Problem 60 - A Graphical Approach to College Algebra

Question

Find all \(n\) complex solutions of each equation of the form \(x^{n}=k\) $$x^{3}=64$$

Step-by-Step Solution

Answer

The solutions are 4, \(-2 + 2i\sqrt{3}\), and \(-2 - 2i\sqrt{3}\).

Step 1: Determine the Magnitude and Argument of k

The equation is \(x^3 = 64\). We rewrite 64 in polar form as \(64 e^{i imes 0}\), since its magnitude is 64 and argument is 0.

Step 2: Calculate the Magnitude of x

We need to find the magnitudes of \(x\) such that \(x^3 = 64\). The magnitude of \(x\) must be the cube root of the magnitude of \(k\), hence \(|x| = \sqrt[3]{64} = 4\).

Step 3: Find the Principal Argument of x

The principal argument for \(x\) is determined by dividing the argument of \(k\) by 3. Therefore, \(\text{Arg}(x) = \frac{0}{3} = 0\).

Step 4: Calculate All Possible Arguments for x

The general formula for the arguments of \(x\) in polar form is given by \(\theta = \frac{0 + 2m\pi}{3}\), where \(m = 0, 1, 2\).

Step 5: Express Each Solution in Polar Form

Substitute \(m = 0, 1, 2\) into the general formula to find all solutions. For \(m = 0\), \(\theta = 0\), so \(x = 4 e^{i0} = 4\). For \(m = 1\), \(\theta = \frac{2\pi}{3}\), so \(x = 4 e^{i \frac{2\pi}{3}} = -2 + 2i\sqrt{3}\). For \(m = 2\), \(\theta = \frac{4\pi}{3}\), so \(x = 4 e^{i \frac{4\pi}{3}} = -2 - 2i\sqrt{3}\).

Step 6: Write the Solutions in Rectangular Form

Convert each complex number from polar to rectangular form, if not done already in step 5.The solutions obtained are: \(x = 4, x = -2 + 2i\sqrt{3}, x = -2 - 2i\sqrt{3}\).

Key Concepts

Polar Form

When dealing with complex numbers, representing them in polar form is often advantageous, especially for multiplication and finding roots. In polar form, a complex number is represented as:

• Magnitude (or absolute value) \(|z|\)
• Argument (the angle \( \theta \) from the positive x-axis)

A complex number \( z = x + yi \) can be expressed in polar form as \( z = |z| e^{i\theta} \), where \( e^{i\theta} \) represents the rotational contribution of the angle. The transformation makes it easier to compute powers and roots of complex numbers. For instance, in our exercise, the number 64 was rewritten as \( 64e^{i \cdot 0} \), indicating no rotation from the positive real axis.

Magnitude and Argument

The magnitude and argument of a complex number provide insights into its distance from the origin and direction on the complex plane, respectively.
The magnitude of a complex number \( z = x + yi \) is calculated using the formula \(|z| = \sqrt{x^2 + y^2}\).
The argument, usually denoted by \( \text{Arg}(z) \), is the angle \( \theta \) formed with the positive x-axis.
Finding the cube roots, as required in the exercise, involves dividing the argument of the given number by the root degree. For example:

• If \( \text{Arg}(k) = 0 \) for \( k = 64 \), then \( \text{Arg}(x) = \frac{0}{3} = 0 \) for its cube roots, initially.

This shows how the division affects the angle, helping us find the root's orientation.

Cube Roots

Finding cube roots in the complex plane involves dividing the circle into equal parts, corresponding to each root.
The formula for finding the roots of a complex number \( k \) with degree \( n \) (like cube roots when \( n = 3 \)) states:

• The potential argument \( \theta \) for each root is \( \frac{\text{Arg}(k) + 2m\pi}{n} \), taking \( m = 0, 1, 2, \ldots, n-1 \).

Using this formula, the arguments for the cube roots of 64 are calculated by replacing \( m \) appropriately:

• When \( m = 0 \), \( \theta = 0 \)
• When \( m = 1 \), \( \theta = \frac{2\pi}{3} \)
• When \( m = 2 \), \( \theta = \frac{4\pi}{3} \)

This method divides the completion of the circle into three parts, allowing for the identification of each cube root.

Rectangular Form

After calculating roots in polar form, converting them back to rectangular form offers a more visually recognizable format for many calculations. A complex number in polar format, \( re^{i\theta} \), can be expressed in rectangular form as \( x + yi \) using the identities:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)

For example:

• With \( r = 4 \) and \( \theta = 0 \), the rectangular form is \( 4 + 0i \), which simplifies to simply 4.
• For \( \theta = \frac{2\pi}{3} \), we get \( x = -2 \) and \( y = 2i\sqrt{3} \).
• And for \( \theta = \frac{4\pi}{3} \), it becomes \(-2 - 2i\sqrt{3} \).

This way, complex numbers can be more easily analyzed and interpreted, given their more familiar representation.