Rotation And Geometric Applications

Q: What is the rotation formula for complex numbers?

If z₁, z₂, z₃ are three points with α being the angle at vertex z₁: (z₃ − z₁)/(z₂ − z₁) = (|z₃ − z₁|/|z₂ − z₁|) · e^(iα). It gives both the ratio of magnitudes and the rotation angle in one equation.

Q: What is the section formula for complex numbers?

If C divides AB internally in ratio m:n: z = (nz₁ + mz₂)/(m + n). Externally: z = (nz₁ − mz₂)/(m − n). Midpoint: z = (z₁ + z₂)/2.

Q: How to find the centroid of a triangle using complex numbers?

If the vertices are z₁, z₂, z₃, the centroid is G = (z₁ + z₂ + z₃)/3 — the average of the three vertices. This is the simplest of the four triangle centres.

Q: What is the incentre formula in complex numbers?

I = (a·z₁ + b·z₂ + c·z₃)/(a + b + c), where a = |z₂ − z₃|, b = |z₃ − z₁|, c = |z₁ − z₂| are the side lengths opposite to each vertex.

Q: How do you check if three complex numbers are collinear?

Three complex numbers z₁, z₂, z₃ are collinear if there exist real numbers l, m, n (not all zero) with lz₁ + mz₂ + nz₃ = 0 and l + m + n = 0. Equivalently, a determinant with z, z̄, 1 columns vanishes.

Q: How to rotate a complex number by 90°?

Rotating a complex number z by 90° anti-clockwise about the origin gives iz (multiplication by i = e^(iπ/2)). Clockwise gives −iz. About a general point z₀: rotated point is z₀ + i(z − z₀).

Q: What are the coordinates of a square's vertices in complex form?

If two adjacent vertices are z₁, z₂: opposite vertex is z₃ = z₁ + (z₂ − z₁)(1 + i) and fourth vertex is z₄ = z₁ + i(z₂ − z₁) — derived by rotating AB by π/4 and π/2 respectively.

MathsComplex NumbersFor JEE aspirants

Rotation of a complex number means turning it about a point by a given angle. If $z$ and $z^{'}$ are complex numbers, then $ar g (z / z^{'})$ is the angle through which $O z^{'}$ must be rotated to lie along $O z$ . Combined with the section formula, rotation gives a powerful tool for solving geometric problems in the Argand plane - locating vertices of squares and triangles, finding centroids, incentres, and orthocentres, and testing collinearity. This concept covers all core geometrical applications of complex numbers for JEE.

Key Formulas - Quick Reference

Rotation formula (triangle): $\frac{z _{3} - z _{1}}{z _{2} - z _{1}} = \frac{∣ z _{3} - z _{1} ∣}{∣ z _{2} - z _{1} ∣} \cdot e^{i α}$ where $α$ = angle $∠ B A C$
Section formula (internal, $m : n$ ): $z = \frac{n z _{1} + m z _{2}}{m + n}$
Section formula (external, $m : n$ ): $z = \frac{n z _{1} - m z _{2}}{m - n}$
Midpoint: $z = \frac{z _{1} + z _{2}}{2}$
Centroid of $△$ : $G = \frac{z _{1} + z _{2} + z _{3}}{3}$
Incentre of $△$ : $I = \frac{a z _{1} + b z _{2} + c z _{3}}{a + b + c}$ (where $a, b, c$ are side lengths)
Circumcentre: $O = \frac{z _{1} sin 2 A + z _{2} sin 2 B + z _{3} sin 2 C}{sin 2 A + sin 2 B + sin 2 C}$
Orthocentre: $P = \frac{z _{1} tan A + z _{2} tan B + z _{3} tan C}{tan A + tan B + tan C}$
Collinearity: $z_{1}, z_{2}, z_{3}$ collinear iff $\exists l, m, n \neq = 0$ with $l z_{1} + m z_{2} + n z_{3} = 0$ and $l + m + n = 0$

1. Concept of Rotation

Definition If

z

and

z^{'}

are two complex numbers, then

ar g (z / z^{'})

is the angle through which

O z^{'}

must be rotated (in the counter-clockwise direction) to align with

O z

\frac{z}{z ^{'}} = \frac{∣ z ∣ e ^{i θ}}{∣ z ^{'} ∣ e ^{i θ^{'}}} = \frac{∣ z ∣}{∣ z ^{'} ∣} e^{i (θ - θ^{'})} = \frac{∣ z ∣}{∣ z ^{'} ∣} e^{i α}

Rotation about a Vertex of a Triangle

Let $z_{1}, z_{2}, z_{3}$ be vertices $A, B, C$ of a triangle (in counter-clockwise sense). Drawing $O P$ and $O Q$ parallel and equal to $A B$ and $A C$ respectively:

