Problem 36 - Contemporary Precalculus

Question

Solve the given equation in the complex number system. $$x^{6}+729=0$$

Step-by-Step Solution

Answer

Now, express the right side in polar form using De Moivre's theorem: $$x^6=(-1)^1 \cdot 3^6[\cos(180^{\circ})+i\sin (180^{\circ})]$$ #Step 2: Take the 6th root of both sides We will take the 6th root of both sides of the equation to solve for x: $$x=\sqrt[6]{(-1)^1 \cdot 3^6[\cos(180^{\circ})+i\sin(180^{\circ})]}$$ Now, we can apply De Moivre's theorem to the right side: $$x=3[\cos\left(\frac{180^{\circ}}{6}\right)+i\sin\left(\frac{180^{\circ}}{6}\right)]$$ #Step 3: Find all solutions for x Since we are solving a degree-6 equation, there should be 6 complex solutions for x. These solutions can be derived by uniformly increasing the angle within the cosine and sine terms by \(360^{\circ}/6 = 60^{\circ}\). The six solutions can be written as: $$x_k=3\left[\cos\left(\frac{180^{\circ}+60^{\circ} k}{6}\right)+i\sin\left(\frac{180^{\circ}+60^{\circ} k}{6}\right)\right], k=0,1,2,3,4,5$$ Short Answer: The six complex solutions for the equation \(x^6+729=0\) are given by: $$x_k=3\left[\cos\left(\frac{180^{\circ}+60^{\circ} k}{6}\right)+i\sin\left(\frac{180^{\circ}+60^{\circ} k}{6}\right)\right], k=0,1,2,3,4,5$$

Step 1: Rewrite the equation using De Moivre's theorem

Rewrite the equation into a form that utilizes De Moivre's theorem, which states \((r[\cos(\theta)+i\sin(\theta)])^n = r^n[\cos(n\theta)+i\sin(n\theta)]\). Our goal is to rewrite the equation so that we can leverage the theorem to find the solutions. $$x^6=-3^6$$ We can rewrite the right side of the equation into its polar form: $$x^6=(-1+0i)3^6$$

Step 2: Take the 6th root of both sides

We will take the 6th root of both sides of the equation to solve for x: $$x=\sqrt[6]{(-1)^1 \cdot 3^6[\cos(180^{\circ})+i\sin(180^{\circ})]}$$ Now, we can apply De Moivre's theorem to the right side: $$x=3[\cos\left(\frac{180^{\circ}}{6}\right)+i\sin\left(\frac{180^{\circ}}{6}\right)]$$

Step 3: Find all solutions for x

Since we are solving a degree-6 equation, there should be 6 complex solutions for x. These solutions can be derived by uniformly increasing the angle within the cosine and sine terms by \(360^{\circ}/6 = 60^{\circ}\). The six solutions can be written as: $$x_k=3\left[\cos\left(\frac{180^{\circ}+60^{\circ} k}{6}\right)+i\sin\left(\frac{180^{\circ}+60^{\circ} k}{6}\right)\right], k=0,1,2,3,4,5$$ Short Answer: The six complex solutions for the equation \(x^6+729=0\) are given by: $$x_k=3\left[\cos\left(\frac{180^{\circ}+60^{\circ} k}{6}\right)+i\sin\left(\frac{180^{\circ}+60^{\circ} k}{6}\right)\right], k=0,1,2,3,4,5$$

Key Concepts

De Moivre's Theorem

De Moivre's Theorem is a powerful tool when working with complex numbers in polar form, especially in relation to raising complex numbers to powers. It greatly simplifies the process by breaking it down into manageable pieces. This theorem tells us that for any complex number in polar form \( r(\cos(\theta) + i\sin(\theta)) \), when raised to the power of \( n \), it can be expressed as

• \( r^n(\cos(n\theta) + i\sin(n\theta)) \).

This means you raise the magnitude \( r \) to the power of \( n \), and multiply the angle \( \theta \) by \( n \).
For students solving complex equations, this theorem is invaluable for simplifying the task of finding complex roots. Instead of performing lengthy calculations, you can directly compute powers of complex numbers using their polar forms.

Polar Form of Complex Numbers

The polar form of a complex number is an alternative way to express complex numbers, which relies on their magnitude and angle rather than the real and imaginary parts. Any complex number \( z = a + bi \) can be represented as

• \( r(\cos(\theta) + i\sin(\theta)) \),

where \( r \) is the magnitude given by \( \sqrt{a^2 + b^2} \), and \( \theta \) is the argument (angle) of the complex number, determined by \( \tan^{-1}(\frac{b}{a}) \).
This form is particularly useful for multiplication, division, and powering of complex numbers as it neatly separates their magnitude and phase angle components. Converting a complex number to its polar form can often make algebraic operations much simpler and more intuitive.
In the context of our original problem, the equation \( x^6 = (-1+0i)3^6 \) fully benefits from polar transformation to determine the roots using De Moivre's Theorem.

Roots of Complex Equations

Finding the roots of complex equations is a common problem and, when using polar form, becomes a task of finding magnitude and angle patterns. When solving for roots, recognize that each degree power introduces multiple solutions. A sixth-degree polynomial, for example, will have six roots.
The roots of the equation derived using De Moivre's Theorem can be placed in the form:

• \( r^{1/n}(\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n})) \)

where \( k \) is each integer from 0 to \( n-1 \). These \( k \) values are essential because they systematically introduce every distinct root within the cycle of one circle revolution (\( 2\pi \)), ensuring all potential angles are evaluated.
This pattern generates all roots evenly spaced along a circle in the complex plane, each at an equal distance, demonstrating the striking symmetry inherent to roots of unity in complex analysis.