Problem 22 - Contemporary Precalculus

Question

Find the nth roots in polar form. $$16\left(\cos \frac{\pi}{7}+i \sin \frac{\pi}{7}\right) ; \quad n=5$$

Step-by-Step Solution

Answer

Answer: The 5th roots of the given complex number in polar form are: 1. $$2\left(\cos \frac{\frac{\pi}{7}+2\cdot0\cdot\pi}{5}+i \sin \frac{\frac{\pi}{7}+2\cdot0\cdot\pi}{5}\right)$$. 2. $$2\left(\cos \frac{\frac{\pi}{7}+2\cdot1\cdot\pi}{5}+i \sin \frac{\frac{\pi}{7}+2\cdot1\cdot\pi}{5}\right)$$. 3. $$2\left(\cos \frac{\frac{\pi}{7}+2\cdot2\cdot\pi}{5}+i \sin \frac{\frac{\pi}{7}+2\cdot2\cdot\pi}{5}\right)$$. 4. $$2\left(\cos \frac{\frac{\pi}{7}+2\cdot3\cdot\pi}{5}+i \sin \frac{\frac{\pi}{7}+2\cdot3\cdot\pi}{5}\right)$$. 5. $$2\left(\cos \frac{\frac{\pi}{7}+2\cdot4\cdot\pi}{5}+i \sin \frac{\frac{\pi}{7}+2\cdot4\cdot\pi}{5}\right)$$.

Step 1: 1. Express the complex number in polar form

The given complex number is already in polar form. We have the modulus as 16 and the argument as \(\frac{\pi}{7}\) (i.e., the angle of the complex number): $$16\left(\cos \frac{\pi}{7}+i \sin \frac{\pi}{7}\right)$$.

Step 2: 2. Apply De Moivre's theorem to find the nth roots

According to De Moivre's theorem, the kth root (with k = 0, 1, 2, ..., n-1) of a complex number in polar form is given by: $$\sqrt[n]{r}\left(\cos \frac{\theta+2k\pi}{n}+i \sin \frac{\theta+2k\pi}{n}\right)$$, where r is the modulus and θ is the argument. In our case, n = 5, r = 16, and θ = \(\frac{\pi}{7}\).

Step 3: 3. Calculate the modulus and argument for each nth root

For our problem, the modulus of each root is: $$\sqrt[5]{16} = 2$$ and the angle for the kth root will be: $$\frac{\frac{\pi}{7}+2k\pi}{5}$$.

Step 4: 4. Write down the nth roots in polar form

We will now write down the 5 roots in polar form using the modulus and argument calculated in step 3: 1. For k = 0: $$2\left(\cos \frac{\frac{\pi}{7}+2\cdot0\cdot\pi}{5}+i \sin \frac{\frac{\pi}{7}+2\cdot0\cdot\pi}{5}\right)$$. 2. For k = 1: $$2\left(\cos \frac{\frac{\pi}{7}+2\cdot1\cdot\pi}{5}+i \sin \frac{\frac{\pi}{7}+2\cdot1\cdot\pi}{5}\right)$$. 3. For k = 2: $$2\left(\cos \frac{\frac{\pi}{7}+2\cdot2\cdot\pi}{5}+i \sin \frac{\frac{\pi}{7}+2\cdot2\cdot\pi}{5}\right)$$. 4. For k = 3: $$2\left(\cos \frac{\frac{\pi}{7}+2\cdot3\cdot\pi}{5}+i \sin \frac{\frac{\pi}{7}+2\cdot3\cdot\pi}{5}\right)$$. 5. For k = 4: $$2\left(\cos \frac{\frac{\pi}{7}+2\cdot4\cdot\pi}{5}+i \sin \frac{\frac{\pi}{7}+2\cdot4\cdot\pi}{5}\right)$$. These are the 5 roots of the given complex number in polar form.

Key Concepts

Polar Form of Complex Numbers

The polar form of complex numbers makes calculations much easier, especially when dealing with operations like multiplication, division, and finding roots. In polar form, a complex number is represented as a combination of its modulus and its argument. This is expressed as:

• Modulus: The distance from the origin to the point in the complex plane. For a complex number in polar form, it is denoted as \( r \).
• Argument: The angle from the positive real axis to the line segment representing the complex number. It is denoted as \( \theta \).
• Polar Form: Written as \( r(\cos \theta + i \sin \theta) \), sometimes abbreviated as \( r \text{cis} \theta \).

When given a complex number in standard form \( a + bi \), it can be converted to polar form by calculating the modulus \( r = \sqrt{a^2 + b^2} \) and the argument \( \theta = \tan^{-1}(\frac{b}{a}) \). In this exercise, the complex number \( 16(\cos \frac{\pi}{7} + i \sin \frac{\pi}{7}) \) is already provided in polar form, with a modulus of 16 and an argument of \( \frac{\pi}{7} \).

De Moivre's Theorem

De Moivre's Theorem is a powerful tool for simplifying the computation of powers and roots of complex numbers in polar form. The theorem states that:

• For a complex number \( z = r(\cos \theta + i \sin \theta) \), its nth power is \( z^n = r^n (\cos(n\theta) + i \sin(n\theta)) \).

This formula allows us to easily compute the powers by raising the modulus to the nth power and multiplying the argument by n. When finding nth roots, we use the inverse process:

• The modulus of the root is \( \sqrt[n]{r} \), and
• The argument of the kth root is \( \frac{\theta + 2k\pi}{n} \), where \( k \) is an integer that takes on values from 0 to \( n-1 \).
• This provides \( n \) distinct roots, distributed evenly on the complex plane.

In the provided exercise, De Moivre's Theorem helps us find the 5th roots of the complex number by calculating new arguments and keeping the modulus consistent for all roots.

Complex Number Calculations

Calculating with complex numbers involves a good understanding of their algebraic and geometric representations. The polar form simplifies multiplication and division due to the properties of trigonometric functions. Here's a quick overview:

• Multiplication: Multiply the moduli and add the arguments.
• Division: Divide the moduli and subtract the arguments.

These operations are much simpler in polar form than in rectangular (a + bi) form, especially when handling exponential forms or finding roots. For the nth roots, using polar representation makes it efficient since we simply take the nth root of the modulus, and adjust the argument using the De Moivre's method discussed earlier. These methods reduce complicated calculations into more manageable steps. In this exercise, the calculations are straightforward, requiring converting angles to fit within the standard range of angles and finding specific expressions for each distinct root.

Modulus and Argument

Understanding the modulus and the argument is crucial when working with complex numbers in polar form. Here’s a deeper look into these key components:

• Modulus: Also known as the absolute value of the complex number, the modulus \( r \) is a measure of its distance from the origin (0,0) in the complex plane. It's calculated with \( r = \sqrt{a^2 + b^2} \), where \( a \) and \( b \) are the real and imaginary parts, respectively.
• Argument: It is the angle between the positive x-axis and the line representing the complex number, moving counterclockwise. Measured in radians, the argument helps in describing the direction of the complex number.

In the process of finding roots, calculating the modulus involves taking the nth root, while the argument is divided by \( n \) as per De Moivre's Theorem. In the exercise, the modulus was simply \( \sqrt[5]{16} = 2 \), and the argument for each root was adjusted by adding multiples of \( 2\pi \) and dividing by 5.