary?

z is real if and only if z = z̄
z is purely imaginary if and only if z = -z̄ (or z + z̄ = 0)

3.3 Modulus (Absolute Value)

Definition

The modulus (or absolute value) of z = a + ib is denoted by |z| and is defined as:

|z| = \sqrt{a^2 + b^2} = \sqrt{z \cdot \bar{z}}

Geometrically: Distance of z from the origin in the Argand plane.

3.4 Properties of Modulus

Essential Properties (Must Memorize!)

1. Non-negative:

|z| \geq 0

2. Modulus of Conjugate:

|\bar{z}| = |z|

3. Product Rule:

|z_1 \cdot z_2| = |z_1| \cdot |z_2|

4. Division Rule:

\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}

5. Power Rule:

\[|z^n| = |z|^n\]

6. Multiplicative Inverse:

z^{-1} = \frac{\bar{z}}{|z|^2}

🔥 Triangle Inequality (Most Important!)

||z_1| - |z_2|| \leq |z_1 \pm z_2| \leq |z_1| + |z_2|

Equality conditions:

|z₁ + z₂| = |z₁| + |z₂| when arg(z₁) = arg(z₂) (same direction)
|z₁ - z₂| = ||z₁| - |z₂|| when arg(z₁) = arg(z₂) (same direction)

📝 Solved Example 5 (JEE Main 2022 Type)

Question: If |z - 3 + 2i| = 4, find the maximum value of |z|.

Solution:

Using triangle inequality:

\[|z| = |z - (3-2i) + (3-2i)|\]

|z| \leq |z - (3-2i)| + |3-2i|

|z| \leq 4 + \sqrt{9+4}

|z| \leq 4 + \sqrt{13}

\text{Maximum value of } |z| = 4 + \sqrt{13}

Argand Plane

The Argand plane provides a geometric representation of complex numbers, making it easier to visualize and solve many problems.

4.1 Representation in Argand Plane

Geometric Representation

A complex number z = a + ib is represented as the point (a, b) in the Argand plane.

X-axis: Real axis (represents real part)
Y-axis: Imaginary axis (represents imaginary part)
Complex number z = a + ib → Point P(a, b)
Also represented as position vector OP

4.2 Distance Formula

The distance between two complex numbers z₁ = a₁ + ib₁ and z₂ = a₂ + ib₂ is:

|z_1 - z_2| = \sqrt{(a_1-a_2)^2 + (b_1-b_2)^2}

This is the same as the distance formula in coordinate geometry!

4.3 Section Formula

The point dividing the line joining z₁ and z₂ in the ratio m:n is:

Internal Division:

z = \frac{mz_2 + nz_1}{m + n}

External Division:

z = \frac{mz_2 - nz_1}{m - n}

Special Case - Midpoint:

z = \frac{z_1 + z_2}{2}

4.4 Argument of Complex Number

Definition

The argument of z = a + ib (denoted arg(z) or θ) is the angle that the line OP makes with the positive real axis.

\tan \theta = \frac{b}{a} = \frac{\text{Im}(z)}{\text{Re}(z)}

⚠️ Principal Argument (Very Important!)

The principal argument lies in (-π, π]. To find it, consider the quadrant:

Quadrant	a	b	Principal Argument
I	a > 0	b > 0	θ = tan⁻¹(b/a)
II	a < 0	b > 0	θ = π - tan⁻¹(\|b/a\|)
III	a < 0	b < 0	θ = -π + tan⁻¹(\|b/a\|)
IV	a > 0	b < 0	θ = -tan⁻¹(\|b/a\|)

📝 Solved Example 6

Question: Find the principal argument of z = -1 + i√3

Solution:

Here a = -1 and b = √3

Since a < 0 and b > 0, z is in Quadrant II

\tan^{-1}\left|\frac{b}{a}\right| = \tan^{-1}\left|\frac{\sqrt{3}}{-1}\right| = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}

For Quadrant II:

\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}

\text{arg}(-1 + i\sqrt{3}) = \frac{2\pi}{3}

4.5 Properties of Argument

Essential Properties

1. Product:

\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)

2. Division:

\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)

3. Conjugate:

\arg(\bar{z}) = -\arg(z)

4. Power:

\arg(z^n) = n \cdot \arg(z)

Note:

All argument properties are modulo 2π (results should be adjusted to principal value range)

Polar Form & Euler's Formula

The polar form is the most elegant way to represent complex numbers. It makes multiplication, division, and finding powers much easier.

