Problem 50 - Contemporary Precalculus

Question

What do you think are the solutions of \(x^{n-1}+x^{n-2}+\cdots+x^{3}+x^{2}+x+1=0 ?\) (Sce Exer- cises \(47-49 .)\)

Step-by-Step Solution

Answer

Answer: The solutions are given by \(x_k = 1^\frac{1}{n} \cdot \left[\cos\left(\frac{2\pi k}{n}\right) + i \cdot \sin\left(\frac{2\pi k}{n}\right)\right], \quad k = 0, 1, 2, ..., n-1\), where \(x_k\) are the complex roots of the equation and \(k\) is an integer.

Step 1: Rewrite the equation as a geometric series

Notice that the given equation is a geometric series with a common ratio of \(x\) and \(n\) terms. We can rewrite the equation as: \(1 + x + x^2 + x^3 + \cdots + x^{n-2} + x^{n-1} = 0\)

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

We will use the formula for the sum of a geometric series: Sum \(= \frac{1 - x^n}{1 - x}\) Plug in the terms of our equation to get: \(\frac{1 - x^n}{1 - x} = 0\)

Step 3: Solve for \(x\)

To find the solutions for \(x\), multiply both sides of the equation by \((1-x)\): \(1 - x^n = 0\) Now, add \(x^n\) to both sides: \(x^n = 1\) To find the roots of this equation, remember that \(x^n\) can be written in polar form: \(x^n = 1 \cdot (cos(2 \pi k {\space} ) + i \cdot sin(2 \pi k))\), where \(k = 0, 1, 2, ..., n-1\). Taking the nth root of both sides, we get: \(x = 1^\frac{1}{n} \cdot \left[\cos\left(\frac{2\pi k}{n}\right) + i \cdot \sin\left(\frac{2\pi k}{n}\right)\right]\)

Step 4: Find all the solutions for \(x\)

Now, we can find all the complex roots of the given equation by plugging in the values of \(k\) from \(0\) to \(n-1\): \(x_k = 1^\frac{1}{n} \cdot \left[\cos\left(\frac{2\pi k}{n}\right) + i \cdot \sin\left(\frac{2\pi k}{n}\right)\right], \quad k = 0, 1, 2, ..., n-1\) These are the solutions of the given equation.

Key Concepts

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 common ratio. Consider the series: \(1 + x + x^2 + x^3 + \cdots + x^{n-1}\). Here, each term is obtained by multiplying the previous term by \(x\), which is the common ratio. This expression is indeed a geometric series with \(n\) terms.

One of the crucial properties of geometric series is that they have a simple formula for their sum. If \(r\) is the common ratio and \(n\) is the total number of terms, the sum \(S\) of the series can be expressed as:

• \(S = \frac{1-r^n}{1-r}\) when \(r eq 1\)

In the problem at hand, understanding this formula allows us to succinctly rewrite the given equation as \( \frac{1-x^n}{1-x} = 0\). By grasping the concept of geometric series, one can easily transform and solve seemingly complex polynomial equations.

nth Roots of Unity

The nth roots of unity are the complex numbers that satisfy the equation \(x^n = 1\). These numbers play a significant role in various areas of mathematics, including number theory and geometry. In essence, these roots are the solutions to the equation \(x^n = 1\) and can be graphically represented as points on the unit circle in the complex plane.

Each nth root of unity can be expressed using the formula:

• \(x = \cos\left(\frac{2\pi k}{n}\right) + i \cdot \sin\left(\frac{2\pi k}{n}\right)\)

Where \(k\) ranges from 0 to \(n-1\). This allows each root to be evenly spaced around the unit circle at angles that are evenly divided by the integer \(n\). The property that these roots sum to zero and multiply to 1 is particularly useful for solving various equations, like our given polynomial equation of geometric series. By finding these roots, we determine all possible solutions.

Polar Form

The polar form of a complex number expresses it in terms of an angle and a magnitude. Any complex number \(x\) can be represented as \(r \cdot (\cos(\theta) + i \cdot \sin(\theta))\), where \(r\) is the modulus (or magnitude) and \(\theta\) is the argument (or angle) of the complex number.

When dealing with equations or expressions involving powers of complex numbers, particularly nth roots, the polar form becomes exceptionally convenient. For example, expressing \(x^n = 1\) using polar coordinates leads to solutions that are easily interpretable as points on the unit circle, each separated by equal angles. This makes it a powerful tool for visualizing and solving equations involving powers of complex numbers.

In the context of the problem above, converting the equation to polar form allowed us to take the nth root of unity, breaking down the complex number calculations into simpler trigonometric angles and magnitudes. This geometric interpretation of complex multiplication simplifies the process of solving polynomial equations, converting them into matters of rotation and scaling on the complex plane.