Problem 60 - A Graphical Approach to Precalculus with Limits

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 \(x = 4\), \(x = -2 + 2i\sqrt{3}\), and \(x = -2 - 2i\sqrt{3}\).

Step 1: Identify the form of the equation

The given equation is \(x^3 = 64\). This is a complex equation of the form \(x^n = k\) where \(n=3\) and \(k=64\).

Step 2: Express \(k\) in polar form

The number \(k = 64\) can be expressed in polar form as \(64e^{i0}\), because \(64 = 64 \times e^{i0}\).

Step 3: Determine the polar form of the principal root

To solve \(x^3 = 64\), we find the principal cube root of the polar form of 64. The cube root of 64 is 4, so the principal cube root in polar form is \(4e^{i0/3} = 4 \cdot e^{i0} = 4\).

Step 4: Calculate all cube roots using De Moivre's Theorem

The solutions can be calculated using De Moivre's Theorem: the \(n\) solutions are given by \(r^{1/n} \cdot \text{cis} \left(\frac{\theta + 2k\pi}{n} \right)\) for \(k = 0, 1, 2\). Here, \(r=64\), \(n=3\), and \(\theta = 0\).

Step 5: Substitute for each \(k\)

Substituting \(k = 0\), we get the first solution: \[ x_0 = 4 \cdot e^{i(\frac{0 + 2\pi \cdot 0}{3})} = 4 \cdot e^{i0} = 4 \]For \(k = 1\), the second solution is: \[ x_1 = 4 \cdot e^{i(\frac{0 + 2\pi \cdot 1}{3})} = 4 \cdot e^{i\frac{2\pi}{3}} = 4 \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -2 + 2i\sqrt{3} \]For \(k = 2\), the third solution is: \[ x_2 = 4 \cdot e^{i(\frac{0 + 2\pi \cdot 2}{3})} = 4 \cdot e^{i\frac{4\pi}{3}} = 4 \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -2 - 2i\sqrt{3} \]

Step 6: List all solutions

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

Key Concepts

Polar Form

When dealing with complex numbers, one useful way to express them is in polar form. This form represents a complex number as a combination of a magnitude and an angle. The general polar form of a complex number is given by:\[ z = r \cdot e^{i\theta} \]where:

• \( r \) is the magnitude (or modulus) of the complex number,
• \( \theta \) is the angle (or argument) in radians.

For example, the number 64 can be expressed in polar form as \( 64 \cdot e^{i0} \) since its magnitude is 64 and the angle associated with it is 0 radians. This polar representation becomes especially useful when performing operations like multiplication and division on complex numbers or when finding powers and roots, as it simplifies calculations significantly.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool in complex number calculations, particularly for finding powers and roots of complex numbers. The theorem states that for any complex number in polar form \( z = r \cdot e^{i\theta} \) and any integer \( n \), the power of the complex number can be expressed as:
\[ z^n = r^n \cdot e^{in\theta} \]
This theorem simplifies the process of taking powers by multiplying the argument \( \theta \) by \( n \) and raising the magnitude \( r \) to the \( n \)th power.
When finding roots, such as cube roots in this exercise, the theorem helps determine the different solutions by distributing the angles evenly around the circle. For cube roots of \( r \cdot e^{i\theta} \), the solutions are:
\[ r^{1/3} \cdot e^{i(\frac{\theta + 2k\pi}{3})} \]
where \( k = 0, 1, 2 \). Each value of \( k \) gives us one distinct cube root, showing the elegance and simplicity this theorem brings to solving such equations.

Cube Roots

Cube roots of a complex number can be found by determining three evenly spaced angles around the unit circle, thanks to the nature of complex numbers. In the equation \( x^3 = 64 \), solving for the cube roots involves finding three roots of unity, which are points on the complex plane where the cube of each point returns to 1.
Let's take the polar form of 64, \( 64 \cdot e^{i0} \). To find the cube roots, we:

• Calculate the principal root. The cube root of 64 is 4, so this is the magnitude of our solutions: \( 4 \).
• Find the angles by dividing full rotation \( 2\pi \) by 3, giving \( \frac{2\pi}{3} \).

Each cube root corresponds to an angle \( \frac{2k\pi}{3} \) for \( k = 0, 1, 2 \).

• The first root: \( 4 \cdot e^{i0} = 4 \).
• The second root: \( 4 \cdot e^{i\frac{2\pi}{3}} = -2 + 2i\sqrt{3} \).
• The third root: \( 4 \cdot e^{i\frac{4\pi}{3}} = -2 - 2i\sqrt{3} \).

These roots show how complex numbers spread solutions symmetrically around the unit circle, making it intuitive to visualize and solve.