Problem 56 - Precalculus

Question

Find all complex solutions to the given equations. $$x^{7}+\pi^{14}=0$$

Step-by-Step Solution

Answer

The complex solutions are \(x_k = \pi^2 e^{i \frac{\pi + 2k\pi}{7}}\) for \(k = 0, 1, 2, 3, 4, 5, 6\).

Step 1: Rewrite the Equation

Rewrite the equation in the form where the polynomial equals zero. The original equation is given by \(x^7 + \pi^{14} = 0\). This can be rewritten as \(x^7 = -\pi^{14}\).

Step 2: Identify the Modulus and Argument

To solve for the complex roots, express the right side in polar form. Since \(-\pi^{14}\) lies on the negative real axis in the complex plane, its modulus is \(\pi^{14}\), and its argument is \(\pi\) as it is located on the negative x-axis.

Step 3: Use the Root Formula for Complex Numbers

For a number \(re^{i\theta}\), the \(n^{th}\) roots are given by \(r^{1/n} e^{i(\theta + 2k\pi)/n}\) for \(k = 0, 1, ..., n-1\). Here, \(r = \pi^{14}\), \(n = 7\), and \(\theta = \pi\).

Step 4: Calculate the Modulus of the Roots

Calculate the modulus of the roots: \((\pi^{14})^{1/7} = \pi^{14/7} = \pi^2\).

Step 5: Calculate the Arguments of the Roots

Calculate the arguments of each root: \(\theta_k = \frac{\pi + 2k\pi}{7}\) for each integer \(k = 0, 1, 2, 3, 4, 5, 6\).

Step 6: Write Down Each Complex Root

Combine the modulus and argument to express the roots in their polar form: \[x_k = \pi^2 e^{i\theta_k} = \pi^2 \left(\cos\left(\frac{\pi + 2k\pi}{7}\right) + i\sin\left(\frac{\pi + 2k\pi}{7}\right)\right) \] for \(k = 0, 1, 2, 3, 4, 5, 6\).

Key Concepts

Polar Form

Complex numbers can be represented in a unique format called the polar form. This is particularly useful when solving equations involving roots, like in our problem with complex solutions for \(x^7 + \pi^{14} = 0\). In polar form, a complex number is expressed as \(re^{i\theta}\), where \(r\) is the modulus and \(\theta\) is the argument.
To convert a complex number to polar form, follow these steps:

• Calculate the modulus \(r\), which is the distance from the origin to the point in the complex plane. For example, the modulus of \(-\pi^{14}\) is \(\pi^{14}\).
• Determine the argument \(\theta\), which is the angle formed with the positive x-axis. Here, \(-\pi^{14}\) has an argument of \(\pi\), as it lies on the negative real axis.

This conversion is crucial for finding roots and performing other operations involving complex numbers.

Modulus and Argument

The modulus and argument are foundational concepts in understanding polar form. Let's dive deeper:

• Modulus: This is the magnitude or absolute value of a complex number. Think of it as the length of the vector in the complex plane. It is calculated as \(r = \sqrt{a^2 + b^2}\) for a complex number \(a + bi\). In our case, \(-\pi^{14}\), the modulus simplifies as \(\pi^{14}\).
• Argument: The argument is the angle \(\theta\) the vector makes with the positive real axis. It is calculated usually using trigonometric functions. For \(-\pi^{14}\), the argument \(\theta = \pi\).

Understanding these two components simplifies multiplying, dividing, and raising complex numbers to powers, since these operations are much more straightforward in polar form.

Complex Roots

Finding complex roots involves an interesting process utilizing polar form. We're looking at the equation \(x^7 = -\pi^{14}\), requiring us to find the 7th roots of \(-\pi^{14}\). First, we express \(-\pi^{14}\) in polar form; we have \(\pi^{14}e^{i\pi}\).
To find the roots:

• Use the formula \(r^{1/n} e^{i(\theta + 2k\pi)/n}\), where \(r\) is the modulus, \(n\) is the degree of the root, and \(k = 0, 1, 2, \ldots, n-1\).
• In our example, the modulus \(r\) becomes \(\pi^2\) since \(\pi^{14/7} = \pi^2\).
• The argument for each root \(\theta_k = \frac{\pi + 2k\pi}{7}\). Calculate these for each \(k = 0, 1, 2, 3, 4, 5, 6\).

After calculation, each root can be represented in the form \(\pi^2 \left(\cos\left(\frac{\pi + 2k\pi}{7}\right) + i\sin\left(\frac{\pi + 2k\pi}{7}\right)\right)\). This method showcases the elegance of using polar form to uncover complex solutions efficiently.