Problem 18 - Algebra and Trigonometry with Analytic Geometry

Question

Find the three cube roots of \(64 i\).

Step-by-Step Solution

Answer

The cube roots of \(64i\) are \(2\sqrt{3} + 2i\), \(-2\sqrt{3} + 2i\), and \(-4i\).

Step 1: Express in Polar Form

First, we need to express the complex number \(64i\) in polar form. A complex number \(a + bi\) in polar form is \(re^{i\theta}\), where \(r = \sqrt{a^2 + b^2}\) and \(\theta = \tan^{-1}(\frac{b}{a})\). Here, \(a = 0\) and \(b = 64\), so \(r = \sqrt{0^2 + 64^2} = 64\). The angle \(\theta = \tan^{-1}(\frac{64}{0})\) leads \(\theta = \frac{\pi}{2}\) since it's purely imaginary in the positive direction.

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

To find the cube roots, we use De Moivre's Theorem. For a complex number \(re^{i\theta}\), the nth roots are given by: \( \sqrt[n]{r} \cdot e^{i(\frac{\theta + 2k\pi}{n})}\), where \(n = 3\) and \(k = 0, 1, 2\). Here, \(r = 64\), \(\theta = \frac{\pi}{2}\), so the cube roots are \(64^{1/3} e^{i(\frac{\frac{\pi}{2} + 2k\pi}{3})}\).

Step 3: Calculate the Roots

First, compute \(64^{1/3} = 4\). Now, calculate the three roots for each \(k\). For \(k = 0\), root 1 is \(4e^{i(\frac{\pi}{6})}\). For \(k = 1\), root 2 is \(4e^{i(\frac{\pi}{2} + \frac{2\pi}{3})}\), which simplifies to \(4e^{i\frac{5\pi}{6}}\). For \(k = 2\), root 3 is \(4e^{i(\frac{\pi}{2} + \frac{4\pi}{3})}\), simplifying to \(4e^{i\frac{3\pi}{2}}\).

Step 4: Convert Back to Rectangular Form

Now, we convert these roots back to the rectangular form (\(a + bi\)). Root 1: \(4e^{i\frac{\pi}{6}} = 4(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}) = 4(\frac{\sqrt{3}}{2} + \frac{i}{2}) = 2\sqrt{3} + 2i\). Root 2: \(4e^{i\frac{5\pi}{6}} = 4(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}) = 4(-\frac{\sqrt{3}}{2} + \frac{i}{2}) = -2\sqrt{3} + 2i\). Root 3: \(4e^{i\frac{3\pi}{2}} = 4(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}) = 4(0 - i) = -4i\).

Key Concepts

Cube Roots

When we talk about cube roots in the realm of complex numbers, we are actually finding three distinct solutions. For any complex number, say \( z \), the cube roots are the numbers which when multiplied by themselves three times will give back \( z \). In our case, the task is to find the cube roots of \( 64i \).
The process reveals some interesting properties of complex numbers:

• Each root can be represented in a geometric sense as vertices of a triangle that fits perfectly around the origin on the complex plane.
• The roots are equally spaced on the circle on the complex plane because they all have the same modulus and their arguments differ by \( \frac{2\pi}{3} \).
• Finding the cube roots involves solving for the roots using a specialized version of nth root formulas rather than the simple arithmetic method we use for real numbers.

This requires converting the complex number into a special form first—for which we use the polar form.

Polar Form

The polar form of complex numbers expresses the number with a radius and an angle instead of a rectangular coordinate system. It's an elegant way to express complex numbers because it aligns extremely well with operations like exponentiation and finding roots, particularly for visualization on the complex plane.
To express a complex number \( a + bi \) in polar form, you calculate:

• The modulus \( r = \sqrt{a^2 + b^2} \), representing distance from the origin, and
• The argument \( \theta = \tan^{-1}(\frac{b}{a}) \), indicating direction as the angle made with the positive real axis.

For \( 64i \):

• The modulus is \( 64 \) because \( r = \sqrt{0^2 + 64^2} \)
• The argument is \( \frac{\pi}{2} \) because being purely imaginary, the angle is 90 degrees or radian \( \frac{\pi}{2} \).

Using this, \( 64i \) becomes \( 64e^{i\frac{\pi}{2}} \). This form streamlines the process of finding the roots using the powerful De Moivre's Theorem.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool used to calculate powers and roots of complex numbers in polar form. It states that for any complex number given in polar form \( (r e^{i\theta}) \) and an integer \( n \), the formula \((r e^{i\theta})^n = r^n e^{in\theta}\) holds.
This theorem simplifies exponentiation and root extraction of complex numbers. For root extraction, if \( n \) is replaced with \( \frac{1}{n}\), such as for cube roots, each root is:

• Given by \( \sqrt[n]{r} \cdot e^{i(\frac{\theta + 2k\pi}{n})} \)
• Calculate multiple roots by varying \( k \), where \( k \) is an integer that runs from 0 to \( n-1 \) for \( n \) distinct roots.

For our task, \( 64e^{i\frac{\pi}{2}} \) has cube roots calculated as \( 4 \cdot e^{i(\frac{\frac{\pi}{2} + 2k\pi}{3})} \). Vary \( k \) to obtain the three different cube roots. This method is efficient and leverages the symmetry in geometrical construction of complex numbers.

Rectangular Form

Finally, let's talk about the rectangular form of complex numbers. This is the conventional representation of complex numbers of the form \( a + bi \), where:

• \( a \) is the real part, indicating the number's projection on the horizontal axis, and
• \( b \) is the imaginary part, indicating the projection on the vertical axis.

Once you've calculated the polar form roots, it's often helpful to convert back to rectangular form to understand exact components of the number.
For example, with the feedback of transformation from polar to rectangular:

• \( 4e^{i\frac{\pi}{6}} \) equals \( 2\sqrt{3} + 2i \)
• \( 4e^{i\frac{5\pi}{6}} \) equals \(-2\sqrt{3} + 2i \)
• \( 4e^{i\frac{3\pi}{2}} \) translates to \(-4i \)

This form is often used for direct algebraic operations and gives a clear view of the positions on the complex plane.