5.1 Polar Form

Definition

A complex number z = a + ib can be written in polar form as:

z = r(\cos\theta + i\sin\theta)

where:

r = |z| = √(a² + b²) is the modulus
θ = arg(z) is the argument

Conversion Formulas:

Rectangular to Polar:

r = √(a² + b²)

θ = tan⁻¹(b/a)

Polar to Rectangular:

a = r cos θ

b = r sin θ

5.2 Euler's Formula

🔥 The Most Beautiful Formula in Mathematics!

e^{i\theta} = \cos\theta + i\sin\theta

Using Euler's formula, the polar form becomes:

z = re^{i\theta}

This is called the exponential form of a complex number.

💡 Special Cases of Euler's Formula

θ = π (Euler's Identity):

e^{i\pi} + 1 = 0

θ = π/2:

e^{i\pi/2} = i

θ = -π/2:

e^{-i\pi/2} = -i

θ = 2π:

e^{2\pi i} = 1

5.3 Multiplication and Division in Polar Form

Let z₁ = r₁e^(iθ₁) and z₂ = r₂e^(iθ₂)

Multiplication:

z_1 \cdot z_2 = r_1r_2 e^{i(\theta_1+\theta_2)}

Moduli multiply, arguments add

Division:

\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1-\theta_2)}

Moduli divide, arguments subtract

📝 Solved Example 7

Question: Express z = 1 + i in polar and exponential form.

Solution:

Step 1: Find modulus

r = |z| = \sqrt{1^2 + 1^2} = \sqrt{2}

Step 2: Find argument

Since a = 1 > 0 and b = 1 > 0, z is in Quadrant I

\theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}

Polar Form:

z = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)

Exponential Form:

z = \sqrt{2}e^{i\pi/4}

De Moivre's Theorem

De Moivre's theorem is one of the most powerful tools in complex numbers. It simplifies calculations involving powers and roots of complex numbers.

6.1 Statement of De Moivre's Theorem

🔥 De Moivre's Theorem

For any complex number z = r(cos θ + i sin θ) and integer n:

z^n = r^n(\cos n\theta + i\sin n\theta)

In exponential form:

(re^{i\theta})^n = r^n e^{in\theta}

💡 Quick Tip

For unit complex numbers (|z| = 1, so r = 1):

(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta

This is the most commonly used form in JEE problems!

📝 Solved Example 8 (JEE Advanced Type)

Question: Find (1 + i)¹⁰

Solution:

Step 1: Convert to polar form

|1 + i| = √2, arg(1 + i) = π/4

1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)

Step 2: Apply De Moivre's Theorem

(1+i)^{10} = (\sqrt{2})^{10}\left(\cos\frac{10\pi}{4} + i\sin\frac{10\pi}{4}\right)

= 32\left(\cos\frac{5\pi}{2} + i\sin\frac{5\pi}{2}\right)

Step 3: Simplify (5π/2 = 2π + π/2 = π/2)

= 32\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)

= 32(0 + i \cdot 1)

(1+i)^{10} = 32i

6.2 Applications: Finding cos nθ and sin nθ

Expansion Method

Using De Moivre's theorem and binomial expansion:

(\cos\theta + i\sin\theta)^n = \sum_{k=0}^{n} \binom{n}{k} \cos^{n-k}\theta \cdot (i\sin\theta)^k

Comparing real and imaginary parts:

cos nθ = sum of all terms with even powers of i (real parts)
sin nθ = sum of all terms with odd powers of i (imaginary parts, divided by i)

📝 Solved Example 9

Question: Express cos 3θ in terms of cos θ.

Solution:

Using De Moivre's theorem:

\cos 3\theta + i\sin 3\theta = (\cos\theta + i\sin\theta)^3

Expand using binomial theorem:

= \cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3

= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta

Comparing real parts:

\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta

= \cos^3\theta - 3\cos\theta(1-\cos^2\theta)

\cos 3\theta = 4\cos^3\theta - 3\cos\theta

Roots of Unity

Roots of unity are complex numbers that satisfy xⁿ = 1. They are crucial for JEE and appear in many algebraic and geometric problems.

