/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 If \(1=z_{0}, z_{1}, \ldots, z_{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If \(1=z_{0}, z_{1}, \ldots, z_{n-1}\) are the \(n\) th roots of unity, prove that \(\left(z-z_{1}\right)\left(z-z_{2}\right) \cdots\left(z-z_{n-1}\right)=1+z+z^{2}+\cdots+z^{n-1}\)

Short Answer

Expert verified
The polynomial \((z-z_1)(z-z_2)\cdots(z-z_{n-1}) = 1 + z + z^2 + \cdots + z^{n-1}\) is the sum of a geometric series representing the nth roots of unity.

Step by step solution

01

Understanding Nth Roots of Unity

The nth roots of unity are the solutions to the equation \(z^n = 1\). They can be expressed as \(z_k = e^{2\pi ik/n}\) for \(k = 0, 1, 2, \ldots, n-1\). Notably, the roots are evenly spaced on the unit circle in the complex plane. The root \(z_0 = 1\) and the rest are \(z_1, z_2, \ldots, z_{n-1}\).
02

Express Polynomial with Roots

We can form a polynomial whose roots are the nth roots of unity (excluding \(1\)). This polynomial is \((z-z_1)(z-z_2)\cdots(z-z_{n-1})\). The degree of this polynomial is \(n-1\), since it excludes \(z_0 = 1\).
03

Recognize Geometric Series

The sum \(1+z+z^2+\cdots+z^{n-1}\) can be recognized as a geometric series. This series can be summed up using the formula for the sum of a geometric series: \(S = \frac{a(r^n - 1)}{r - 1}\) where \(a = 1\) and \(r = z\). Here, since \(z^n = 1\), it simplifies to \(\frac{z^n - 1}{z - 1} = \frac{1 - 1}{z - 1} = 0\) if \(z eq 1\).
04

Apply Symmetry and Unity Root Property

The polynomial \((z-1)(z-z_1)\cdots(z-z_{n-1})\) can be expanded and simplified due to the nature of roots of unity and symmetry. From the identity \(z^n - 1 = (z-1)(z-z_1)(z-z_2)\cdots(z-z_{n-1})\), dividing by \(z - 1\) (since for \(z eq 1\), it doesn't impact), provides \((z-z_1)(z-z_2)\cdots(z-z_{n-1}) = 1 + z + z^2 + \cdots + z^{n-1}\).
05

Conclusion

By combining the result of dividing \(z^n - 1\) by \(z - 1\) with the roots being all roots of unity except \(1\), the initial polynomial expression \((z - z_1)(z - z_2)\cdots(z - z_{n-1})\) matches directly with the geometric series \(1 + z + z^2 + \cdots + z^{n-1}\), thus proving the statement.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Roots of Unity
The concept of "roots of unity" is a fascinating area in complex analysis. These are complex numbers that equate to 1 when raised to a specific power. Let's take the "nth" roots of unity as an example. These numbers are solutions to the equation \(z^n = 1\). They are evenly distributed on the unit circle in the complex plane. Each root can be represented in exponential form as \( z_k = e^{2\pi i k / n} \), where \( k \) is an integer ranging from 0 to \( n-1 \).

Visualize the unit circle as a clock. If \( n \) is 12, each root corresponds to a number on the clock face, starting at \( z_0 = 1 \) (akin to 12 on the clock) and proceeding round clockwise. These roots have a neat property: if you sum them all, the result is zero (except for the root at \( z_0 = 1 \)). This symmetry is vital in simplifying many mathematical equations related to polynomials and series.
Geometric Series
A geometric series 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 ratio. Recognizing when you have a geometric series can greatly simplify the sum of the series. For instance, \(1 + z + z^2 + \cdots + z^{n-1}\) is an example of such a series with the first term \(a = 1\) and the common ratio \(r = z\).

The formula for summing the first \(n\) terms of a geometric series is \(S = \frac{a(r^n - 1)}{r - 1}\). When \(z^n = 1\) (since these terms are the roots of unity), the sum simplifies to \(\frac{1 - 1}{z - 1} = 0\), unless \(z = 1\). However, when addressing the polynomial roots excluding 1, this sum becomes relevant as a way of tying back to the product of the polynomial roots.
Polynomial Roots
Understanding the roots of a polynomial is crucial when solving equations. For a polynomial such as \((z - z_1)(z - z_2)\cdots(z - z_{n-1})\), the roots are \(z_1, z_2, \ldots , z_{n-1}\). These roots are strategically selected to exclude \(z_0 = 1\), focusing on the non-trivial solutions. The polynomial equation \(z^n - 1 = (z-1)(z - z_1)(z - z_2)\cdots(z - z_{n-1})\) brings in every nth root of unity, incorporating symmetry naturally.

Dividing \(z^n - 1\) by \(z - 1\) (since \(z eq 1\)) derives \((z - z_1)(z - z_2)\cdots(z - z_{n-1})\). What is intriguing here is how such roots relate to the geometric series sum \(1 + z + z^2 + \cdots + z^{n-1}\), showcasing the intricate symmetry and interconnection between polynomial roots and these series.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(P(z)=a_{n} z^{n}+a_{n-1} z^{n-1}+\cdots+a_{1} z+a_{0}\) be a polynomial of degree \(n .\) (a) Suppose that \(a_{n}, a_{n-1}, \ldots, a_{1}, a_{0}\) are all real. Show that if \(z_{1}\) is a root of \(P\), then \(\overline{z_{1}}\) is also a root. In other words, the roots must be complex conjugates, something you likely learned without proof in high school. (b) Suppose not all of \(a_{\mathrm{m}}, a_{n-1}, \ldots, a_{1}, a_{0}\) are real. Show that \(P\) has at least one root whose complex conjugate is not a root. Hint: Prove the contrapositive. (c) Find an example of a polynomial that has some roots occurring as complex conjugates, and some not.

Let \(z_{1}\) and \(z_{2}\) be two distinct points in the complex plane, and let \(K\) be a positive real constant that is less than the distance between \(z_{1}\) and \(z_{2}\). (a) Show that the set of points \(\left\\{z:\left|z-z_{1}\right|-\left|z-z_{2}\right|=K\right\\}\) is a hyperbola with foci \(z_{1}\) and \(z_{2}\). (b) Find the equation of the hyperbola with foci \(\pm 2\) that goes through the point \(2+3 i\). (c) Find the equation of the hyperbola with foci \(\pm 25\) that goes through the point \(7+24 i\).

Let \(m\) and \(n\) be positive integers that have no common factor. Show that there are \(n\) distinct solutions to \(w^{n}=z^{m}\) and that they are given by \(w_{k}=r^{m}\left(\cos \frac{m(\theta+2 \pi k)}{n}+i \sin \frac{m(\theta+2 \pi k)}{n}\right)\) for \(k=0,1, \ldots, n-1 .\)

Find all solutions to the following depressed cubics. (a) \(27 x^{3}-9 x-2=0 .\) Hint: Get an equivalent monic polynomial. (b) \(x^{3}-27 x+54=0\)

Show that if \(z \neq 0\), the four points \(z\), \(\bar{z},-z\), and \(-\bar{z}\) are the vertices of a rectangle with its center at the otigin.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.