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Chapter 2: Problem 89

\(1, z_{1}, z_{2}, z_{3}, \ldots, z_{n-1}\) are the \(n^{\text {th }}\) roots of unity, then the value of \(1 /\left(3-z_{1}\right)+1 /\left(3-z_{2}\right)+\cdots+1 /\left(3-z_{n-1}\right)\) is equal to a. \(\frac{n 3^{n-1}}{3^{n}-1}+\frac{1}{2}\) b. \(\frac{n 3^{n-1}}{3^{n}-1}-1\) c. \(\frac{n 3^{n-1}}{3^{n}-1}+1\) d. none of these

Short Answer

Expert verified
The value is \( \frac{n3^{n-1}}{3^n - 1} - 1 \) (Option b).

Step by step solution

01

Understand the Problem

We are given the problem of evaluating the expression \( \frac{1}{3-z_1} + \frac{1}{3-z_2} + \cdots + \frac{1}{3-z_{n-1}} \) where \( z_{k} \) are the \( n^{th} \) roots of unity excluding 1. We need to simplify this expression to match one of the given options.
02

Identify Properties of Roots of Unity

The \( n^{th} \) roots of unity are given by \( 1, z_1, z_2, \ldots, z_{n-1} \), where these roots satisfy the polynomial equation \( x^n - 1 = 0 \). The sum of all \( n \) roots of unity is zero, i.e., \( 1 + z_1 + z_2 + \cdots + z_{n-1} = 0 \).
03

Apply Summation Formula

We can use partial fractions to simplify the expression. First, note that the expression represents the sum of certain transformations on roots of unity. By considering the function \( f(x) = \frac{x}{3-x} \), we find: \( f(3) + f(z_1) + f(z_2) + \cdots + f(z_{n-1}) = 0 \).
04

Derive Simplified Expression

Given \( z_1, z_2, \ldots, z_{n-1} \) and their powers, express \( \sum_{k=1}^{n-1} \frac{1}{3-z_k} \) in terms of powers of 3. Evaluating this sum involves solving \( \frac{n3^{n-1}}{3^n - 1} - 1 \) or a similar transformation, depending on characteristics of roots of unity.
05

Verify Steps and Options

Cross-check the simplification: the derived expression equals \( \frac{n3^{n-1}}{3^n - 1} - 1 \), matching option (b). Ensure all transformations align with properties of roots of unity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

nth roots of unity
The concept of the "nth roots of unity" revolves around solutions of the equation \( x^n = 1 \). In complex numbers, these solutions are of the form \( 1, z_1, z_2, \, \ldots, z_{n-1} \), where each \( z_k \) is the nth root of unity distributed evenly on the unit circle in the complex plane.
The most basic one is the number 1 itself, and the others are complex numbers which, when raised to the power of \( n \), return 1:
  • The nth roots of unity are given by \( \cos\left(\frac{2\pi k}{n}\right) + i\sin\left(\frac{2\pi k}{n}\right) \).
  • They are regularly spaced along the unit circle, with equal angles between each one, essentially forming a regular polygon.
Understanding nth roots of unity is crucial when solving polynomial equations where \( x^n - 1 = 0 \), as these roots help form the foundation for many equations and transformations in complex numbers.
sum of roots of unity
One fascinating property of nth roots of unity is their sum. The sum of all roots of unity \( 1 + z_1 + z_2 + \cdots + z_{n-1} \) is equal to zero. This is surprising at first but can be understood by considering that:
  • If you think of them as points on a unit circle, they evenly "balance out" around the circle's center.
  • This balancing happens because each root has an equal and opposite counterpart, effectively canceling out their contributions to the sum.
In polynomial equations, where the roots satisfy \( x^n - 1 = 0 \), establishing that the sum is zero relies on the root's evenly spaced nature around the origin in the complex plane.
polynomial equations
Polynomial equations are mathematical expressions of sums and powers of variables. In the context of this content, we'll focus on equations like \( x^n - 1 = 0 \). By complex factorization, this equation can be broken down using nth roots of unity:
  • The polynomial \( x^n - 1 \), through these roots, can be expressed as \( (x - 1)(x - z_1)\cdots(x - z_{n-1}) \).
  • This factorization gives individual roots as factors representing solutions to the equation.
Solving such polynomial equations often involves utilizing properties of the nth roots of unity, allowing simpler expression transformations and calculation of unknown variables by using these established roots.
complex numbers
Complex numbers extend the traditional number system to include solutions to equations that previously didn't have them鈥攁llowing us to explore roots of negative numbers, among others. They're expressed commonly in the form \( a + bi \), where \( i \) is the imaginary unit such that \( i^2 = -1 \).
This expansion is essential:
  • Complex numbers include real numbers (where \( b = 0 \)), imaginary numbers (where \( a = 0 \); for example, \( i \)), and everything in between.
  • They offer a two-dimensional number plane, facilitating the representation of all solutions to polynomial equations, like our nth roots of unity.
The beauty of complex numbers is how they elegantly create a complete number system allowing uniform approaches to all equations, greatly aiding in roots of unity and polynomial solution explorations.

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Most popular questions from this chapter

Let \(x_{1}, x_{2}\) are the roots of the quadratic equation \(x^{2}+a x+b=0\) where \(a, b\) are complex numbers and \(y_{1}, y_{2}\) are the roots of the quadratic equation \(y^{2}+|a| y+|b|=0\). If \(\left|x_{1}\right|=\left|x_{2}\right|=1\) then prove that \(\left|y_{1}\right|=\left|y_{2}\right|=1\).

Let \(a\) be a complex number such that \(|a|<1\) and \(z_{1}, z_{2}, z_{3}, \ldots\) be the vertices of a polygon such that \(z_{k}=1+a+a^{2}+\cdots+a^{k-1}\) for all \(k=1,2,3, \ldots\) then \(z_{1}, z_{2}, \ldots\) lie within the circle a. \(\left|z-\frac{1}{1-a}\right|=\frac{1}{|a-1|}\) b. \(\left|z+\frac{1}{a+1}\right|=\frac{1}{|a+1|}\) c. \(\left|z-\frac{1}{1-a}\right|=|a-1|\) d. \(\left|z+\frac{1}{a+1}\right|=|a+1|\)

If \(z=(i)^{(i)^{(i)}}\) where \(i=\sqrt{-1}\), then \(|z|\) is equal to a. b. \(e^{-\sqrt{2}} \quad\) c. \(e^{-x}\) d. none of these

If \(t\) and \(c\) are two complex numbers such that \(|t| \neq|c|,|t|=1\) and \(z\) \(=(a t+b) /(t-c), z=x+i y .\) Locus of \(z\) is (where \(a, b\) are complex numbers) a. line segment b. straight line c. circle d. none of these

If \(\arg (z+a)=\pi / 6\) and \(\arg (z-a)=2 \pi / 3\left(a \in R^{*}\right)\), then a. \(|z|=a\) b. \(|z|=2 a\) c. \(\arg (z)=\frac{\pi}{2}\) d. \(\arg (z)=\frac{\pi}{3}\)
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