7.1 nth Roots of Unity

Definition

The nth roots of unity are the solutions of zⁿ = 1.

z_k = e^{2\pi ik/n} = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}

where k = 0, 1, 2, ..., n-1

Key Points:

There are exactly n distinct nth roots of unity
All roots lie on the unit circle |z| = 1
Roots are equally spaced at angles of 2π/n
They form a regular n-gon in the Argand plane

7.2 Cube Roots of Unity

🔥 Most Important for JEE!

The cube roots of unity are the solutions of z³ = 1:

z₀ = 1

\[1\]

z₁ = ω

\omega = \frac{-1 + i\sqrt{3}}{2}

z₂ = ω²

\omega^2 = \frac{-1 - i\sqrt{3}}{2}

7.3 Properties of ω (Cube Root of Unity)

Essential Properties (Must Memorize!)

1. Sum of roots:

1 + \omega + \omega^2 = 0

2. Product of roots:

1 \cdot \omega \cdot \omega^2 = \omega^3 = 1

3. ω³ = 1:

\omega^3 = 1

4. Conjugate relation:

\bar{\omega} = \omega^2

5. ω and ω² are conjugates:

\omega + \omega^2 = -1

6. Product:

\omega \cdot \omega^2 = \omega^3 = 1

Cyclic Property:

\omega^{3n} = 1, \quad \omega^{3n+1} = \omega, \quad \omega^{3n+2} = \omega^2

📝 Solved Example 10 (JEE Main Type)

Question: Find the value of (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)

Solution:

First, simplify the powers using ω³ = 1:

ω⁴ = ω³ · ω = ω
ω⁸ = ω⁶ · ω² = ω²

So the expression becomes:

(1 + \omega)(1 + \omega^2)(1 + \omega)(1 + \omega^2)

= [(1 + \omega)(1 + \omega^2)]^2

Now, (1 + ω)(1 + ω²):

= 1 + \omega^2 + \omega + \omega^3

= 1 + (\omega + \omega^2) + 1

\[= 1 + (-1) + 1 = 1\]

\text{Answer} = 1^2 = 1

7.4 Properties of nth Roots of Unity

Let α = e^(2πi/n) be a primitive nth root of unity. Then:

1. Sum of all nth roots:

1 + \alpha + \alpha^2 + ... + \alpha^{n-1} = 0

2. Product of all nth roots:

\alpha^0 \cdot \alpha^1 \cdot \alpha^2 \cdot ... \cdot \alpha^{n-1} = (-1)^{n+1}

3. Geometric series formula:

\sum_{k=0}^{n-1} \alpha^k = \frac{\alpha^n - 1}{\alpha - 1} = 0

Geometry in Complex Plane

Complex numbers provide elegant solutions to geometric problems. Understanding these applications is crucial for JEE Advanced.

8.1 Geometric Interpretations

Key Geometric Meanings

1. |z₁ - z₂|:

Distance between points z₁ and z₂

2. arg(z₁ - z₂):

Angle that line z₁z₂ makes with positive real axis

3. |z - z₀| = r:

Circle with center z₀ and radius r

4. |z - z₁| = |z - z₂|:

Perpendicular bisector of line segment joining z₁ and z₂

8.2 Rotation Formula

🔥 Most Important for JEE Advanced!

To rotate point z about point z₀ by angle θ anticlockwise:

z' - z_0 = (z - z_0) \cdot e^{i\theta}

Or equivalently:

z' = z_0 + (z - z_0)(\cos\theta + i\sin\theta)

📝 Solved Example 11

Question: The point z = 2 + 3i is rotated about the origin by 90° anticlockwise. Find the new position.

Solution:

Here z₀ = 0 (origin), θ = 90° = π/2

z' = z \cdot e^{i\pi/2}

z' = (2 + 3i) \cdot (\cos 90° + i\sin 90°)

z' = (2 + 3i) \cdot (0 + i)

z' = (2 + 3i) \cdot i

\[z' = 2i + 3i^2 = 2i - 3\]

\[z' = -3 + 2i\]

8.3 Collinearity Condition

Three points z₁, z₂, z₃ are collinear if and only if:

\begin{vmatrix} z_1 & \bar{z_1} & 1 \\ z_2 & \bar{z_2} & 1 \\ z_3 & \bar{z_3} & 1 \end{vmatrix} = 0

Or equivalently:

\frac{z_3 - z_1}{z_2 - z_1} \text{ is purely real}

8.4 Equation of Line

1. Line through z₁ and z₂:

\frac{z - z_1}{\bar{z} - \bar{z_1}} = \frac{z_2 - z_1}{\bar{z_2} - \bar{z_1}}

2. General form:

\bar{a}z + a\bar{z} + b = 0

where a is complex and b is real

8.5 Equation of Circle

1. Circle with center z₀ and radius r:

\[|z - z_0| = r\]

