Problem 87 - Precalculus Mathematics for Calculus

Question

Find the indicated roots, and graph the roots in the complex plane. The cube roots of \(i\)

Step-by-Step Solution

Answer

The cube roots of \(i\) are \(\frac{\sqrt{3}}{2} + i\frac{1}{2}\), \(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\), and \(-i\). They form an equilateral triangle on the complex plane.

Step 1: Express Complex Number in Polar Form

The first step in finding the cube roots of a complex number is to express the complex number in polar form. The complex number given is \(i\), which can be written as \(0 + i\). In polar form,\( i = e^{i\frac{\pi}{2}}\) because its modulus is 1 (distance from origin) and its argument is \(\frac{\pi}{2}\) (90 degrees above the real axis).

Step 2: Apply De Moivre's Theorem for Cube Roots

De Moivre's Theorem states that if \(z = r e^{i\theta}\), then the \(n\)th roots are given by \(z_k = r^{1/n} e^{i(\theta + 2k\pi)/n}\) for \(k = 0, 1, ..., n-1\). For cube roots (=3), the roots of \(i = e^{i\frac{\pi}{2}}\) are \(z_k = 1 e^{i(\frac{\pi}{2} + 2k\pi)/3}\) for \(k = 0, 1, 2\).

Step 3: Calculate Each Cube Root

Calculate the three roots one by one:- For \(k = 0\): \(z_0 = e^{i\frac{\pi}{6}} = \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} = \frac{\sqrt{3}}{2} + i\frac{1}{2}\).- For \(k = 1\): \(z_1 = e^{i\frac{5\pi}{6}} = \cos \frac{5\pi}{6} + i\sin \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + i\frac{1}{2}\).- For \(k = 2\): \(z_2 = e^{i\frac{3\pi}{2}} = \cos \frac{3\pi}{2} + i\sin \frac{3\pi}{2} = 0 - i\).

Step 4: Graph the Roots on the Complex Plane

Now, plot the roots on the complex plane:- Root \(z_0\) at \(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\).- Root \(z_1\) at \(\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\).- Root \(z_2\) at \((0, -1)\). These three points should form an equilateral triangle centered at the origin on the complex plane.

Key Concepts

Polar form

The polar form is a powerful way to represent complex numbers, making it easier to perform various operations like multiplication and finding roots. Each complex number can be represented in the form of \( r \times e^{i\theta} \), where \( r \) is the modulus and \( \theta \) is the argument (angle) of the complex number.

When you have a complex number like \( i \), it's initially written as \( 0 + i = i \). In polar form, this translates to \( i = e^{i\frac{\pi}{2}} \), because:

• The modulus \( r \) is 1, which is the distance from the origin to the point on the complex plane.
• The argument \( \theta \) is \( \frac{\pi}{2} \) radians, which equates to 90 degrees, indicating its position on the imaginary axis.

Representing complex numbers in polar form simplifies many math operations, especially when you need to find roots or powers.

De Moivre's Theorem

De Moivre's Theorem is a crucial tool in complex number calculus. It relates complex numbers in polar form to exponentiation and makes finding roots manageable. According to this theorem, if you have a complex number \( z = r e^{i\theta} \), then its \( n \)th roots can be calculated as:

• \( z_k = r^{1/n} e^{i(\theta + 2k\pi)/n} \)
• Here, \( k = 0, 1, 2, \, ..., \, n-1 \)

For our example, the number \( i = e^{i\frac{\pi}{2}} \) is simplified when finding cube roots. When you apply the formula:

\[ z_k = 1^{1/3} e^{i(\frac{\pi}{2} + 2k\pi)/3} \]

You calculate the roots for \( k = 0, 1, \text{and} \, 2 \) as follows:

• For \( k = 0 \), \( z_0 = e^{i\frac{\pi}{6}} \)
• For \( k = 1 \), \( z_1 = e^{i\frac{5\pi}{6}} \)
• For \( k = 2 \), \( z_2 = e^{i\frac{3\pi}{2}} \)

This theorem simplifies the process by providing a systematic approach to calculate the roots.

Complex plane graphing

Graphing complex numbers on the complex plane helps visualize solutions and provides insight into their geometric relationships. The complex plane, also known as the Argand plane, is a valuable tool in representing complex numbers where:

• The horizontal axis represents real numbers.
• The vertical axis represents imaginary numbers.

When graphing the cube roots of \( i \), you plot the coordinates of the calculated roots:

• Root \( z_0 \) at \( \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \)
• Root \( z_1 \) at \( \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \)
• Root \( z_2 \) at \( (0, -1) \)

These points form an equilateral triangle centered at the origin. Such geometric interpretations, especially symmetrical formations like equilateral triangles, help understand the nature of solutions in a visual and intuitive way.