Problem 64 - A Graphical Approach to College Algebra

Question

Find all \(n\) complex solutions of each equation of the form \(x^{n}=k\) $$x^{3}=-27$$

Step-by-Step Solution

Answer

The solutions are \(\frac{3}{2} + i \frac{3\sqrt{3}}{2}, -3, \frac{3}{2} - i \frac{3\sqrt{3}}{2}.\)

Step 1: Identify the Equation's Form

Recognize that the equation is in the form of \(x^n = k\) with \(n = 3\) and \(k = -27\). We need to find all complex solutions of this equation.

Step 2: Convert the Constant to Polar Form

Convert \(-27\) into polar form. Since \(-27\) lies on the negative real axis, it can be written in polar form as \(27\text{cis}(\pi)\).

Step 3: Use De Moivre's Theorem

According to De Moivre's Theorem, the solutions to \(x^3 = 27\text{cis}(\pi)\) are given by:\[x = 27^{1/3} \text{cis}\left( \frac{\pi + 2k\pi}{3} \right)\quad \text{for } k = 0, 1, 2\]

Step 4: Calculate the Magnitude

The magnitude (or modulus) of each solution is \(27^{1/3} = 3\) since the cube root of 27 is 3.

Step 5: Calculate Each Solution

- For \(k = 0\), the angle is \(\frac{\pi}{3}\), yielding the solution \(3\text{cis}\left(\frac{\pi}{3}\right)\).- For \(k = 1\), the angle is \(\pi\), yielding the solution \(3\text{cis}(\pi)\).- For \(k = 2\), the angle is \(\frac{5\pi}{3}\), yielding the solution \(3\text{cis}\left(\frac{5\pi}{3}\right)\).

Step 6: Express Solutions in Rectangular Form

Convert each solution from polar form to rectangular form:- \(3\text{cis}\left(\frac{\pi}{3}\right) = 3\left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = \frac{3}{2} + i \frac{3\sqrt{3}}{2}\)- \(3\text{cis}(\pi) = -3\)- \(3\text{cis}\left(\frac{5\pi}{3}\right) = 3\left(\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = \frac{3}{2} - i \frac{3\sqrt{3}}{2}\)

Key Concepts

De Moivre's Theorem

De Moivre's Theorem is a powerful tool for solving equations of the form \(x^n = k\). It connects complex numbers and trigonometry by allowing complex number powers and roots to be calculated in a straightforward way using polar coordinates.
Here's how it works: if a complex number is expressed in polar form as \( r \text{cis}(\theta) \), then raising it to a power \( n \) is simplified to \( r^n \text{cis}(n\theta) \).

• "cis" stands for cosine plus imaginary sine, or \( \cos(\theta) + i \sin(\theta) \).
• When finding roots (such as cube roots), De Moivre's theorem helps us find solutions by dividing the angle by \( n \) and distributing the solutions circularly around the complex plane.

In our example, solving \( x^3 = -27 \), the theorem helps in expressing the root as \( 27^{1/3} \text{cis}\left(\frac{\pi + 2k\pi}{3}\right) \). Each value of \( k \) gives us a different root by adjusting the angle.

Polar Form

Polar form gives us a way to express complex numbers using a magnitude and an angle, which can make complex calculations more intuitive.
Each complex number is represented as \( r \text{cis}(\theta) \) or \( r( \cos(\theta) + i \sin(\theta) ) \), where:

• \( r \) is the magnitude or modulus of the number, representing the distance from the origin in the complex plane.
• \( \theta \) is the angle, called the argument, measured from the positive real axis.

In this framework, prior to using De Moivre's Theorem, we must convert \(-27\) to polar form. We see \( \left| -27 \right| = 27 \) and since \(-27\) is located on the negative real axis, the angle \( \theta \) is \( \pi \). Hence, in polar form, this becomes \( 27 \text{cis}(\pi) \).

Converting into polar form simplifies taking roots since multiplying powers involves multiplying the angles, which can be quite handy when solving complex equations.

Cube Roots of Unity

The cube roots of unity refer to the solutions of the equation \(x^3 = 1\). These roots hold a special structure due to symmetry in the complex plane, and they provide insight when finding any cubic root of a number.

• The solutions are \(1\), \(\omega = \text{cis}\left(\frac{2\pi}{3}\right)\), and \(\omega^2 = \text{cis}\left(\frac{4\pi}{3}\right)\).
• These roots lie equally spaced on the unit circle in the complex plane, dividing it into three equal parts.

In our exercise, even though we are not solving \(x^3 = 1\), the structure of cube roots gives us a pattern. It helps in understanding how to approach finding cube roots of other numbers, like \(-27\).
In such cases, our cube root calculations take on similar symmetry and spacing in the complex plane, modified by the modulus (radius) or the original number \(-27\). Here, each root is spaced by an additional \(\frac{2\pi}{3}\) radians, manifesting in:

• \(\text{cis}\left(\frac{\pi}{3}\right)\)
• \(\text{cis}(\pi)\)
• \(\text{cis}\left(\frac{5\pi}{3}\right)\)

This not only visualizes the roots but also helps in predicting subsequent roots in more advanced computation.