2. General form:

z\bar{z} + \bar{a}z + a\bar{z} + b = 0

Center = -a, Radius = √(|a|² - b)

3. Circle with diameter z₁ and z₂:

\text{arg}\left(\frac{z - z_1}{z - z_2}\right) = \pm\frac{\pi}{2}

Or: (z - z₁)(z̄ - z̄₂) + (z - z₂)(z̄ - z̄₁) = 0

Locus Problems

Locus problems are favorites in JEE Advanced. They test your understanding of geometric interpretation of complex equations.

9.1 Standard Locus Equations

Equation	Locus
\|z - z₀\| = r	Circle with center z₀, radius r
\|z - z₁\| = \|z - z₂\|	Perpendicular bisector of z₁z₂
\|z - z₁\| + \|z - z₂\| = 2a	Ellipse with foci z₁, z₂ (if 2a > \|z₁ - z₂\|)
\|z - z₁\| - \|z - z₂\| = 2a	Hyperbola with foci z₁, z₂
\|z - z₁\| / \|z - z₂\| = k	Circle (Apollonius circle) if k ≠ 1, line if k = 1
arg(z - z₀) = θ	Ray from z₀ making angle θ with real axis
arg(z - z₁) - arg(z - z₂) = π/2	Major arc of circle with diameter z₁z₂
Re(z) = c	Vertical line x = c
Im(z) = c	Horizontal line y = c

📝 Solved Example 12 (JEE Advanced Type)

Question: Find the locus of z if |z - 2i| / |z + 2i| = 2

Solution:

Let z = x + iy

\frac{|z - 2i|}{|z + 2i|} = 2

\[|z - 2i| = 2|z + 2i|\]

\[|x + i(y-2)| = 2|x + i(y+2)|\]

\sqrt{x^2 + (y-2)^2} = 2\sqrt{x^2 + (y+2)^2}

Squaring both sides:

\[x^2 + (y-2)^2 = 4[x^2 + (y+2)^2]\]

\[x^2 + y^2 - 4y + 4 = 4x^2 + 4y^2 + 16y + 16\]

\[-3x^2 - 3y^2 - 20y - 12 = 0\]

x^2 + y^2 + \frac{20y}{3} + 4 = 0

x^2 + \left(y + \frac{10}{3}\right)^2 = \frac{100}{9} - 4 = \frac{64}{9}

\text{Circle with center } (0, -10/3) \text{ and radius } 8/3

Advanced Topics

10.1 Triangle Inequality Extended

For n complex numbers:

|z_1 + z_2 + ... + z_n| \leq |z_1| + |z_2| + ... + |z_n|

Equality holds when:

All complex numbers have the same argument (same direction)

10.2 nth Roots of a Complex Number

The nth roots of z = r(cos θ + i sin θ) are:

z_k = r^{1/n}\left[\cos\frac{\theta + 2\pi k}{n} + i\sin\frac{\theta + 2\pi k}{n}\right]

where k = 0, 1, 2, ..., n-1

Key Points:

There are exactly n distinct nth roots
All roots lie on a circle of radius r^(1/n)
Roots are equally spaced at angles of 2π/n

📝 Solved Example 13

Question: Find all cube roots of 8i.

Solution:

Step 1: Convert 8i to polar form

8i = 8(0 + i·1) = 8(cos π/2 + i sin π/2)

r = 8, θ = π/2

Step 2: Apply nth root formula with n = 3

z_k = 8^{1/3}\left[\cos\frac{\pi/2 + 2\pi k}{3} + i\sin\frac{\pi/2 + 2\pi k}{3}\right]

z_k = 2\left[\cos\frac{\pi + 4\pi k}{6} + i\sin\frac{\pi + 4\pi k}{6}\right]

Step 3: Find all three roots (k = 0, 1, 2)

k = 0:

z_0 = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = \sqrt{3} + i

k = 1:

z_1 = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) = 2\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -\sqrt{3} + i

k = 2:

z_2 = 2\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = 2(0 - i) = -2i

\text{Cube roots of } 8i: \quad \sqrt{3} + i, \quad -\sqrt{3} + i, \quad -2i

10.3 Logarithm of Complex Number

For z = re^(iθ):

