The complex numbers z1, z2 and z3 satisfying (z1-z3)/(z2-z3) = (1-i√3)/2 are the vertices of a trian

Q: Wait, why do we use the ratio $\frac{z_1-z_3}{z_2-z_3}$ to find the triangle's shape? Can't we just use the distance formula?

You absolutely could use the distance formula, but it would be like using a sledgehammer to crack a nut! In the complex plane, the ratio $\frac{z_1-z_3}{z_2-z_3}$ is a powerful tool because it encodes both the relative lengths (modulus) and the relative orientation (argument) of the vectors in one single expression. It is the language of rotation and scaling, making it the most elegant way to solve geometry problems involving complex numbers.

Q: The argument is $-\frac{\pi}{3}$, which is $-60^\circ$. Does the negative sign mean the triangle is somehow 'negative' or invalid?

Not at all! The negative sign simply indicates the direction of rotation—clockwise. In geometry, when we talk about the interior angle of a triangle, we care about the magnitude of the rotation, which is $60^\circ$. The negative sign just tells you that to get from the vector $\vec{z_3z_2}$ to $\vec{z_3z_1}$, you rotate clockwise by $60^\circ$. The triangle itself is perfectly real and valid.

Q: I found that the triangle is isosceles because the sides are equal. Do I need to do anything else?

You are halfway there! Being isosceles is a great start, but in JEE Advanced, you must always check if the triangle has additional properties. An equilateral triangle is a special type of isosceles triangle. By using the angle information (the argument), you prove that the included angle is $60^\circ$, which forces the other two angles to be $60^\circ$ as well. Always check the angles to be sure!

Question

The complex numbers $z_1, z_2$ and $z_3$ satisfying $\frac{z_1-z_3}{z_2-z_3} = \frac{1-i\sqrt{3}}{2}$ are the vertices of a triangle which is

Options:

(A) of area zero

(B) right-angled isosceles

(C) equilateral

(D) obtuse-angled isosceles

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the complex plane, a vast, two-dimensional canvas where numbers are coordinates in space. You have three points, $z_1$, $z_2$, and $z_3$, forming the vertices of a triangle.

At first glance, the equation

$$ \frac{z_1-z_3}{z_2-z_3} = \frac{1-i\sqrt{3}}{2} $$

might look like a dry algebraic constraint. However, we can interpret this as a geometric map.

### The Vector Perspective

In the world of complex numbers, the difference $z_1-z_3$ is a vector representing the directed line segment starting at $z_3$ and ending at $z_1$. Similarly, $z_2-z_3$ is the vector from $z_3$ to $z_2$.

When we write their ratio, $\frac{z_1-z_3}{z_2-z_3}$, we are essentially asking: "How do I transform the vector $\vec{z_3z_2}$ into the vector $\vec{z_3z_1}$?"

This ratio is a complex number, and every complex number has two faces: its magnitude (how much it stretches) and its argument (how much it rotates).

### The Polar Transformation

Let us look at the right-hand side: $\frac{1}{2} - i\frac{\sqrt{3}}{2}$. To understand its geometric soul, we must convert it to polar form, $r e^{i	heta}$.

The magnitude $r$ is calculated as follows:

$$ r = \sqrt{\left(\frac{1}{2}ight)^2 + \left(-\frac{\sqrt{3}}{2}ight)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1 $$

This is a profound realization! Because the magnitude is $1$, there is no scaling. The vector $\vec{z_3z_2}$ is not stretched or compressed; it is merely rotated.

Now, for the angle $	heta$. Since the real part is positive and the imaginary part is negative, we are in the fourth quadrant. The angle $	heta$ is:

$$ 	heta = \arctan(-\sqrt{3}) = -\frac{\pi}{3} = -60^\circ $$

### The Synthesis

We now have our equation in its most revealing form:

$$ \frac{z_1-z_3}{z_2-z_3} = 1 \cdot e^{-i\frac{\pi}{3}} $$

This single equation tells us two things. First, by taking the modulus of both sides, we get $|z_1-z_3| = |z_2-z_3|$. This means the distance from $z_3$ to $z_1$ is equal to the distance from $z_3$ to $z_2$, confirming our triangle is **isosceles**.

Second, by taking the argument, we see that the angle between these two equal sides is $60^\circ$. An isosceles triangle with an included angle of $60^\circ$ is a special creature.

If the angle at the vertex is $60^\circ$, the remaining $120^\circ$ must be split equally between the other two angles, making them $60^\circ$ each. All three angles are $60^\circ$.

We have arrived at the perfect symmetry of an **equilateral triangle**. It is not just a triangle; it is a balanced, harmonious structure. You have successfully decoded the complex plane!

Animated Solution for Mathematics - Complex Numbers: 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

Select Answer:

Visualized Solution

The Complex Triangle

The Rotation Concept

Analyzing the Right Hand Side

Magnitude of the Ratio

Argument of the Ratio

Euler Form Representation

Equating Magnitudes

Isosceles Triangle Confirmed

Equating Arguments

The Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Vector Perspective

The Polar Transformation

The Synthesis

Similar Questions

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 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:

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 $b = \dots$

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 $Z_{2} = \dots, Z_{3} = \dots$

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 $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}$ .

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 :

Let O be the origin, the point A be $z_{1} = 3 + 22 i,$ the point $B (z_{2})$ be such that $3 ∣ z_{2} ∣ = ∣ z_{1} ∣$ and $a r g (z_{2}) = a r g (z_{1}) + \frac{π}{6}$ . Then

Show that the area of the triangle on the Argand diagram formed by the complex numbers $z, i z$ and $z + i z$ is $\frac{1}{2} ∣ z ∣^{2}$ .

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