Problem 73 - A Complete Resource Book in Mathematics for JEE Main

Question

Let ' \(a\) ' be a complex number such that \(|a|<1\) and \(z_{1}\), \(z_{2}, \ldots, z_{n}\) be the vetices of a polygon such that \(z_{k}=1+\) \(a+a^{2}+\ldots+a^{k}\), then the vertices of the polygon lie within the circle (A) \(|z|=\frac{1}{|1-a|}\) (B) \(|\mathrm{z}-a|=\frac{1}{|1-a|}\) (C) \(\left|z-\frac{1}{1-a}\right|=\frac{1}{|1-a|}\) (D) None of these

Step-by-Step Solution

Answer

Option (A): The vertices lie within the circle \(|z| = \frac{1}{|1-a|}\).

Step 1: Understand the Geometric Series

We are given that the vertices are defined by \(z_k = 1 + a + a^2 + ... + a^k\). This is a geometric series where \(|a| < 1\), allowing us to use the formula for the sum of a finite geometric series: \( S_n = \frac{1-a^{n+1}}{1-a} \).

Step 2: Compute the Expression for zk

Using the geometric series sum formula, we have \(z_k = \frac{1 - a^{k+1}}{1-a}\). This makes \(z_k\) the sum of the first \(k\) terms of the series.

Step 3: Determine the Limit of zk

As \( n \to \infty \), the term \(a^{n+1} \to 0\) due to \(|a| < 1\). Thus, the expression for the vertices approaches \(\frac{1}{1-a}\) as \(k\) becomes large.

Step 4: Analyze the Magnitude for Circle Condition

Observe that \(|z_k| = \left|\frac{1 - a^{k+1}}{1-a}\right| \leq \left| \frac{1}{1-a} \right|\) since \(|a^{k+1}| < 1\). This indicates that all the vertices \(z_k\) lie within the circle of radius \(\frac{1}{|1-a|}\) centered at the origin.

Step 5: Check Options

Given the analysis, the polygon vertices are within the circle described by \(|z| = \frac{1}{|1-a|}\), which corresponds to option (A). The other options involve conditions not met by our findings.

Key Concepts

Geometric series

A geometric series is a sum of terms that each multiply by a common factor as you progress through the sequence. In the context of complex numbers, the vertex positions of a polygon can sometimes be described by a geometric series. When evaluating a geometric series, the common ratio (factor) is critical. Here, the complex number \( a \) serves as this ratio.
For a geometric series where \( |a| < 1 \), which means each successive term becomes smaller in magnitude, we can often simplify calculations by applying the geometric series sum formula. This situation is beneficial when dealing with infinities, like reaching the limit of the polygon's vertices as in the exercise.

Geometric series sum formula

The geometric series sum formula provides a quick way to find the total of a sum for terms involving a constant ratio. Specifically, for a finite geometric series with first term \( a_1 \) and common ratio \( r \) (where \( |r| < 1 \)), the sum of the first \( n \) terms is given by:

• \( S_n = \frac{a_1(1-r^n)}{1-r} \)

In this exercise, this formula is crucial for calculating the position of polygon vertices described by \( z_k = 1 + a + a^2 + \ldots + a^k \). Applying the formula in this case gives: \( z_k = \frac{1-a^{k+1}}{1-a} \).
This sum formula reveals that as \( k \) increases, the series approaches the limit \( \frac{1}{1-a} \), provided that \( |a| < 1 \). This analysis is essential for understanding how the vertices behave as the polygon grow.

Polygon vertices

In geometry, vertices are the corner points of a polygon. When vertices are represented in the complex plane, each vertex can be seen as a complex number. In the problem at hand, the vertices \( z_1, z_2, \ldots, z_n \) are derived from a geometric series sum, where each \( z_k \) is determined by the formula \( z_k = \frac{1-a^{k+1}}{1-a} \).
The underlying geometric concept is that how these complex numbers (vertices) relate to each other indicates the overall shape and potential bounding conditions, such as the circle condition, which is a crucial aspect of this exercise.

Circle condition

The circle condition is a vital element in complex number analysis and is applied here to determine if the vertices of a polygon lie within a specific circle in the complex plane. This condition involves understanding the magnitude of a complex number and its relation to a geometric boundary or circle.
For the polygon vertices \( z_k \) given in the exercise, it's established that \(|z_k| = \left|\frac{1-a^{k+1}}{1-a}\right|\). The series gradually converges towards \( \frac{1}{1-a} \) due to the shrinking magnitude of \( a^{k+1} \).
The problem concludes that these vertices must be within a circle of radius \( \frac{1}{|1-a|} \), centered at the origin. This is validated by noting that the length of the segment from the origin to any vertex \( z_k \) does not exceed \( \frac{1}{|1-a|} \), thereby falling within the boundaries of the circle described by the condition \(|z| = \frac{1}{|1-a|}\).