Problem 69 - College Trigonometry

Question

Find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. the three cube roots of \(z=64\)

Step-by-Step Solution

Answer

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

Step 1: Convert to Polar Form

To find the cube roots of \(z=64\), we first express \(64\) in polar form. Since \(64\) is a real number, it is located on the real axis, making its magnitude \(r=64\) and its angle (argument) \( \theta = 0 \). Thus, in polar form we have \(z = 64(\cos 0 + i \sin 0)\).

Step 2: Find Cube Roots in Polar Form

For the cube roots, we use the formula for nth roots of a complex number: \( z_k = r^{1/n} \left( \cos \frac{\theta + 2\pi k}{n} + i \sin \frac{\theta + 2\pi k}{n} \right) \) where \( n = 3 \) and \( k = 0, 1, 2 \). Here, \( r = 64 \) and \( \theta = 0 \), so \( r^{1/3} = 4 \). The roots then are:

Step 3: Compute Cube Roots for k=0,1,2

- For \(k=0\): \(z_0 = 4\left( \cos 0 + i \sin 0 \right) = 4\) - For \(k=1\): \(z_1 = 4\left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) \) - For \(k=2\): \(z_2 = 4\left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right) \)

Step 4: Convert to Rectangular Form

Convert each of the polar forms to rectangular form: - For \(z_0\): Already in rectangular form, \(4 + 0i\)- For \(z_1\): \(4\left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) = 4(-\frac{1}{2} + i \frac{\sqrt{3}}{2}) = -2 + 2i\sqrt{3}\)- For \(z_2\): \(4\left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right) = 4(-\frac{1}{2} - i \frac{\sqrt{3}}{2}) = -2 - 2i\sqrt{3}\)

Key Concepts

Polar Form

The polar form of a complex number is a way to represent complex numbers on a coordinate plane using a radius and an angle. It is especially helpful when dealing with powers, roots, and trigonometric functions. Essentially, any complex number can be expressed as \[ z = r(\cos \theta + i \sin \theta) \] where \( r \) is the magnitude (or absolute value) of the complex number and \( \theta \) is the argument, which is the angle formed with the positive real axis.

\( r \) is calculated as the square root of the sum of the squares of the real and imaginary parts of the complex number.
• \( r = \sqrt{x^2 + y^2} \)
• \( \theta \) is found by taking the \( \arctan \left( \frac{y}{x} \right) \) , giving you an angle in the complex plane.

In the case of the number \( 64 \), it is a real number on the real axis, so \( \theta = 0 \) and magnitude \( r = 64 \). Therefore, its polar form is \( 64(\cos 0 + i \sin 0) \).Polar form is widely utilized to simplify the process of finding roots, including cube roots, adding efficiency and precision to the calculations.

Rectangular Form

The rectangular form, also known as Cartesian form, is a more common representation of complex numbers. It appears as \( z = x + yi \), where \( x \) is the real part and \( yi \) is the imaginary part. This form is typically used when working with addition, subtraction, and multiplication of complex numbers.

• \( x \) represents the horizontal position on the complex plane, i.e., the real component.
• \( yi \) represents the vertical position on the complex plane, i.e., the imaginary component, where \( i \) is the imaginary unit \( \sqrt{-1} \).

After finding the roots in polar form, converting them to rectangular form can give a clearer picture. For example, one of the cube roots for \( 64 \) was initially \( 4 \), appearing as \( 4 + 0i \) in rectangular form. Another root like \( 4(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}) \) converts to \(-2 + 2i\sqrt{3} \). Polar to rectangular conversion is achieved using:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

This conversion aids in visualizing and working with complex numbers in practical situations.

Cube Roots

Finding cube roots of complex numbers is a fascinating and frequently encountered challenge in mathematics. The formula used for finding the \( n \)-th roots, including cube roots, is\[ z_k = r^{1/n} \left( \cos \frac{\theta + 2\pi k}{n} + i \sin \frac{\theta + 2\pi k}{n} \right) \] where \( n \) is the number of roots you are seeking, and \( k \) ranges from 0 to \( n-1 \).

• \( r^{1/n} \) is the principal root, or the real part of the root for a given \( n \).
• For each value of \( k \), you calculate one specific root with a unique angle.

When finding cube roots of \( 64 \), we set \( n = 3 \) and calculate for \( k = 0, 1, 2 \). The real part \( r^{1/3} = 4 \) in this case. The roots identified are

• \( z_0 = 4 \left( \cos 0 + i \sin 0 \right) = 4 \)
• \( z_1 = 4 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) \)
• \( z_2 = 4 \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right) \)

Each root explores a different portion of the complex plane, revealing the balanced nature of these mathematical structures.