Problem 45 - College Algebra with Corequisite Support

Question

For the following exercises, use a calculator to help answer the questions. Show that a solution of \(x^{6}+1=0\) is \(\frac{\sqrt{3}}{2}+\frac{1}{2} i\).

Step-by-Step Solution

Answer

The correct solution is \(\frac{\sqrt{3}}{2} - \frac{1}{2}i\).

Step 1: Identify the equation

The given equation is \(x^6 + 1 = 0\). This equation can be rewritten as \(x^6 = -1\), indicating that we are looking for the sixth roots of \(-1\).

Step 2: Find the sixth roots of -1

To find the sixth roots of \(-1\), we use the fact that \(-1\) in polar form is represented as \(-1 = e^{i\pi}(\cos\pi + i\sin\pi)\). The general formula for the roots is then \(x_k = e^{i(\pi + 2k\pi)/6}\) for \(k = 0, 1, 2, 3, 4, 5\).

Step 3: Compute \(x_5\) for a specific value

We will find one of these roots specifically for \(k = 5\). Substituting \(k = 5\) into the formula \(x_k = e^{i(\pi + 2k\pi)/6}\), we get \(x_5 = e^{i(\pi + 10\pi)/6} = e^{i11\pi/6}\).

Step 4: Convert to rectangular form

We use Euler's formula to convert \(x_5 = e^{i11\pi/6}\) into rectangular form: \(x_5 = \cos\left(\frac{11\pi}{6}\right) + i\sin\left(\frac{11\pi}{6}\right)\). Calculating these, we find that \(\cos\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2}\) and \(\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}\). Thus, \(x_5 = \frac{\sqrt{3}}{2} - \frac{1}{2}i\).

Step 5: Verify and Correct (if needed)

The solution of the equation \(x^6 + 1 = 0\) given in the problem as \(x = \frac{\sqrt{3}}{2} + \frac{1}{2}i\) does not seem to match the root we calculated, suggesting either an issue with the problem statement or the conversion of the negative. Our computation shows that for \(k = 5\), the solution is \(\frac{\sqrt{3}}{2} - \frac{1}{2}i\), not plus.

Key Concepts

Polar Coordinates

Complex numbers can be represented using polar coordinates. This involves expressing them in terms of a magnitude, known as the modulus, and a direction, represented by an angle called the argument. These coordinates provide a nice alternative to rectangular coordinates. To understand this better, any complex number can be written in the form \[ z = r( ext{cos} \theta + i \text{sin} \theta) \]where \(r\) is the modulus and \(\theta\) is the argument of the complex number.

• **Modulus**, \(r\), is the distance of the complex number from the origin on the complex plane.
• **Argument**, \(\theta\), is the angle made by the line representing the complex number with the positive real axis.

This form is particularly useful for multiplying and dividing complex numbers. It also simplifies finding powers and roots of complex numbers by using formulas derived from these polar expressions. Converting back to polar coordinates at the end of calculating powers or roots ensures clarity of the results.

Euler's Formula

Euler's Formula provides a beautiful relationship between complex numbers and exponential functions. It states that for any real number \(\theta\), \[ e^{i\theta} = \text{cos}(\theta) + i\text{sin}(\theta) \].This formula allows complex numbers to be expressed in exponential form, making computations far more manageable, especially when dealing with roots of complex numbers or when raising complex numbers to powers.For example, when solving for the roots of \(-1\), we often write \(-1\) in its exponential form using Euler's formula:

• \(-1\) can be expressed as \(e^{i \pi}\).

Using Euler’s formula simplifies multiplying and dividing complex numbers, and makes computations involving rotations much cleaner and more intuitive.

Roots of Unity

The concept of the roots of unity arises in finding the roots of any number, in this case, \(-1\). Specifically, the **nth roots of a complex number** are the solutions to the equation \(z^n = 1\). These roots are spaced evenly on the unit circle in the complex plane.For example, to find the sixth roots of \(-1\):

• First, express \(-1\) in polar form as it is convenient for such calculations: \(-1 = e^{i \pi}\).
• Then the sixth roots of \(-1\) are given by \[ x_k = e^{i(\frac{\pi + 2k\pi}{6})} \] for \(k = 0, 1, 2, ..., 5\).

Each specific \(k\) value corresponds to a unique root, allowing us to find all the sixth roots by iterating through these values. Understanding and calculating roots of unity is fundamental in various applications such as signal processing and Fourier transforms.

Rectangular Form Conversion

In complex numbers, we often convert between polar and rectangular (or Cartesian) forms. Rectangular form represents a complex number as \(a + bi\), where \(a\) and \(b\) are real numbers.Converting from polar to rectangular form uses the trigonometric values of the angle involved. For our example, to find the rectangular form of a sixth root of \(-1\):

• Start with the polar form \(x_k = e^{i\theta}\).
• Apply Euler’s formula: \[ x_k = \text{cos}(\theta) + i\text{sin}(\theta) \]
• Substitute the angle values using known trigonometric identities. If \(\theta\) is \(\frac{11\pi}{6}\), we substitute to find:
  • \(\text{cos}\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2} \)
  • \(\text{sin}\left(\frac{11\pi}{6}\right) = -\frac{1}{2} \)

This conversion is crucial because many problems are solved more easily in one form or the other, necessitating familiarity with interchanging these representations.