Problem 18 - Contemporary Precalculus

Question

Find the nth roots in polar form. $$64\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) ; \quad n=2$$

Step-by-Step Solution

Answer

Question: Determine the two square roots of the complex number \(64\left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right)\) in polar form. Answer: The two square roots of the complex number \(64\left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right)\) in polar form are: $$z_0 = 8\left(\cos\left(\frac{\pi}{8}\right) + i\sin\left(\frac{\pi}{8}\right)\right)$$ $$z_1 = 8\left(\cos\left(\frac{9\pi}{8}\right) + i\sin\left(\frac{9\pi}{8}\right)\right)$$

Step 1: Identify the polar form of the complex number

The given polar form is: $$64\left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right)$$ We can see that the modulus is \(R = 64\) and the argument is \(\Theta = \frac{\pi}{4}\).

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

Recall the De Moivre's theorem to find the nth roots: $$z^n = R^n\left(\cos(n\Theta) + i \sin(n\Theta)\right)$$ In this case, we want to find the square roots, so n = 2: $$z^2 = 64^2\left(\cos 2\left(\frac{\pi}{4}\right) + i \sin 2\left(\frac{\pi}{4}\right)\right)$$

Step 3: Simplify the equation and solve for z

First, simplify the modulus: $$R' = \sqrt[2]{64} = 8$$ Now, divide the argument by n: $$\Theta ' = \frac{\Theta}{2} = \frac{\frac{\pi}{4}}{2} = \frac{\pi}{8}$$ Finally, find the roots using the new modulus and argument: $$z_0 = 8\left(\cos\left(\frac{\pi}{8}+0 \cdot \frac{2\pi}{2}\right) + i\sin\left(\frac{\pi}{8}+0 \cdot \frac{2\pi}{2}\right)\right) = 8\left(\cos\left(\frac{\pi}{8}\right) + i\sin\left(\frac{\pi}{8}\right)\right)$$ $$z_1 = 8\left(\cos\left(\frac{\pi}{8}+1 \cdot \frac{2\pi}{2}\right) + i\sin\left(\frac{\pi}{8}+1 \cdot \frac{2\pi}{2}\right)\right) = 8\left(\cos\left(\frac{9\pi}{8}\right) + i\sin\left(\frac{9\pi}{8}\right)\right)$$

Step 4: Write the final answer

The complex number \(64\left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right)\) has two square roots in polar form: $$z_0 = 8\left(\cos\left(\frac{\pi}{8}\right) + i\sin\left(\frac{\pi}{8}\right)\right)$$ $$z_1 = 8\left(\cos\left(\frac{9\pi}{8}\right) + i\sin\left(\frac{9\pi}{8}\right)\right)$$

Key Concepts

Polar Form

The polar form of complex numbers is a way to represent them in terms of their magnitude and direction. This is especially useful for multiplication and division of complex numbers. In polar form, a complex number is expressed as:

• Magnitude (also called the modulus), represented by \( R \)
• Argument (also called the angle \( \theta \)), showing its direction

The polar form is written as \( R (\cos \theta + i \sin \theta) \). This format helps visualize complex numbers as points in a plane:

• Magnitude \( R \) indicates the distance from the origin to the point
• Angle \( \theta \) is measured counterclockwise from the positive x-axis

Through polar form, complex numbers can be easily manipulated, especially when raising to powers or extracting roots.

Complex Numbers

Complex numbers are numbers that include both a real part and an imaginary part, typically written in the form \( a + bi \). Here:

• \( a \) is the real part
• \( b \) is the imaginary part

The imaginary unit \( i \) is defined as the square root of \( -1 \). Complex numbers can be represented graphically on the complex plane:

• The real part corresponds to the x-axis
• The imaginary part corresponds to the y-axis

Complex numbers are essential in engineering, physics, and applied mathematics. They allow solving equations that involve square roots of negative numbers and provide a complete solution to polynomial equations.

De Moivre's Theorem

De Moivre's theorem is a fundamental relation in complex number theory, particularly in polar form calculations. This theorem relates complex numbers to trigonometry:

• It states that \( (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) \)

This is particularly useful for raising complex numbers to any power \( n \). For nth roots, the theorem helps in determining the roots by raising the complex number to the power of \( 1/n \). Using De Moivre's theorem simplifies calculations:

• No need to convert back to rectangular form for calculations
• Keeps the relationship between magnitude and angle intact

It's applied in finding multiple roots of complex numbers, such as calculating square roots in the example.

Square Roots

Square roots of complex numbers can be found using the concept of roots in polar form. When finding the square roots, one aims to discover two numbers that when squared will result in the original complex number. In polar form:

• Find the square root of the modulus
• Divide the angle by 2

Thus, if \( z = R (\cos \theta + i \sin \theta) \), its square roots are expressed as:

• \( \sqrt{R} \left( \cos \left(\frac{\theta}{2} + k \cdot \pi \right) + i \sin \left(\frac{\theta}{2} + k \cdot \pi \right) \right) \)
• Where \( k = 0, 1 \) since square roots give two solutions

The process simplifies finding roots by focusing on basic arithmetic with angles and square roots without dealing with complex multiplications.