$P$ corresponds to $z_{2} - z_{1}$
$Q$ corresponds to $z_{3} - z_{1}$

Rotation Formula

\frac{z _{3} - z _{1}}{z _{2} - z _{1}} = \frac{∣ z _{3} - z _{1} ∣}{∣ z _{2} - z _{1} ∣} \cdot e^{i α} = \frac{C A}{B A} \cdot e^{i α}

where

α

∠ B A C

, the angle through which

A B

must rotate anti-clockwise to align with

A C

Figure 8: Rotation of vector AB to AC by angle α at vertex A(z₁) in the Argand plane

Clockwise Rotation

Equivalently, rotating clockwise by $(2 π - α)$ :

\frac{z _{3} - z _{1}}{z _{2} - z _{1}} = \frac{∣ z _{3} - z _{1} ∣}{∣ z _{2} - z _{1} ∣} \cdot e^{- i (2 π - α)}

Sign Convention Anti-clockwise rotation ⇒ positive angle. Clockwise rotation ⇒ negative angle. Get this wrong and answers flip in sign.

Solved Example 1

Consider a square

A B C D

with vertices

z_{1}, z_{2}, z_{3}, z_{4}

. Express

z_{3}

and

z_{4}

in terms of

z_{1}

and

z_{2}

Solution:

Rotating $A B$ about $A$ by $π /4$ to reach $A C$ : Since $∣ A C ∣/∣ A B ∣ = 2$ (diagonal of unit square):

\frac{z _{3} - z _{1}}{z _{2} - z _{1}} = 2 (cos \frac{π}{4} + i sin \frac{π}{4}) = 2 \cdot \frac{1 + i}{2} = 1 + i

\Rightarrow z_{3} = z_{1} + (z_{2} - z_{1}) (1 + i)

Rotating $A B$ about $A$ by $π /2$ to reach $A D$ : Since $∣ A D ∣/∣ A B ∣ = 1$ :

\frac{z _{4} - z _{1}}{z _{2} - z _{1}} = e^{iπ /2} = i \Rightarrow z_{4} = z_{1} + i (z_{2} - z_{1})

Solved Example 2

∣ z - 3∣ = 3

, show that

\frac{z - 6}{z} = i tan (ar g z)

Solution:

$∣ z - 3∣ = 3$ is a circle with centre $(3, 0)$ and radius $3$ - it passes through origin $O$ and the point $A (6, 0)$ .

Since $O A$ is the diameter, $∠ O P A = π /2$ for any point $P (z)$ on the circle (angle in semicircle).

Applying rotation formula at $P$ : rotating $P A$ (which is $6 - z$ ) to $P O$ (which is $- z$ ) by $π /2$ :

\frac{z - 6}{z - 0} = \frac{∣ z - 6∣}{∣ z ∣} \cdot e^{iπ /2} = \frac{P A}{P O} \cdot i = i tan θ

where $θ = ar g z$ . Hence $\frac{z - 6}{z} = i tan (ar g z)$ .

2. Section Formula

Internal Division Let

z_{1}, z_{2}

represent points

A, B

. If

C (z)

divides

A B

internally in ratio

m : n

(i.e.

A C / C B = m / n

z = \frac{n z _{1} + m z _{2}}{m + n}

External Division If

C (z)

divides

A B

externally in ratio

m : n

z = \frac{n z _{1} - m z _{2}}{m - n}

Midpoint (Special Case,

m = n

)

z = \frac{z _{1} + z _{2}}{2}

Derivation via Rotation

At point $C$ , the vectors $C A$ and $C B$ point in opposite directions (angle $π$ ). Applying rotation:

\frac{z _{1} - z}{z _{2} - z} = \frac{∣ z _{1} - z ∣}{∣ z _{2} - z ∣} e^{iπ} = \frac{m}{n} \cdot (- 1)

Solving: $n (z_{1} - z) = - m (z_{2} - z) \Rightarrow z = \frac{n z _{1} + m z _{2}}{m + n}$ .

3. Centres of a Triangle

For a triangle with vertices $z_{1}, z_{2}, z_{3}$ (in anti-clockwise sense), the four important centres are given by:

Centroid

G = \frac{z _{1} + z _{2} + z _{3}}{3}

Derivation: Midpoint of $B C$ is $D = (z_{2} + z_{3}) /2$ . Centroid divides $A D$ in ratio $2 : 1$ :

G = \frac{1 \cdot z _{1} + 2 \cdot ( z _{2} + z _{3} ) /2}{2 + 1} = \frac{z _{1} + z _{2} + z _{3}}{3}

Incentre

I = \frac{a z _{1} + b z _{2} + c z _{3}}{a + b + c}

where

a = ∣ z_{2} - z_{3} ∣

b = ∣ z_{3} - z_{1} ∣

c = ∣ z_{1} - z_{2} ∣

are the side lengths opposite to vertices

z_{1}, z_{2}, z_{3}

Circumcentre

O = \frac{z _{1} sin 2 A + z _{2} sin 2 B + z _{3} sin 2 C}{sin 2 A + sin 2 B + sin 2 C}

