Rotation Theorem (The "Coni" Method) for JEE Main – Master Guide

Introduction

The Rotation Theorem transforms geometric problems into simple algebra. For triangles, squares, and polygons, it's often the fastest JEE approach. Also called the Coni Method (COmplex-NImber method).

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The Fundamental Formula

For rotation about point $z_0$ by angle $\theta$ (anticlockwise): $\boxed{\frac{z_{\text{final}} - z_0}{z_{\text{initial}} - z_0} = e^{i\theta}}$

Immediate Consequences:

1. Rotating $z_1$ about $z_0$ by $\theta$ to get $z_2$: $z_2 = z_0 + (z_1 - z_0)e^{i\theta}$

2. Magnitude preserved: $|z_2 - z_0| = |z_1 - z_0|$

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Essential Rotation Factors (Memorize!)

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Equilateral Triangles – 4 Key Formulas

For vertices $A(z_1)$, $B(z_2)$, $C(z_3)$:

1. Rotation Formula (Most Useful): $z_3 - z_1 = (z_2 - z_1)e^{\pm i\pi/3}$ (Sign depends on orientation)

2. Symmetric Condition: $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$

3. With Origin as One Vertex: If $z_1 = 0$, then: $z_2^2 + z_3^2 = z_2z_3 \quad \text{and} \quad z_3 = z_2 e^{\pm i\pi/3}$

4. Centroid/Circumcenter Relations: If centroid at origin: $z_1 + z_2 + z_3 = 0$
If circumcenter at origin: $z_1^2 + z_2^2 + z_3^2 = 0$

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Squares – Quick Formulas

For consecutive vertices $z_1, z_2, z_3, z_4$:

1. Adjacent Side Rotation: $z_2 - z_1 = i(z_4 - z_1)$

2. Diagonal Rotation: $z_3 - z_1 = i(z_4 - z_2)$

3. Center Property: $\frac{z_1 + z_3}{2} = \frac{z_2 + z_4}{2}$ (Point where diagonals bisect)

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Isosceles Right Triangles

For right angle at $z_3$ with equal legs $z_1z_3$ and $z_2z_3$: $\frac{z_1 - z_3}{z_2 - z_3} = \pm i$ (Plus for anticlockwise, minus for clockwise)

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JEE PYQs Solved with Rotation

PYQ 1: JEE Main 2021

Let $z_1, z_2$ be roots of $z^2 + az + 12 = 0$ forming equilateral triangle with origin. Find $|a|$.

Solution (60 seconds): For vertices $0, z_1, z_2$ to be equilateral: $z_1^2 + z_2^2 = z_1z_2$

From equation: $z_1 + z_2 = -a$, $z_1z_2 = 12$

Also: $z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1z_2 = a^2 - 24$

Equate: $a^2 - 24 = 12$ $a^2 = 36$, $|a| = 6$

Answer: $6$

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PYQ 2: JEE Main 2019

If $z$ lies on circle $|z| = 1$, and $\frac{z-1}{z+1}$ is purely imaginary, prove $|z| = 1$.

Rotation Insight: $\frac{z-1}{z+1}$ purely imaginary ⇒ $\arg\left(\frac{z-1}{z+1}\right) = \pm\pi/2$

This means angle between vectors $(z-1)$ and $(z+1)$ is $90°$.

By geometry, $z$ lies on circle with diameter endpoints at $-1$ and $1$ ⇒ Circle: $|z| = 1$

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PYQ 3: JEE Main Pattern

Vertices $z_1, z_2, z_3$ of equilateral triangle with circumradius 2. If $z_1 = 1+i\sqrt{3}$, find $z_2, z_3$.

Rotation Method: Since circumcenter at origin, $z_1, z_2, z_3$ are rotations of each other by $120°$:

$z_2 = z_1 \cdot \omega = (1+i\sqrt{3})\left(\frac{-1+i\sqrt{3}}{2}\right) = -2$

$z_3 = z_1 \cdot \omega^2 = (1+i\sqrt{3})\left(\frac{-1-i\sqrt{3}}{2}\right) = 1-i\sqrt{3}$

Answer: $z_2 = -2$, $z_3 = 1-i\sqrt{3}$

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PYQ 4: Classic Problem

$ABCD$ is a square with $A(z_1)$ and $B(z_2)$ adjacent vertices. Find $D$ in terms of $z_1, z_2$.

