Wait, why do we use $e^{i\theta}$ instead of just adding angles?

Great question! While you can think of it as adding angles, using $e^{i\theta}$ is the algebraic way to perform that rotation. In the complex plane, multiplying by $e^{i\theta}$ is a rotation operator. It is much more powerful than just adding angles because it allows you to treat the entire complex number as a single entity, making the math elegant and error-free.

JEE Advanced 1994

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: Suppose $Z_{1}, Z_{2}, Z_{3}$ are the vertices of an equilateral triangle inscribed in the circle $∣ Z ∣ = 2$ . If $Z_{1} = 1 + i 3$ then $Z_{2} = \dots, Z_{3} = \dots$

Visualized Solution

Visualizing the Circle and $Z_{1}$

The given circle equation is $∣ Z ∣ = 2$ , representing a circle centered at the origin with radius $2$ .
The first vertex is given as $Z_{1} = 1 + i 3$ .
Let's verify its distance from the origin: $∣ Z_{1} ∣ = 1^{2} + (3)^{2} = 4 = 2$ .
This confirms that $Z_{1}$ lies perfectly on the boundary of the circle.

Converting $Z_{1}$ to Euler Form

To perform rotations easily, we convert $Z_{1}$ to its Euler form: $Z_{1} = r e^{i θ}$ .
The modulus is $r = 2$ .
The argument is $θ = tan^{- 1} (\frac{3}{1}) = \frac{π}{3}$ radians ( $6 0^{\circ}$ ).
Thus, the Euler form is $Z_{1} = 2 e^{i \frac{π}{3}}$ .

Geometry of the Inscribed Triangle

An equilateral triangle is inscribed in the circle $∣ Z ∣ = 2$ .
The three vertices are equally spaced around the center.
The angle subtended by each side at the center is $\frac{2 π}{3}$ radians ( $12 0^{\circ}$ ).
Therefore, we can find the other vertices by rotating $Z_{1}$ by multiples of $\frac{2 π}{3}$ .

Setting up Rotation for $Z_{2}$

To find $Z_{2}$ , we rotate $Z_{1}$ counter-clockwise by $\frac{2 π}{3}$ radians.
In the complex plane, counter-clockwise rotation by angle $ϕ$ is achieved by multiplying by $e^{i ϕ}$ .
The rotation operator is $e^{i \frac{2 π}{3}}$ .
The setup is: $Z_{2} = Z_{1} \cdot e^{i \frac{2 π}{3}}$ .

Computing $Z_{2}$ in Euler Form

Substitute $Z_{1} = 2 e^{i \frac{π}{3}}$ into the setup: $Z_{2} = 2 e^{i \frac{π}{3}} \cdot e^{i \frac{2 π}{3}}$ .
Add the exponents: $Z_{2} = 2 e^{i (\frac{π}{3} + \frac{2 π}{3})}$ .
Simplify the exponent: $Z_{2} = 2 e^{iπ}$ .

Converting $Z_{2}$ to Cartesian Form

Use Euler's formula: $e^{iπ} = cos (π) + i sin (π)$ .
We know that $cos (π) = - 1$ and $sin (π) = 0$ .
Substitute these values: $Z_{2} = 2 (- 1 + 0 i) = - 2$ .

Setting up Rotation for $Z_{3}$

To find $Z_{3}$ , we rotate $Z_{2}$ counter-clockwise by another $\frac{2 π}{3}$ radians.
The rotation operator is again $e^{i \frac{2 π}{3}}$ .
The setup is: $Z_{3} = Z_{2} \cdot e^{i \frac{2 π}{3}}$ .

Computing $Z_{3}$ in Euler Form

Substitute $Z_{2} = 2 e^{iπ}$ into the setup: $Z_{3} = 2 e^{iπ} \cdot e^{i \frac{2 π}{3}}$ .
Add the exponents: $Z_{3} = 2 e^{i (π + \frac{2 π}{3})}$ .
Simplify the exponent: $Z_{3} = 2 e^{i \frac{5 π}{3}}$ .

Converting $Z_{3}$ to Cartesian Form

Use Euler's formula: $Z_{3} = 2 (cos (\frac{5 π}{3}) + i sin (\frac{5 π}{3}))$ .
Evaluate the trigonometric values: $cos (\frac{5 π}{3}) = \frac{1}{2}$ and $sin (\frac{5 π}{3}) = - \frac{3}{2}$ .
Simplify: $Z_3 = 2\left(\frac{1}{2} - i\frac{\sqrt{3}}{2} ight) = 1 - i\sqrt{3}$.

Verification and Centroid Check

The three vertices are: $Z_{1} = 1 + i 3$ , $Z_{2} = - 2$ , and $Z_{3} = 1 - i 3$ .
Let's check their sum: $Z_{1} + Z_{2} + Z_{3} = (1 + i 3) + (- 2) + (1 - i 3) = 0$ .
Since the sum is zero, the centroid $\frac{Z _{1} + Z _{2} + Z _{3}}{3}$ is at the origin $(0, 0)$ , confirming our solution is correct.