\log z = \log r + i(\theta + 2n\pi)

where n is any integer. The principal value (n = 0) is:

\text{Log } z = \log|z| + i \cdot \text{Arg}(z)

10.4 Square Root Formula

For z = a + ib, the square roots are:

\sqrt{a + ib} = \pm\left[\sqrt{\frac{|z| + a}{2}} + i \cdot \text{sign}(b) \cdot \sqrt{\frac{|z| - a}{2}}\right]

where |z| = √(a² + b²) and sign(b) = +1 if b ≥ 0, -1 if b < 0

📝 Solved Example 14

Question: Find √(3 + 4i)

Solution:

Here a = 3, b = 4

|z| = \sqrt{9 + 16} = 5

Using the formula:

\sqrt{3 + 4i} = \pm\left[\sqrt{\frac{5+3}{2}} + i\sqrt{\frac{5-3}{2}}\right]

= \pm\left[\sqrt{4} + i\sqrt{1}\right]

\sqrt{3 + 4i} = \pm(2 + i)

10.5 Important Identities

Must Know Identities

1. Factorization:

x^n - 1 = (x-1)(x-\omega)(x-\omega^2)...(x-\omega^{n-1})

2. For cube roots:

x^3 - 1 = (x-1)(x-\omega)(x-\omega^2)

3. Sum formula:

x^3 + 1 = (x+1)(x+\omega)(x+\omega^2)

4. Useful identity:

a^3 + b^3 + c^3 - 3abc = (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)

🎯 JEE Strategy for Complex Numbers

Time allocation: Spend 5-6 days mastering this chapter (12-15% JEE weightage)
Focus areas: Modulus-Argument (30%), Roots of Unity (25%), Geometry (25%), Algebra (20%)
Practice: Minimum 150 problems covering all types
Visualization: Always draw Argand plane diagrams
Shortcuts: Memorize ω properties and rotation formulas

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Modulus and Argument: 30%
✓ Algebra of Complex Numbers: 25%
✓ Cube Roots of Unity: 20%
✓ Locus Problems: 15%
✓ De Moivre's Theorem: 10%

JEE Advanced (Last 5 Years)

✓ Geometry & Locus: 35%
✓ Roots of Unity: 25%
✓ Rotation & Transformation: 20%
✓ De Moivre & Powers: 15%
✓ Miscellaneous: 5%

Weightage Analysis

JEE Main: 12-16 marks (3-4 questions)

JEE Advanced: 15-24 marks (4-6 questions)

Difficulty Level: Medium to Difficult

Time Required: 5-6 hours practice

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

Find the value of i⁵⁷ + i⁵⁸ + i⁵⁹ + i⁶⁰.
If z = 3 + 4i, find |z| and arg(z).
Express (2 + 3i)/(1 - i) in the form a + ib.
Find the conjugate and modulus of z = (1 + i)/(1 - i).
If z = 1 + i, find z⁴ using De Moivre's theorem.
Prove that 1 + ω + ω² = 0 where ω is a cube root of unity.
Find the square roots of 5 + 12i.
If |z - 3| = |z + 3|, find the locus of z.

Level 2: Intermediate (JEE Main/Advanced)

If |z₁| = |z₂| = |z₃| = 1 and z₁ + z₂ + z₃ = 0, find |z₁² + z₂² + z₃²|.
Find all values of (−1 + i√3)^(2/3).
If arg(z - 1) = arg(z + 3i) = π/4, find z.
Prove that |z₁ + z₂|² + |z₁ - z₂|² = 2(|z₁|² + |z₂|²).
Find the value of (1 + ω - ω²)⁷ + (1 - ω + ω²)⁷.
If |z - 2i| ≤ 2, find the maximum value of |z + 3 + 4i|.
Find the locus of z if |z - 1 - i| = |z + 1 + i|.
Rotate z = 2 + i about the point 1 + i by 45° anticlockwise.

Level 3: Advanced (JEE Advanced/Olympiad)

If z₁, z₂, z₃ are vertices of an equilateral triangle inscribed in |z| = 1, prove z₁ + z₂ + z₃ = 0.
Find all complex numbers z such that z² = z̄.
If α = e^(2πi/7), find the value of α + α² + α⁴.
Prove that ∑(k=0 to n-1) cos(2πk/n) = 0 for n ≥ 2.
Find the locus of z if Re(z + 1)/(z - 1) = 0, z ≠ 1.
If |z - 4| + |z + 4| = 10, find the locus of z.
Find all cube roots of unity ω such that (1 + ω)⁷ = A + Bω, and find A and B.
If z^n = (z + 1)^n, find all values of z for n = 2, 3.