Rotation Solution: $D$ is obtained by rotating $A$ about $B$ by $-90°$ (or $A$ about midpoint by $90°$):

Actually simpler: $AD = i \cdot AB$

So: $D - A = i(B - A)$ $D = A + i(B - A) = (1-i)A + iB$

Check: If $A=0$, $B=1$, then $D = i$ ✓

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Step-by-Step Problem Solving

Type 1: Equilateral Triangle Problems

Given: Two vertices, find third Step 1: Identify center (often midpoint or one vertex) Step 2: Use $z_3 - z_{\text{center}} = (z_1 - z_{\text{center}})e^{\pm i\pi/3}$ Step 3: Solve algebraically

Type 2: Square Problems

Given: Two vertices, find others Step 1: Determine if given vertices are adjacent or diagonal Step 2: Apply rotation by $90°$ appropriately Step 3: Use center property for verification

Type 3: Rotation About Origin

Simplest case: $z' = z \cdot e^{i\theta}$ Use for inscribed polygons in $|z|=R$ circles

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Common Pitfalls & Checks

1. Orientation matters: $\pm$ sign in rotation
2. Center selection: Rotations are about a point, choose wisely
3. Order of vertices: Consecutive order affects rotation direction
4. Verify: Check lengths remain equal after rotation

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Advanced Applications

1. Regular n-gon Inscribed in $|z|=R$

Vertices: $z_k = R \cdot e^{i(2\pi k/n + \theta_0)}$ for $k=0,1,...,n-1$

2. Similar Triangles

If $\triangle ABC \sim \triangle PQR$, then: $\frac{c-a}{b-a} = \frac{r-p}{q-p}$ (ratio of corresponding sides with rotation)

3. Collinearity Check

Points $z_1, z_2, z_3$ collinear if: $\frac{z_3 - z_1}{z_2 - z_1} \ \text{is real}$

4. Perpendicularity Check

Lines $z_1z_2$ and $z_3z_4$ perpendicular if: $\frac{z_2 - z_1}{z_4 - z_3} \ \text{is purely imaginary}$

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Quick Reference Table

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Practice Problems

Problem 1: $z_1=2+3i$, $z_2=5+i$ form adjacent vertices of a square. Find other vertices.

Solution: $AB = z_2 - z_1 = 3-2i$ $AD = i \cdot AB = i(3-2i) = 2+3i$ $D = A + AD = (2+3i)+(2+3i)=4+6i$ $C = D + AB = (4+6i)+(3-2i)=7+4i$

Answer: $C=7+4i$, $D=4+6i$

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Problem 2: Equilateral triangle with vertices $0$, $1$, and $z$. Find $z$.

Solution: $z - 0 = (1-0)e^{\pm i\pi/3}$ $z = e^{\pm i\pi/3} = \frac{1}{2}(1 \pm i\sqrt{3})$

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Problem 3: $z_1=1$, $z_2=2i$ are ends of hypotenuse of isosceles right triangle. Find third vertex.

Solution: Midpoint $M = \frac{1+2i}{2}$ Third vertex $z_3 = M \pm i\left(\frac{z_2-z_1}{2}\right)$ $= \frac{1+2i}{2} \pm i\left(\frac{2i-1}{2}\right)$

Two solutions: $z_3 = \frac{1}{2}(3+i)$ or $\frac{1}{2}(-1+3i)$

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Exam Strategy

1. Identify figure type immediately
2. Choose simplest center (often origin or given vertex)
3. Apply rotation formula directly
4. Check with magnitude if uncertain
5. Time target: 1-2 minutes for rotation problems

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Pro Tip: For origin-centered problems, remember:

• Multiplication by $e^{i\theta}$ = rotation by $\theta$
• Multiplication by $i$ = 90° rotation
• Multiplication by $\omega$ = 120° rotation

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Related: Complex Geometry | Cube Roots of Unity | Inequalities