Problem 73 - A Complete Resource Book in Mathematics for JEE Main 2017

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) \(|z-a|=\frac{1}{|1-a|}\) (C) \(\left|z-\frac{1}{1-a}\right|=\frac{1}{|1-a|}\) (D) None of these

Step-by-Step Solution

Answer

The vertices lie within the circle given by option (C): \(\left|z - \frac{1}{1-a}\right| = \frac{1}{|1-a|}\).

Step 1: Understand the expression for zk

The given expression for \(z_k\) is the sum of a geometric series: \(z_k = 1 + a + a^2 + \ldots + a^k\). This is a geometric progression with the first term \(1\) and common ratio \(a\).

Step 2: Apply the formula for the sum of a geometric series

The sum \(S_k\) of the first \(k+1\) terms of a geometric series where \(|a| < 1\) is given by: \[ S_k = \frac{1-a^{k+1}}{1-a} \] Thus, \(z_k = \frac{1-a^{k+1}}{1-a}\).

Step 3: Explore the behavior of zk as k approaches infinity

As \(k\) goes to infinity, \(a^{k+1}\) approaches zero because \(|a| < 1\), therefore \[ z_k \to \frac{1}{1-a} \] This shows that all vertices \(z_k\) approach \(\frac{1}{1-a}\) as \(k\) becomes large.

Step 4: Determine the range for zk

Given \(z_k = \frac{1-a^{k+1}}{1-a}\), the expression highlights that \(z_k\) will lie within a disk centered at \(\frac{1}{1-a}\) with radius \(\frac{|a^{k+1}|}{|1-a|}\). As \(k\) increases, \(|a^{k+1}|\) approaches 0, concentrating \(z_k\) around \(\frac{1}{1-a}\).

Step 5: Compare options to find the correct circle

Analyze the options:- Option (C): \(\left|z - \frac{1}{1-a}\right| = \frac{1}{|1-a|}\), correctly describes a circle centered at \(\frac{1}{1-a}\) with radius \(\frac{1}{|1-a|}\).- The other options do not match this geometric behavior as the circle's properties described above.

Key Concepts

Geometric Progression

Geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Let's break it down with a simple example. Consider the sequence: 2, 4, 8, 16. Here, each term is multiplied by 2 to get the next term. That's the essence of a geometric progression.
When dealing with complex numbers, such as in this exercise where we use a complex number 'a' with \(|a| < 1\), the sequence becomes: 1, a, a², a³, ..., aᵏ. Imagine plotting these terms on the complex plane; they form a spiral shrinking toward the origin if \(|a| < 1\).

• First Term: This is the term from which the progression starts (usually denoted as \(a_1\)). Here, it's 1.
• Common Ratio: A repeated factor for each consecutive term. In our case, this is 'a'.

Understanding geometric progression with complex numbers is crucial as it sets the ground for summing series and similar operations.

Sum of Series

The sum of a series in geometric progression is a powerful tool that tells us how to sum infinitely or finitely many terms. When the common ratio \(|a|\) is less than one, it becomes especially interesting as the sum tends towards a finite number. For example, if we want to find the sum of the series: \(1 + a + a^2 + \ldots + a^k\),where the magnitude of \(a\) is less than one, we can use the formula:\[S_k = \frac{1 - a^{k+1}}{1 - a}\] Here, \(S_k\) is the sum of the first \(k + 1\) terms. As \(k\) grows larger and approaches infinity, the geometric series simplifies further because \(a^{k+1}\) approaches zero. Thus, \[S = \frac{1}{1 - a}\]So, even an infinite series of this kind has a sum, which is a remarkable result in mathematics. This concept can be applied in various mathematical, physical, and engineering fields.

Polygon Vertices

In geometry, the vertices of a polygon are the corner points where two sides meet. In the context of complex numbers and geometric progressions, each term of the series \(z_k = 1 + a + a^2 + \ldots + a^k\) represents a point (or vertex) in a particular pattern on the complex plane. This polygon formed has its vertices at these points. For instance, imagine a point starting at 1 and each subsequent point is determined by adding powers of \(a\). Each \(z_k\) is a vertex.

• These points will be located in a path that curves towards a limit as \(k\) increases, forming interesting shapes.
• The overall configuration of the vertices depends on 'a' and its properties in the complex plane.

By analyzing the positions of these vertices, we are equipped to understand complex geometries through the lens of complex numbers.

Convergence

Convergence is an essential concept when dealing with infinite sequences or series. It describes how a sequence approaches a specific value as it progresses towards infinity. In terms of geometric progression, convergence occurs when the series approaches a finite limit.
In the current problem, since \(|a|<1\), the series \(z_k = 1 + a + a^2 + \ldots + a^k\) converges to \(\frac{1}{1-a}\) as \(k\) approaches infinity.

• As \(k\) becomes very large, \(a^{k+1}\) tends to zero, contributing less to the total sum.
• Thus, \(z_k\) gets closer and closer to \(\frac{1}{1-a}\), making it the convergence limit.

Understanding convergence in complex numbers allows us to perform calculations and understand behaviors where intuition alone might not suffice, making it a cornerstone of more advanced mathematical study.