00:00 / 00:00

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Geometry of the Complex Plane

Imagine you are standing at the origin of the complex plane. You have a circle of radius

2

drawn around you, and on this circle, there is a point

Z_{1} = 1 + i 3

This point is not just a coordinate; it is the first vertex of an equilateral triangle. Our goal is to find the other two vertices,

Z_{2}

and

Z_{3}

. This problem is a beautiful exercise in symmetry and the power of complex numbers.

Phase 1

Visualizing the Playground

First, let us verify our starting point. The modulus of

Z_{1}

is:

∣ Z_{1} ∣ = 1^{2} + (3)^{2} = 4 = 2

It sits perfectly on the circle

∣ Z ∣ = 2

. To make our lives easier, we convert

Z_{1}

into Euler form. The modulus is

r = 2

, and the argument is

θ = tan^{- 1} (3 /1) = π /3

Thus,

Z_{1} = 2 e^{iπ /3}

. This form is our key to unlocking the rotation.

Phase 2

The Magic of Rotation

An equilateral triangle inscribed in a circle has a special property: its vertices are equally spaced. The angle subtended by each side at the center is

36 0^{\circ} /3 = 12 0^{\circ}

, or

2 π /3

radians.

In the complex plane, rotating a point by an angle

ϕ

is as simple as multiplying by

e^{i ϕ}

. This is the 'rotation operator.'

To find

Z_{2}

, we rotate

Z_{1}

2 π /3

counter-clockwise:

Z_{2} = Z_{1} \cdot e^{i 2 π /3}

Substituting our Euler form:

Z_{2} = 2 e^{iπ /3} \cdot e^{i 2 π /3} = 2 e^{i (π /3 + 2 π /3)} = 2 e^{iπ}

Using Euler's formula,

e^{iπ} = cos (π) + i sin (π) = - 1

. Therefore,

Z_{2} = 2 (- 1) = - 2

. The second vertex is simply

- 2

on the real axis.

Phase 3

Finding the Final Vertex

Now for

Z_{3}

. We rotate

Z_{2}

by another

2 π /3

radians:

Z_{3} = Z_{2} \cdot e^{i 2 π /3} = 2 e^{iπ} \cdot e^{i 2 π /3} = 2 e^{i (5 π /3)}

Converting this back to Cartesian form:

Z_{3} = 2 (cos (5 π /3) + i sin (5 π /3))

Since

5 π /3

is in the fourth quadrant,

cos (5 π /3) = 1/2

and

sin (5 π /3) = - 3 /2

. Thus:

Z_{3} = 2 (1/2 - i 3 /2) = 1 - i 3

Phase 4

The Final Verification

We have our vertices:

Z_{1} = 1 + i 3

Z_{2} = - 2

, and

Z_{3} = 1 - i 3

. Let us check the centroid:

Z_{1} + Z_{2} + Z_{3} = (1 + i 3) + (- 2) + (1 - i 3) = 0

The sum is zero, confirming the centroid is at the origin. The symmetry is perfect. You have just mastered the art of complex rotation!

Common Traps & FAQs

Wait, why do we use

e^{i θ}

instead of just adding angles?

Great question! While you can think of it as adding angles, using

e^{i θ}

is the algebraic way to perform that rotation. In the complex plane, multiplying by

e^{i θ}

is a rotation operator. It is much more powerful than just adding angles because it allows you to treat the entire complex number as a single entity, making the math elegant and error-free.

Can we use the centroid formula

Z_{1} + Z_{2} + Z_{3} = 0

directly?

Absolutely! For an equilateral triangle inscribed in a circle centered at the origin, the centroid is indeed at the origin. So,

Z_{1} + Z_{2} + Z_{3} = 0

is a valid property. You could use this to find the other vertices if you already knew one, but you would still need to know the relative positions of the vertices. The rotation method is more direct for finding the specific coordinates.

Is there a way to solve this using coordinates

(x, y)

instead of complex numbers?

Yes, you could use rotation matrices from coordinate geometry, but that is significantly more tedious. You would have to rotate the point

(1, 3)

12 0^{\circ}

using the matrix

(cos (12 0^{\circ}) sin (12 0^{\circ}) - sin (12 0^{\circ}) cos (12 0^{\circ}))

. Complex numbers are essentially a shortcut for this exact matrix operation, making them the preferred tool for JEE problems involving rotations.

