Problem 113 - A Complete Resource Book in Mathematics for JEE Main

Question

Suppose, \(z_{1}, z_{2}, z_{3}\) are the vertices of an equilateral triangle inscribed in the circle \(|z|=2 .\) If \(z_{1}=1+i \sqrt{3}\) then \(z_{2}\) and \(z_{3}\) are equal to (A) \(-2,1-i \sqrt{3}\) (B) \(2,1-i \sqrt{3}\) (C) \(-2,1+i \sqrt{3}\) (D) None of these

Step-by-Step Solution

Answer

Option (A) is correct: \(z_2 = -2\) and \(z_3 = 1 - i\sqrt{3}\).

Step 1: Calculate the magnitude of \(z_1\)

Given that \(z_1 = 1 + i\sqrt{3}\), we start by calculating its magnitude using the formula for the modulus of a complex number: \[ |z_1| = \sqrt{(\text{real part})^2 + (\text{imaginary part})^2} \]Substituting the real part \(1\) and the imaginary part \(\sqrt{3}\):\[ |z_1| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \] Since \(|z_1| = 2\), it satisfies the condition \(|z| = 2\). This confirms \(z_1\) lies on the circle.

Step 2: Determine the angle between vertices

For an equilateral triangle inscribed in a circle, the angle between the vertices from the circle's center is \(120^\circ\) or \(\frac{2\pi}{3}\) radians. Thus, each vertex can be obtained by rotating the previous vertex by \(\frac{2\pi}{3}\) radians.

Step 3: Determine \(z_2\) using rotation

To find \(z_2\), rotate \(z_1\) by \(\frac{2\pi}{3}\) radians. The formula for this rotation is:\[ z_2 = z_1 \times e^{i\frac{2\pi}{3}} \]First, determine \(e^{i\frac{2\pi}{3}}\):\[ e^{i\frac{2\pi}{3}} = \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \]Multiply it with \(z_1 = 1 + i\sqrt{3}\):\[ z_2 = (1 + i\sqrt{3})(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) \]Expand this multiplication:\[ z_2 = (-\frac{1}{2})*1 - \frac{1}{2}(i\sqrt{3}) + i\frac{\sqrt{3}}{2}(1) + i^2\frac{3}{2} \]Since \(i^2 = -1\):\[ z_2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2} + i\frac{\sqrt{3}}{2} - \frac{3}{2} \]Cancelling out the imaginary terms:\[ z_2 = -2 \]

Step 4: Determine \(z_3\) using another rotation

Similarly, to find \(z_3\), rotate \(z_2\) by another \(\frac{2\pi}{3}\) radians.Using \(z_2 = -2\), the same formula applies:\[ z_3 = z_2 \times e^{i\frac{2\pi}{3}} \]We find:\[ z_3 = -2 \times \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) \]\[ z_3 = 1 - i\sqrt{3} \]

Step 5: Check each answer option

Review the results in the context of the provided options:- Option (A) matches exactly with \(z_2 = -2\) and \(z_3 = 1 - i\sqrt{3}\).Thus, option (A) is the correct choice.

Key Concepts

Equilateral triangle

An equilateral triangle is a special type of polygon where all three sides are of equal length and each internal angle is precisely 60 degrees. In this particular problem, we are dealing with an equilateral triangle where the vertices are complex numbers located on a circle. The symmetry and equal-length properties of equilateral triangles play a useful role in geometric constructions and calculations. In the complex plane, these triangles maintain their properties and can be manipulated using complex arithmetic. With an equilateral triangle inscribed in a circle, the center of the circle acts as a common vertex from which three equal angles are drawn towards each vertex of the triangle. This symmetry allows us to use rotations to determine the other vertices based on one known vertex.

Inscribed circle

An inscribed circle, also known as an incircle, is a circle nestled inside a polygon that touches all its sides. Here, the problem focuses on an equilateral triangle inscribed within a circle, where each vertex of the triangle touches the circle's circumference. This geometric configuration simplifies the location of the vertices since all are equidistant from the circle's center, equal to the circle's radius in value. For any point on the inscribed circle in a complex plane, the modulus (or magnitude) will consistently remain the same, equal to the circle's radius. When a triangle like this is inscribed in a circle, particularly one with vertices as complex numbers, it opens avenues for using complex operations such as rotation.

Rotation in complex plane

Rotation in the complex plane is a transformation advantageous when dealing with geometric figures like polygons. Rotations in this context can be performed using multiplication by a complex number of unit modulus. For our case, because the vertices of the triangle are equally spaced in a circle of unit modulus, we can rotate a vertex by multiplying it by a complex number representing this rotation. To rotate a given vertex by an angle \(\theta\), use the expression \(e^{i\theta}\). As given in the exercise, to rotate by 120 degrees (or \(\frac{2\pi}{3}\) radians), we use \(e^{i\frac{2\pi}{3}} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\). This specific rotation changes a complex number's position in the plane, creating a new vertex still equal in modulus from the first vertex. By applying these rotations successively from one vertex to derive the others, we leverage the strength of complex number operations in simplifying what would otherwise be a complicated geometric task.

Modulus of complex number

The modulus of a complex number, analogous to the length of a vector, represents how far a complex number is from the origin in the complex plane. Formally, for a complex number \(z = a + bi\), the modulus \(|z|\) is given by \(\sqrt{a^2 + b^2}\). It captures the notion of distance in complex analysis, much like absolute value captures it in real numbers. In the context of our triangle inscribed on a circle of radius 2, each vertex's position is determined such that its modulus equals the circle's radius. As verified in the solution steps, with \(z_1 = 1 + i\sqrt{3}\), calculating its modulus results in \(2\), the radius of our inscribed circle, confirming its rightful placement on the circle. Consistent modulus values are significant as they ensure all vertices of the equilateral triangle remain equidistant from the circle's center, maintaining regularity and symmetry inherent in both circles and triangles.