Problem 50 - A Graphical Approach to Precalculus with Limits

Question

Find all complex solutions for each equation. Leave your answers in trigonometric form. $$x^{3}+i=0$$

Step-by-Step Solution

Answer

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

Step 1: Rewrite the Equation

First, rewrite the equation in the form suitable for solving: \[ x^3 = -i \]

Step 2: Express in Polar Form

Express \(-i\) in polar form. The point \(-i\) is on the negative imaginary axis, which can be represented as \(-i = e^{-i\frac{\pi}{2}}\).In polar form, \( -i= 1(\cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2}))\).

Step 3: Use De Moivre's Theorem

Apply De Moivre's Theorem to solve for the cube roots: If \(z = re^{i\theta}\), then \(z^{1/3} = r^{1/3} e^{i\frac{\theta + 2k\pi}{3}}\) for \(k = 0, 1, 2\).

Step 4: Calculate the First Solution

Set \(k = 0\) and calculate the first solution:\[ x_0 = e^{i\frac{-\pi/2 + 2(0)\pi}{3}} = e^{-i\pi/6} = \cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6}) \] which simplifies to: \( x_0 = \frac{\sqrt{3}}{2} - \frac{1}{2}i \).

Step 5: Calculate the Second Solution

Set \(k = 1\) and calculate the second solution: \[ x_1 = e^{i\frac{-\pi/2 + 2(1)\pi}{3}} = e^{i3\pi/6} = e^{i\pi/2} = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}) \] which simplifies to: \( x_1 = i \).

Step 6: Calculate the Third Solution

Set \(k = 2\) and calculate the third solution: \[ x_2 = e^{i\frac{-\pi/2 + 2(2)\pi}{3}} = e^{i7\pi/6} = \cos(\frac{7\pi}{6}) + i\sin(\frac{7\pi}{6}) \] which simplifies to: \( x_2 = -\frac{\sqrt{3}}{2} - \frac{1}{2}i \).

Step 7: Present Solutions in Trigonometric Form

List all solutions in trigonometric form:1. \( x_0 = \frac{\sqrt{3}}{2} - \frac{1}{2}i \)2. \( x_1 = i \)3. \( x_2 = -\frac{\sqrt{3}}{2} - \frac{1}{2}i \)

Key Concepts

Polar Form

Polar Form is a way to represent complex numbers by expressing them in terms of their magnitude (also called modulus) and an angle known as the argument. This form is extremely useful for tasks like multiplying complex numbers or finding powers and roots of complex numbers.

In Polar Form, a complex number is represented as:

• Magnitude \( r \): Distance from the origin to the point in the complex plane.
• Angle \( \theta \): Angle with respect to the positive x-axis.
• Expression: \( re^{i\theta} \), or alternatively, \( r(\cos\theta + i\sin\theta) \).

For example, in the context of our exercise, the complex number \(-i\) is located on the negative imaginary axis. It has a magnitude of 1 and an angle of \(-\frac{\pi}{2}\) radians, written in polar form as \( 1(\cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2})) \). This transformation helps simplify operations like finding cube roots, as seen in the solution process.

De Moivre's Theorem

De Moivre's Theorem is a foundational tool in complex number analysis, common for working with powers and roots of complex numbers. The theorem states:

• If \( z = re^{i\theta} \), then \( z^n = r^n e^{in\theta} \).
• For roots, \( z^{1/n} = r^{1/n} e^{i(\theta + 2k\pi)/n} \) for \( k = 0, 1, ..., n-1 \).

This makes computation straightforward because it separates the magnitude and the angle, allowing for easy scaling and cycling of the angle. In the exercise, De Moivre's Theorem was applied to find the cube roots of \(-i\) by dividing the angle by 3 and considering the three possible values of \( k \), yielding three distinct solutions distributed evenly around the circle.

Trigonometric Form

Trigonometric Form is another way to express complex numbers, especially useful for visualization and solving operations like multiplication and division.

In this form, a complex number is expressed as:

• \( r(\cos\theta + i\sin\theta) \)

This is similar to Polar Form in structure and is ideal for tasks where sines and cosines give meaningful interpretations — for instance, finding cube roots in our exercise. By handling both magnitude and direction separately, it cleverly divides the task of "turning around" the complex plane based on the periodic properties of trigonometric functions.

Cube Roots

Finding Cube Roots of a complex number involves identifying its three distinct values that satisfy the cube root equation \( x^3 = z \).

The approach employs:

• Converting the complex number into Polar or Trigonometric Form.
• Using De Moivre's Theorem to systematically calculate each root.
• Applying the formula \( x = r^{1/3}e^{i(\theta + 2k\pi)/3} \) for \( k = 0, 1, 2 \).

This method ensures that the roots are evenly spaced on the complex plane, forming an equilateral triangle. In our exercise, aligning \(-i\) along the negative imaginary axis resulted in identified cube roots \( x_0, x_1, \) and \( x_2 \). Each is a straightforward reflection of how De Moivre's Theorem effectively partitions the circle into three equal arcs, capturing all possible cube roots efficiently.