Animated Solution for Mathematics - Complex Numbers: Suppose $Z_{1}, Z_{2}, Z_{3}$ are the vertices of an equilateral triangle inscribed in the circle $∣ Z ∣ = 2$ . If $Z_{1} = 1 + i 3$ then $Z_{2} = \dots, Z_{3} = \dots$

Visualized Solution

Visualizing the Circle and $Z_{1}$

Converting $Z_{1}$ to Euler Form

Geometry of the Inscribed Triangle

Setting up Rotation for $Z_{2}$

Computing $Z_{2}$ in Euler Form

Converting $Z_{2}$ to Cartesian Form

Setting up Rotation for $Z_{3}$

Computing $Z_{3}$ in Euler Form

Converting $Z_{3}$ to Cartesian Form

Verification and Centroid Check

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Geometry of the Complex Plane

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

If $a, b, c$ , are the numbers between 0 and 1 such that the points $z_{1} = a + i, z_{2} = 1 + bi$ and $z_{3} = 0$ form an equilateral triangle, then $a = \dots$ and $b = \dots$

The complex numbers $z_{1}, z_{2}$ and $z_{3}$ satisfying $\frac{z _{1} - z _{3}}{z _{2} - z _{3}} = \frac{1 - i 3}{2}$ are the vertices of a triangle which is

Let the complex number $z_{1}, z_{2}, z_{3}$ be the vertices of an equilateral triangle. Let $z_{0}$ be the circumcentre of the triangle. Then prove that $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}$ .

If one the vertices of the square circumscribing the circle $∣ z - 1∣ = 2$ is $2 + 3 i$ . Find the other vertices of the square.

If $z_{1}, z_{2}, z_{3} \in C$ are the vertices of an equilateral triangle, whose centroid is $z_{0}$ , then $\sum_{k = 1}^{3} (z_{k} - z_{0})^{2}$ is equal to

Prove that the complex numbers $z_{1}, z_{2}$ and the origin form an equilateral triangle only if $z_{1}^{2} + z_{2}^{2} - z_{1} z_{2} = 0$ .

Let $Z_{1}$ and $Z_{2}$ be two roots of the equation $Z^{2} + a Z + b = 0$ , $Z$ being complex. Further, assume that the origin, $Z_{1}$ and $Z_{2}$ form an equilateral triangle. Then

Let a complex number be $w = 1 - 3 i$ . Let another complex number $z$ be such that $∣ z w ∣ = 1$ and $ar g (z) - ar g (w) = \frac{π}{2}$ . Then the area of the triangle with vertices origin, $z$ and $w$ is equal to:

The area of the triangle with vertices $A (z), B (i z)$ and $C (z + i z)$ is :

Visualized Solution

Visualizing the Circle and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z1​

Converting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z1​ to Euler Form

Geometry of the Inscribed Triangle

Setting up Rotation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z2​

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z2​ in Euler Form

Converting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z2​ to Cartesian Form

Setting up Rotation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z3​

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z3​ in Euler Form

Converting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z3​ to Cartesian Form

Verification and Centroid Check

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Geometry of the Complex Plane

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a,b,c, are the numbers between 0 and 1 such that the points z1​=a+i,z2​=1+bi and z3​=0 form an equilateral triangle, then a=… and b=…

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The complex numbers z1​,z2​ and z3​ satisfying z2​−z3​z1​−z3​​=21−i3​​ are the vertices of a triangle which is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one the vertices of the square circumscribing the circle ∣z−1∣=2​ is 2+3​i. Find the other vertices of the square.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​,z2​,z3​∈C are the vertices of an equilateral triangle, whose centroid is z0​, then ∑k=13​(zk​−z0​)2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove that the complex numbers z1​,z2​ and the origin form an equilateral triangle only if z12​+z22​−z1​z2​=0.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Z1​ and Z2​ be two roots of the equation Z2+aZ+b=0, Z being complex. Further, assume that the origin, Z1​ and Z2​ form an equilateral triangle. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the triangle with vertices A(z),B(iz) and C(z+iz) is :

Visualizing the Circle and $Z_{1}$

Converting $Z_{1}$ to Euler Form

Setting up Rotation for $Z_{2}$

Computing $Z_{2}$ in Euler Form

Converting $Z_{2}$ to Cartesian Form

Setting up Rotation for $Z_{3}$

Computing $Z_{3}$ in Euler Form

Converting $Z_{3}$ to Cartesian Form

If $a, b, c$ , are the numbers between 0 and 1 such that the points $z_{1} = a + i, z_{2} = 1 + bi$ and $z_{3} = 0$ form an equilateral triangle, then $a = \dots$ and $b = \dots$

The complex numbers $z_{1}, z_{2}$ and $z_{3}$ satisfying $\frac{z _{1} - z _{3}}{z _{2} - z _{3}} = \frac{1 - i 3}{2}$ are the vertices of a triangle which is

If one the vertices of the square circumscribing the circle $∣ z - 1∣ = 2$ is $2 + 3 i$ . Find the other vertices of the square.

If $z_{1}, z_{2}, z_{3} \in C$ are the vertices of an equilateral triangle, whose centroid is $z_{0}$ , then $\sum_{k = 1}^{3} (z_{k} - z_{0})^{2}$ is equal to

Prove that the complex numbers $z_{1}, z_{2}$ and the origin form an equilateral triangle only if $z_{1}^{2} + z_{2}^{2} - z_{1} z_{2} = 0$ .

Let $Z_{1}$ and $Z_{2}$ be two roots of the equation $Z^{2} + a Z + b = 0$ , $Z$ being complex. Further, assume that the origin, $Z_{1}$ and $Z_{2}$ form an equilateral triangle. Then

The area of the triangle with vertices $A (z), B (i z)$ and $C (z + i z)$ is :