Problem 49 - A Graphical Approach to Precalculus with Limits

Question

Find all complex solutions for each equation. Leave your answers in trigonometric form. $$x^{3}+1=0$$

Step-by-Step Solution

Answer

The complex solutions are: \( x = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}, x = \cos \pi + i \sin \pi, x = \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \).

Step 1: Rewrite the Equation

Rewrite the given equation as a standard polynomial to see what type of solutions to expect.The given equation is:\[ x^3 + 1 = 0 \]We can rewrite it as:\[ x^3 = -1 \]

Step 2: Express the Right Side in Trigonometric Form

The right side of the equation, \(-1\), can be expressed in trigonometric form using Euler's formula.Since the complex number \(-1\) can be represented in the complex plane, it has a magnitude of 1 and an argument (angle) of \(\pi\) radians (equivalent to 180 degrees):\[ -1 = 1 \times (\cos \pi + i \sin \pi) \]

Step 3: Use De Moivre's Theorem

To find the cube roots of \(-1\), De Moivre's Theorem is used. It allows us to find the \(n\)-th roots of a complex number which in this case is the cube roots.Using De Moivre's Theorem, the roots are given by:\[ x = \sqrt[3]{1} \left( \cos \frac{(\pi + 2k\pi)}{3} + i \sin \frac{(\pi + 2k\pi)}{3} \right) \] where \(k = 0, 1, 2\).

Step 4: Calculate the Roots

Now calculate for each value of \(k\):1. For \(k = 0\): - \[ x = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2} \]2. For \(k = 1\): - \[ x = \cos \pi + i \sin \pi = -1 + i(0) = -1 \]3. For \(k = 2\): - \[ x = \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} = \frac{1}{2} - i\frac{\sqrt{3}}{2} \]

Step 5: Write Final Answer in Trigonometric Form

Compile all the complex roots in trigonometric form as found in Step 4:1. \( x = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \)2. \( x = -1 + 0i = \cos \pi + i \sin \pi \)3. \( x = \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \)

Key Concepts

Trigonometric Form

The trigonometric form of a complex number is a way to express complex numbers using trigonometry. It uses polar coordinates to represent the magnitude (or modulus) and direction (angle or argument) of the complex number. This form is quite useful, especially when dealing with powers and roots of complex numbers.

When we write a complex number in trigonometric form, it looks like this:

• \[ z = r(\cos \theta + i \sin \theta) \]

Here, \( r \) is the modulus of the complex number, which is the distance from the origin to the point \((a, b)\) in the complex plane. The angle \( \theta \) (also called the argument) is measured from the positive x-axis to the line joining the origin with the point.

The trigonometric form becomes particularly useful when you want to multiply or divide complex numbers, or to find their powers or roots, as products and powers are easier to handle in polar coordinates than in rectangular (a + bi) form.

De Moivre's Theorem

De Moivre's Theorem is a fundamental principle in complex numbers that provides a straightforward method for finding powers and roots of complex numbers. The theorem states that for any complex number written in trigonometric form, \( z = r(\cos \theta + i \sin \theta) \), and an integer \( n \), we can calculate

• \[ z^n = r^n (\cos(n\theta) + i\sin(n\theta) ) \]

This theorem is incredibly powerful as it allows us to easily compute higher powers and roots of complex numbers by simply multiplying the angle \( \theta \) by the power or root we are interested in. This way, we keep our calculations clean and simple. For negative roots, adding or subtracting \( 2\pi \) within the trigonometric functions ensures that the calculations address the correct angles associated with each root solution.

In the context of cube roots, as seen in the given problem, De Moivre's Theorem helps us systematically find all the unique cube roots by rotating through the complex plane at intervals, thus providing a consistent method for determining multiple roots.

Cube Roots

Cube roots of a complex number are the three values that satisfy the equation \( z^3 = w \), where \( w \) is the complex number you are trying to find the roots for. In the original exercise of finding the cube roots of \( -1 \), we observe an intrinsic rotation in the complex plane.

To visualize this, imagine placing the complex number \( -1 \) on a circle with a radius (modulus) of 1 and rotating in the complex plane. Expressing \( -1 \) in trigonometric form as \( 1(\cos \pi + i \sin \pi) \), we use De Moivre's Theorem to systematically identify all three cube roots by calculating angles:

• For \( k = 0\), \( x = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \) corresponds to the first root.
• For \( k = 1\), \( x = \cos \pi + i \sin \pi \) is the second root.
• For \( k = 2\), \( x = \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \) represents the third root.

Each root represents an equal division of the 360-degree circle by three, spreading evenly across the complex plane. By using these roots, we aptly cover all rotations and potential angle adjustments made at \( 120 \)-degree intervals, showing how trigonometric functions guide us through each step of the solution.