Prove that the complex numbers z1, z2 and the origin form an equilateral triangle only if z1² + z2²

Q: Why can't we just use the distance formula $|z_1 - z_2| = |z_1| = |z_2|$? It seems more intuitive.

You absolutely can! Using the distance formula is a valid path, but it often leads to a mess of square roots and conjugates. The rotation method is the 'JEE way'—it's elegant, faster, and builds the intuition you need for more complex problems involving rotations and transformations.

Q: Why is there a $\pm$ sign in the rotation? Does it matter?

Great observation! The $\pm$ sign accounts for the orientation of the triangle. If you rotate $z_1$ by $+\frac{\pi}{3}$ (counter-clockwise), you get $z_2$. If you rotate it by $-\frac{\pi}{3}$ (clockwise), you also get $z_2$ (depending on which vertex is which). Since the final equation $z_1^2 + z_2^2 - z_1z_2 = 0$ is symmetric, the sign doesn't change the final result!

Q: Is this condition $z_1^2 + z_2^2 - z_1z_2 = 0$ only for the origin?

Yes, this specific form assumes the origin is one of the vertices. If the triangle had vertices at $z_1, z_2, z_3$, the condition would be $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$. This is a beautiful generalization that you should definitely memorize for your exam!

Question

Prove that the complex numbers $z_1, z_2$ and the origin form an equilateral triangle only if $z_1^2 + z_2^2 - z_1z_2 = 0$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITians. Today, we are going to peel back the layers of a classic JEE Advanced problem. We are not just solving an equation; we are exploring the profound geometric elegance of the Argand plane.

Imagine the complex plane as a vast, two-dimensional canvas. We have the origin, $O(0)$, acting as our anchor. We have two other complex numbers, $z_1$ and $z_2$, floating in this space.

When we connect these three points, we form a triangle. Because this is an equilateral triangle with one vertex at the origin, the distance from the origin to $z_1$ must equal the distance from the origin to $z_2$, and the angle between these two vectors must be exactly $\frac{\pi}{3}$ radians, or $60^\circ$.

### The Power of Rotation

In the world of complex numbers, we have a secret weapon: the rotation operator. When you multiply a complex number by $e^{i	heta}$, you are rotating it by an angle $	heta$ around the origin.

This means $z_2$ is simply $z_1$ rotated by $\frac{\pi}{3}$. Mathematically, we write this as:

$$z_2 = z_1 e^{\pm i\frac{\pi}{3}}$$

The $\pm$ sign is crucial—it accounts for whether we rotate counter-clockwise or clockwise.

### The Algebraic Dance

Now, let's get our hands dirty with the algebra. Our goal is to reach the condition $z_1^2 + z_2^2 - z_1z_2 = 0$.

Notice that our target equation has no exponential terms. To eliminate the exponential, we consider the ratio:

$$\frac{z_2}{z_1} = e^{\pm i\frac{\pi}{3}}$$

Taking the reciprocal, we have:

$$\frac{z_1}{z_2} = e^{\mp i\frac{\pi}{3}}$$

Now, add these two ratios together:

$$\frac{z_2}{z_1} + \frac{z_1}{z_2} = e^{i\frac{\pi}{3}} + e^{-i\frac{\pi}{3}}$$

Here, the magic happens. Euler's identity tells us that $e^{i	heta} + e^{-i	heta} = 2\cos(	heta)$. With $	heta = \frac{\pi}{3}$, we get:

$$2\cos\left(\frac{\pi}{3}ight) = 2 	imes \frac{1}{2} = 1$$

So, our equation simplifies to:

$$\frac{z_2}{z_1} + \frac{z_1}{z_2} = 1$$

### The Final Triumph

To clear the denominators, multiply the entire equation by $z_1z_2$. This gives us:

$$z_2^2 + z_1^2 = z_1z_2$$

Rearranging the terms, we arrive at the beautiful, symmetric condition:

**$$z_1^2 + z_2^2 - z_1z_2 = 0$$**

This is the hallmark of an equilateral triangle in the complex plane. Remember this result as a testament to the power of rotation and symmetry.

As a bonus, if you ever encounter a triangle with vertices $z_1, z_2, z_3$, the general condition is:

$$z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$$

Keep this in your toolkit, and you will be ready for any challenge the JEE throws at you. Keep practicing, keep visualizing, and most importantly, keep falling in love with the mathematics behind the problems.

Animated Solution for Mathematics - Complex Numbers: Prove that the complex numbers $z_{1}, z_{2}$ and the origin form an equilateral triangle only if $z_{1}^{2} + z_{2}^{2} - z_{1} z_{2} = 0$ .

Visualized Solution

The Argand Plane Setup

Properties of the Equilateral Triangle

The Concept of Rotation

Writing the Rotation Equation

Isolating the Exponential

Eliminating the Imaginary Exponent

Adding the Ratio and its Reciprocal

Applying Euler's Identity

Evaluating the Trigonometric Term

Simplifying the Left Hand Side

Arriving at the Final Condition

Conclusion and Generalization

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Power of Rotation

The Algebraic Dance

The Final Triumph

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