Secret Weapon Reading Time: 15 min JEE Advanced Focus

Rotation Theorem in Complex Numbers - A Secret Weapon

Transform complex geometry problems into simple algebraic equations. Master this powerful technique for JEE Advanced.

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Key Formulas

15min

Mastery Time

Why Rotation Theorem is Your Secret Weapon

The Rotation Theorem in complex numbers is one of the most powerful yet underutilized tools in JEE mathematics. It transforms complex geometric problems involving rotations, angles, and perpendicular lines into simple algebraic equations.

🎯 JEE Advantage

Solves geometry problems 5x faster than coordinate geometry
Eliminates the need for lengthy trigonometric calculations
Provides elegant one-line solutions to complex problems
Works for both JEE Main and Advanced level problems

🚀 Quick Navigation

Basic Concept Key Formulas Applications JEE Problems

1. The Fundamental Concept

What is Rotation in Complex Plane?

In the complex plane, multiplication by $e^{i\theta}$ rotates a complex number by angle $\theta$ counterclockwise about the origin.

Visual Representation:

If $z = re^{i\alpha}$, then:

$$z \cdot e^{i\theta} = re^{i(\alpha + \theta)}$$

This rotates $z$ by angle $\theta$ while maintaining the same magnitude.

Key Insight

Multiplication by $i$ rotates by $90^\circ$ counterclockwise:

$$z \cdot i = z \cdot e^{i\pi/2} \Rightarrow \text{Rotation by } 90^\circ$$

Multiplication by $-i$ rotates by $90^\circ$ clockwise:

$$z \cdot (-i) = z \cdot e^{-i\pi/2} \Rightarrow \text{Rotation by } -90^\circ$$

2. The Three Key Rotation Formulas

Formula 1: Rotation about Origin

$$z_2 = z_1 \cdot e^{i\theta}$$

Point $z_2$ is obtained by rotating $z_1$ by angle $\theta$ about origin.

When to Use:

When rotation is about origin
For problems involving regular polygons centered at origin
For finding vertices of rotated figures

Formula 2: Rotation about Arbitrary Point

$$z_3 = z_0 + (z_1 - z_0)e^{i\theta}$$

Point $z_3$ is obtained by rotating $z_1$ by angle $\theta$ about point $z_0$.

When to Use:

When rotation center is not at origin
For triangle rotation problems
For finding images of points after rotation

Formula 3: Perpendicular Lines Condition

$$\frac{z_3 - z_1}{z_2 - z_1} = \pm i \cdot k \quad (k \in \mathbb{R})$$

Line $z_1z_3$ is perpendicular to line $z_1z_2$.

When to Use:

For proving perpendicularity in geometry problems
For finding foot of perpendicular
For problems involving right angles

3. Powerful Applications

Application 1: Equilateral Triangles

For triangle $z_1z_2z_3$ to be equilateral (counterclockwise orientation):

$$z_2 - z_1 = (z_3 - z_1) \cdot e^{i\pi/3}$$ $$\text{OR}$$ $$z_1 + \omega z_2 + \omega^2 z_3 = 0 \quad \text{where } \omega = e^{i2\pi/3}$$

💡 Pro Tip

For clockwise orientation, use $e^{-i\pi/3}$ instead of $e^{i\pi/3}$.

Application 2: Square Properties

For square $z_1z_2z_3z_4$ (in order):

$$z_2 = z_1 + (z_4 - z_1) \cdot i$$ $$z_3 = z_1 + (z_4 - z_1) \cdot i^2 = z_1 - (z_4 - z_1)$$ $$z_4 = z_1 + (z_4 - z_1) \cdot i^3 = z_1 - (z_4 - z_1) \cdot i$$

Application 3: Regular Polygons

For regular n-gon with vertices $z_0, z_1, z_2, ..., z_{n-1}$:

$$z_k = z_0 + (z_1 - z_0) \cdot \omega^k \quad \text{where } \omega = e^{i2\pi/n}$$

Special Cases:

Equilateral Triangle: $\omega = e^{i2\pi/3}$
Square: $\omega = e^{i\pi/2} = i$
Regular Pentagon: $\omega = e^{i2\pi/5}$
Regular Hexagon: $\omega = e^{i\pi/3}$

4. JEE Level Problems Solved

JEE Advanced 2019 Rotation Theorem

Problem 1: Equilateral Triangle Construction

Let $z_1 = 2 + 3i$ and $z_2 = 3 + 2i$. Find $z_3$ such that $z_1z_2z_3$ forms an equilateral triangle.

Solution using Rotation Theorem:

Step 1: Vector from $z_1$ to $z_2$: $z_2 - z_1 = (3+2i) - (2+3i) = 1 - i$

Step 2: Rotate by $60^\circ$ counterclockwise: Multiply by $e^{i\pi/3} = \frac{1 + i\sqrt{3}}{2}$

Step 3: $(1-i) \cdot \frac{1 + i\sqrt{3}}{2} = \frac{(1-i)(1 + i\sqrt{3})}{2} = \frac{1 + i\sqrt{3} - i + \sqrt{3}}{2}$

Step 4: $= \frac{(1+\sqrt{3}) + i(\sqrt{3}-1)}{2}$

Step 5: $z_3 = z_1 + \frac{(1+\sqrt{3}) + i(\sqrt{3}-1)}{2}$

Final Answer: $z_3 = \frac{5+\sqrt{3}}{2} + i\frac{5+\sqrt{3}}{2}$

JEE Main 2021 Perpendicular Lines

Problem 2: Foot of Perpendicular

Find the foot of perpendicular from point $3 + 4i$ to the line joining $1 + i$ and $2 + 3i$.

Solution using Rotation Theorem:

Step 1: Let foot of perpendicular be $z$

Step 2: Using perpendicular condition: $\frac{z - (3+4i)}{(2+3i) - (1+i)} = ki$ where $k \in \mathbb{R}$

Step 3: $(2+3i) - (1+i) = 1 + 2i$

Step 4: So $z - (3+4i) = ki(1+2i) = k(-2 + i)$

Step 5: Also $z$ lies on line: $z = (1+i) + t(1+2i)$ for some $t \in \mathbb{R}$

Step 6: Solve to get $t = \frac{3}{2}$, $k = -\frac{1}{2}$

Final Answer: $z = \frac{5}{2} + 4i$

🎯 Memory Tips & Tricks

Quick Recall Formulas

$i$ = 90° rotation CCW → Multiply by $i$
$-i$ = 90° rotation CW → Multiply by $-i$
Perpendicular lines → Multiply by $\pm i$
60° rotation → Multiply by $\frac{1 \pm i\sqrt{3}}{2}$
120° rotation → Multiply by $\frac{-1 \pm i\sqrt{3}}{2}$

Common Patterns

Equilateral triangle → Use $e^{\pm i\pi/3}$
Square → Use powers of $i$
Regular hexagon → Use $e^{\pm i\pi/3}$ multiples
Right angle → Multiply by $\pm i$

📝 Practice Problems

Test your understanding with these JEE-level problems:

1. $z_1 = 1 + i$, $z_2 = 2 + 2i$. Find $z_3$ such that $z_1z_2z_3$ is an equilateral triangle.

Hint: Use 60° rotation

2. Prove that triangle with vertices $2+i$, $3+2i$, $1+3i$ is right-angled.

Hint: Use perpendicular condition

3. Find the square whose one diagonal has endpoints $1+2i$ and $3+4i$.

Hint: Rotate by 90° about midpoint

Mastered Rotation Theorem?

You now have a powerful secret weapon for JEE geometry problems!