Get Solutions PDF with Step-by-Step Working

📋 Quick Formula Sheet

Basic Formulas

• i² = -1, i³ = -i, i⁴ = 1
• z·z̄ = |z|²
• z + z̄ = 2Re(z)
• z - z̄ = 2i·Im(z)
• |z₁z₂| = |z₁||z₂|
• arg(z₁z₂) = arg(z₁) + arg(z₂)

Polar & Euler

• z = r(cos θ + i sin θ) = re^(iθ)
• e^(iπ) + 1 = 0
• z^n = r^n(cos nθ + i sin nθ)
• e^(iθ) = cos θ + i sin θ

Cube Roots of Unity (ω)

• ω = (-1 + i√3)/2
• ω² = (-1 - i√3)/2
• 1 + ω + ω² = 0
• ω³ = 1
• ω̄ = ω²

Triangle Inequality

• ||z₁| - |z₂|| ≤ |z₁ ± z₂| ≤ |z₁| + |z₂|
• Rotation: z' - z₀ = (z - z₀)e^(iθ)
• Distance: |z₁ - z₂|

Download Complete Formula Sheet PDF

Year	JEE Main	JEE Advanced	Topic Focus
2024	4 Questions (16 marks)	5 Questions (20 marks)	Locus, Roots of unity, Modulus
2023	3 Questions (12 marks)	6 Questions (24 marks)	De Moivre, Rotation, Geometry
2022	4 Questions (16 marks)	4 Questions (16 marks)	Cube roots, Argument, Powers

📑 Quick Navigation - Complex Numbers

Complex Numbers JEE Main & Advanced 2025-26

Introduction to Complex Numbers

1.1 The Imaginary Unit

Definition of i (iota)

1.2 Powers of i

Cyclic Nature of Powers of i

💡 Quick Memory Trick

📝 Solved Example 1

1.3 Definition of Complex Number

Standard Form

1.4 Equality of Complex Numbers

📝 Solved Example 2

Algebra of Complex Numbers

2.1 Addition and Subtraction

2.2 Multiplication

📝 Solved Example 3

2.3 Division

Method: Multiply by Conjugate

📝 Solved Example 4

2.4 Properties of Operations

Conjugate and Modulus

3.1 Complex Conjugate

Definition

3.2 Properties of Conjugate

Essential Properties (Must Memorize!)

⚠️ Important: When is z Real or Purely Imaginary?

3.3 Modulus (Absolute Value)

Definition

3.4 Properties of Modulus

Essential Properties (Must Memorize!)

🔥 Triangle Inequality (Most Important!)

📝 Solved Example 5 (JEE Main 2022 Type)

Argand Plane

4.1 Representation in Argand Plane

Geometric Representation

4.2 Distance Formula

4.3 Section Formula

4.4 Argument of Complex Number

Definition

⚠️ Principal Argument (Very Important!)

📝 Solved Example 6

4.5 Properties of Argument

Essential Properties

Polar Form & Euler's Formula

5.1 Polar Form

Definition

5.2 Euler's Formula

🔥 The Most Beautiful Formula in Mathematics!

💡 Special Cases of Euler's Formula

5.3 Multiplication and Division in Polar Form

📝 Solved Example 7

De Moivre's Theorem

6.1 Statement of De Moivre's Theorem

🔥 De Moivre's Theorem

💡 Quick Tip

📝 Solved Example 8 (JEE Advanced Type)

6.2 Applications: Finding cos nθ and sin nθ

Expansion Method

📝 Solved Example 9

Roots of Unity

7.1 nth Roots of Unity

Definition

7.2 Cube Roots of Unity

🔥 Most Important for JEE!

7.3 Properties of ω (Cube Root of Unity)

Essential Properties (Must Memorize!)

📝 Solved Example 10 (JEE Main Type)

7.4 Properties of nth Roots of Unity

Geometry in Complex Plane

8.1 Geometric Interpretations

Key Geometric Meanings

8.2 Rotation Formula

🔥 Most Important for JEE Advanced!

📝 Solved Example 11

8.3 Collinearity Condition

8.4 Equation of Line

8.5 Equation of Circle

Locus Problems

9.1 Standard Locus Equations