Problem 24 - A Graphical Approach to Precalculus with Limits

Question

Find the cube roots of each complex number. Leave the answers in trigonometric form. Then graph each cube root as a vector in the complex plane. $$27 i$$

Step-by-Step Solution

Answer

Cube roots are: \( 3 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right), \) \( 3 \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right), \) and \( 3 \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right).\)

Step 1: Convert to Polar Form

The complex number given is \(27i\). This can be expressed in polar form \(r(\cos \theta + i \sin \theta)\). The modulus \(r\) is \(|27i| = 27\) and the argument \(\theta\) for \(27i\) (which lies on the positive imaginary axis) is \(\frac{\pi}{2}\). Hence, \(27i = 27(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})\).

Step 2: Determine Cube Roots Using De Moivre’s Theorem

To find the cube roots, we use De Moivre's theorem, which states that for a complex number \(z = r(\cos \theta + i \sin \theta)\), the cube roots are given by:\[ r^{1/3} \left( \cos \left( \frac{\theta + 2k\pi}{3} \right) + i \sin \left( \frac{\theta + 2k\pi}{3} \right) \right), \quad k = 0, 1, 2 \]For \(27i\), \(r = 27\), so \(r^{1/3} = 3\).

Step 3: Calculate Each Cube Root for k = 0, 1, 2

Substitute \(k = 0, 1, 2\) in the formula:1. For \(k = 0\): \[ 3 \left( \cos \left( \frac{\pi}{6} \right) + i \sin \left( \frac{\pi}{6} \right) \right) = 3 \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = \frac{3\sqrt{3}}{2} + i \frac{3}{2} \]2. For \(k = 1\): \[ 3 \left( \cos \left( \frac{5\pi}{6} \right) + i \sin \left( \frac{5\pi}{6} \right) \right) = 3 \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = -\frac{3\sqrt{3}}{2} + i \frac{3}{2} \]3. For \(k = 2\): \[ 3 \left( \cos \left( \frac{3\pi}{2} \right) + i \sin \left( \frac{3\pi}{2} \right) \right) = 3(0 - i) = -3i \]

Step 4: Present Solutions in Trigonometric Form

Write the three cube roots in trigonometric form:1. \( \frac{3\sqrt{3}}{2} + i \frac{3}{2} = 3 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \)2. \( -\frac{3\sqrt{3}}{2} + i \frac{3}{2} = 3 \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right) \)3. \( -3i = 3 \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right) \)

Key Concepts

Trigonometric Form

The trigonometric form of a complex number is a way of expressing complex numbers using their distance from the origin (modulus) and the angle they make with the positive real axis (argument). It is an alternative to the standard form, which is expressed as \(a + bi\). The trigonometric form helps us to work with complex numbers in a manner similar to vectors. This is particularly useful when dealing with multiplication and powers of complex numbers.

The trigonometric form is given by the expression \(r(\cos \theta + i \sin \theta)\), where:

• \(r\) is the modulus of the complex number, calculated as \(|a + bi| = \sqrt{a^2 + b^2}\).
• \(\theta\) is the argument, which is the angle formed with the positive real axis. It can be calculated using the arctan function.

Using this form becomes especially helpful when combined with De Moivre's Theorem, making it easier to find roots of complex numbers.

Polar Form

The polar form of a complex number is closely related to its trigonometric form. In essence, these two forms are equivalent but are used in different contexts. The polar form represents a complex number using its modulus and argument but emphasizes the geometric interpretation.

For a complex number \(z = a + bi\), its polar form can be represented as \(r \operatorname{cis} \theta\), where "cis" is shorthand for \(\cos \theta + i \sin \theta\). This representation highlights:

• \(r\), the modulus, equivalent to the radius in polar coordinates.
• \(\theta\), the argument, equivalent to the angle.

The polar form is particularly useful in tasks like finding the roots of complex numbers or when visualizing these numbers on the complex plane. When \(27i\) is converted to polar form, we find its modulus to be \(27\) and its argument to be \(\frac{\pi}{2}\). This step sets up the application of De Moivre’s theorem for finding roots.

De Moivre's Theorem

De Moivre's Theorem provides a powerful technique for raising complex numbers to a power and for finding roots of complex numbers, which is essential in solving polynomial equations involving them. This theorem states that for any complex number in trigonometric form \(r(\cos \theta + i \sin \theta)\), and any integer \(n\), the number raised to the power \(n\) is given by:\[ z^n = r^n (\cos(n\theta) + i \sin(n\theta)) \]
However, when finding the \(n\)-th roots, De Moivre’s Theorem adapts to:\[ z^{1/n} = r^{1/n}\left( \cos\left(\frac{\theta + 2k\pi}{n}\right) + i \sin\left(\frac{\theta + 2k\pi}{n}\right)\right) \]for \(k = 0, 1, 2, \ldots, n-1\). Each \(k\) value provides a different root, which are evenly spaced around a circle in the complex plane.

With \(27i\) and \(n=3\), we found the three cube roots are spaced 120 degrees apart on the complex plane.

Complex Plane

The complex plane is a two-dimensional plane where each point represents a complex number. The horizontal axis is known as the real axis, while the vertical axis is called the imaginary axis. This plane offers a visual way of representing complex numbers. By graphing on the complex plane, it's easier to understand relationships and operations involving complex numbers, such as addition or finding roots.

When working with cube roots of complex numbers, like in this exercise, the complex plane allows us to visualize the roots as vectors. These vectors originate from the origin and make distinct angles with the real axis. For \(27i\), the cube roots are plotted on the complex plane as vectors forming a triangle, each spaced at 120 degrees from each other.

• First root at \(\frac{\pi}{6}\).
• Second root at \(\frac{5\pi}{6}\).
• Third root at \(\frac{3\pi}{2}\).

This triangular arrangement highlights the symmetrical nature of roots in the complex plane.