Problem 58 - A Graphical Approach to Precalculus with Limits

Question

Find all \(n\) complex solutions of each equation of the form \(x^{n}=k\) $$x^{3}=-8$$

Step-by-Step Solution

Answer

The complex solutions are: \(1 + i\sqrt{3}\), \(-2\), \(1 - i\sqrt{3}\).

Step 1: Express in Polar Form

First, express \(-8\) in its polar form. The polar form of a complex number \(z\) is given by \(z = r(\cos(\theta) + i\sin(\theta))\), where \(r\) is the magnitude and \(\theta\) is the argument. Since \(-8\) is on the negative real axis, \(r = 8\) and \(\theta = \pi + 2k\pi\), where \(k\) is an integer to account for the full rotation.

Step 2: Apply De Moivre's Theorem

To solve \(x^3 = -8\), we use De Moivre's Theorem, which states \(x = \sqrt[3]{r}(\cos(\frac{\theta + 2m\pi}{3}) + i\sin(\frac{\theta + 2m\pi}{3}))\) for \(m = 0, 1, 2\). Substitute \(r = 8\) and \(\theta = \pi\).

Step 3: Calculate Magnitude and Argument

\(\sqrt[3]{8} = 2\) for the magnitude. For the arguments, calculate: 1. \(m=0\): \(\frac{\pi}{3}\)2. \(m=1\): \(\frac{\pi}{3} + \frac{2\pi}{3} = \pi\)3. \(m=2\): \(\frac{\pi}{3} + \frac{4\pi}{3} = \frac{5\pi}{3}\)

Step 4: Convert Back to Rectangular Form

Convert each polar form solution back to rectangular coordinates using \(\cos\) and \(\sin\):1. \(m=0\): \(2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})) = 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 1 + i\sqrt{3}\)2. \(m=1\): \(2(\cos(\pi) + i\sin(\pi)) = 2(-1 + 0i) = -2\)3. \(m=2\): \(2(\cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3})) = 2(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 1 - i\sqrt{3}\)

Key Concepts

Polar Form of Complex Numbers

Polar form is a way of expressing complex numbers using a distance from the origin and an angle from the positive real axis. This is helpful when dealing with multiplication and raising numbers to powers. We express a complex number in polar form as \( z = r(\cos(\theta) + i\sin(\theta)) \), where:

• \( r \) is the magnitude, calculated as \( \sqrt{a^2 + b^2} \) for a complex number \( a+bi \).
• \( \theta \) is the angle (or argument), which can be found using the arctan function based on the imaginary and real parts.

For the number \(-8\), which lies on the negative real axis, we don't have an imaginary part. Thus:

• The magnitude \( r \) is \( 8 \) since the absolute value of \(-8\) is \( 8 \).
• The angle \( \theta \) with respect to the positive real axis is \( \pi \) radians, acknowledging its position on the left side of the complex plane.

Knowing how to convert complex numbers into polar form makes solving equations like \( x^n = k \) much easier.

De Moivre's Theorem

De Moivre's Theorem provides a powerful way to work with powers and roots of complex numbers when expressed in polar form. This theorem states that if you have a complex number in polar form, \( z = r(\cos(\theta) + i\sin(\theta)) \), then:

• for raising to a power \( n \), \( z^n = r^n(\cos(n\theta) + i\sin(n\theta)) \)
• for finding an \( n^{th} \) root, \( z^{1/n} = \sqrt[n]{r}(\cos(\frac{\theta + 2m\pi}{n}) + i\sin(\frac{\theta + 2m\pi}{n})) \)

In our example, where \( x^3 = -8 \), we leveraged this theorem to find the roots. We started with \(-8\) as \( r = 8 \) and \( \theta = \pi \), then calculated:

• \( \sqrt[3]{8} = 2 \) for magnitude.
• The arguments, \( \frac{\pi}{3}, \pi, \frac{5\pi}{3} \), provide three distinct roots due to their different rotations in the complex plane.

De Moivre's Theorem is a bridge from polar visualization back to real-life calculations.

Rectangular Coordinates

Moving from polar form back to rectangular coordinates means expressing the complex number as \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. This transition is important for interpreting the solutions in a more intuitive form.

• Using the polar forms found earlier, we used \( \cos \) and \( \sin \) functions to determine the coordinates:
• For \( m=0 \): \( 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})) = 1 + i\sqrt{3} \).
• For \( m=1 \): \( 2(-1 + 0i) = -2 \).
• For \( m=2 \): \( 2(\cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3})) = 1 - i\sqrt{3} \).

These conversions use the properties of sine and cosine to translate back to \( a \) and \( b \), allowing us to visualize exactly where these roots lie in the complex plane. Understanding rectangular coordinates is crucial as it is the standard form we use in most applications involving complex numbers.