Orthocentre

P = \frac{z _{1} tan A + z _{2} tan B + z _{3} tan C}{tan A + tan B + tan C}

Note The centroid

G

, circumcentre

O

, and orthocentre

P

are collinear on the Euler line, with

G

dividing

O P

in the ratio

1 : 2

4. Condition for Collinearity

Three complex numbers

z_{1}, z_{2}, z_{3}

are collinear if and only if there exist three real numbers

l, m, n

(not all zero) such that:

l z_{1} + m z_{2} + n z_{3} = 0 and l + m + n = 0

Equivalently, they are collinear iff:

z_{1} \overset{z}{ˉ}_{1} 1 z_{2} \overset{z}{ˉ}_{2} 1 z_{3} \overset{z}{ˉ}_{3} 1 = 0

5. Equation of a Straight Line

Line Through Two Points

z_{1}

and

z_{2}

Using

x = \frac{z + z ˉ}{2}

y = \frac{z - z ˉ}{2 i}

and the two-point form:

\frac{z - z _{1}}{z _{2} - z _{1}} = \frac{z ˉ - z ˉ _{1}}{z ˉ _{2} - z ˉ _{1}}

Or equivalently, in determinant form:

z \overset{z}{ˉ} 1 z_{1} \overset{z}{ˉ}_{1} 1 z_{2} \overset{z}{ˉ}_{2} 1 = 0

Common Mistakes to Avoid

Watch out

Wrong sign for rotation direction. Anti-clockwise = positive angle; clockwise = negative. Getting this backwards flips the entire result.
Confusing $(n z_{1} + m z_{2})$ vs $(m z_{1} + n z_{2})$ in section formula. The coefficient of $z_{2}$ is $m$ (the ratio-part closer to $B$ ), not the coefficient of $z_{1}$ .
Forgetting the ratio direction. $m : n$ vs $n : m$ gives different points. Always identify which vertex divides in what ratio.
Using wrong side-length convention for incentre. $a$ is opposite $z_{1}$ , $b$ opposite $z_{2}$ , $c$ opposite $z_{3}$ . Swapping these gives the wrong point.
Using the collinearity determinant without $\overset{z}{ˉ}$ . Complex-plane collinearity requires the $\overset{z}{ˉ}$ column - omitting it gives a real-plane condition that's not equivalent.

Frequently Asked Questions

What is the rotation formula for complex numbers?

If $z_{1}, z_{2}, z_{3}$ are three points with $α$ being the angle at vertex $z_{1}$ : $\frac{z _{3} - z _{1}}{z _{2} - z _{1}} = \frac{∣ z _{3} - z _{1} ∣}{∣ z _{2} - z _{1} ∣} e^{i α}$ . It gives both the ratio of magnitudes and the rotation angle in one equation.

What is the section formula for complex numbers?

If $C$ divides $A B$ internally in ratio $m : n$ : $z = \frac{n z _{1} + m z _{2}}{m + n}$ . Externally: $z = \frac{n z _{1} - m z _{2}}{m - n}$ . Midpoint: $z = (z_{1} + z_{2}) /2$ .

How to find the centroid of a triangle using complex numbers?

If the vertices are $z_{1}, z_{2}, z_{3}$ , the centroid is $G = \frac{z _{1} + z _{2} + z _{3}}{3}$ - the average of the three vertices. This is the simplest of the four triangle centres.

What is the incentre formula in complex numbers?

$I = \frac{a z _{1} + b z _{2} + c z _{3}}{a + b + c}$ , where $a = ∣ z_{2} - z_{3} ∣$ , $b = ∣ z_{3} - z_{1} ∣$ , $c = ∣ z_{1} - z_{2} ∣$ are the side lengths opposite to each vertex.

How do you check if three complex numbers are collinear?

Three complex numbers $z_{1}, z_{2}, z_{3}$ are collinear if $\exists$ real numbers $l, m, n$ (not all zero) with $l z_{1} + m z_{2} + n z_{3} = 0$ and $l + m + n = 0$ . Equivalently, the determinant $z_{i} \overset{z}{ˉ}_{i} 1 = 0$ .

How to rotate a complex number by 90°?

Rotating a complex number $z$ by $90°$ anti-clockwise about the origin gives $i z$ (multiplication by $i = e^{iπ /2}$ ). Clockwise gives $- i z$ . About a general point $z_{0}$ : rotated point is $z_{0} + i (z - z_{0})$ .

What are the coordinates of a square's vertices in complex form?

If two adjacent vertices are $z_{1}, z_{2}$ : opposite vertex is $z_{3} = z_{1} + (z_{2} - z_{1}) (1 + i)$ and fourth vertex is $z_{4} = z_{1} + i (z_{2} - z_{1})$ - derived by rotating $A B$ by $π /4$ and $π /2$ respectively.

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