Chapter 11: Problem 73
In Exercises \(65-76\), find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. $$ \text { the four fourth roots of } z=16 $$
Short Answer
Expert verified
The four fourth roots of 16 are: \(2+0i, 0+2i, -2+0i, 0-2i\).
Step by step solution
01
Express in Polar Form
First, express the complex number 16 in polar form. A real number can be represented in polar form as \( r \times e^{i\theta} \). Since 16 is positive and on the real axis, its polar form is \( 16 \times e^{i \cdot 0} \). Here, \( r = 16 \) and \( \theta = 0 \).
02
Find the Roots in Polar Form
To find the fourth roots of a complex number in polar form, we use the formula for the nth root: \( z^{1/n} = r^{1/n} \times e^{i (\theta + 2k\pi)/n} \) for \( k = 0, 1, 2, \ldots, n-1 \). For this exercise, \( r = 16 \), \( \theta = 0 \), and \( n = 4 \). Calculate the magnitude: \( r^{1/4} = 16^{1/4} = 2 \). This gives us the roots: \( 2 \times e^{i (0 + 2k\pi)/4} \).
03
Compute Each Root
Now compute the fourth roots by considering four evenly spaced angles: \( \theta_k = \frac{2k\pi}{4} \) for \( k = 0, 1, 2, 3 \).1. For \( k = 0 \), \( \theta_0 = 0 \): Root is \( 2 \times e^{i \cdot 0} = 2 + 0i \).2. For \( k = 1 \), \( \theta_1 = \frac{\pi}{2} \): Root is \( 2 \times e^{i \pi/2} = 0 + 2i \).3. For \( k = 2 \), \( \theta_2 = \pi \): Root is \( 2 \times e^{i \pi} = -2 + 0i \).4. For \( k = 3 \), \( \theta_3 = \frac{3\pi}{2} \): Root is \( 2 \times e^{i 3\pi/2} = 0 - 2i \).
04
Convert to Rectangular Form
Each root obtained in polar form can be expressed as a complex number in the rectangular form by calculating the real and imaginary parts:1. \( 2 \times e^{i \cdot 0} = 2 \rightarrow 2 + 0i \).2. \( 2 \times e^{i \pi/2} = 0 + 2i \).3. \( 2 \times e^{i \pi} = -2 \rightarrow -2 + 0i \).4. \( 2 \times e^{i 3\pi/2} = 0 - 2i \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Form
The polar form is a way to express complex numbers using a magnitude (or length) and an angle. This is especially useful when dealing with complex numbers as it simplifies many mathematical operations like multiplication and finding roots.
- A complex number is usually written as \( z = r \angle \theta \) in polar form, where \( r \) is the magnitude and \( \theta \) is the angle with the positive real axis.
- The angle, \( \theta \), is measured in radians, and \( r \) is always non-negative.
- In our exercise, the number 16 can be represented as \( 16 \times e^{i \cdot 0} \). Here, \( r = 16 \) and \( \theta = 0 \), meaning the number is purely real and non-negative, lying on the positive real axis.
Polar form is particularly helpful in finding roots because it distributes angles evenly, making calculations systematic and structured.
- A complex number is usually written as \( z = r \angle \theta \) in polar form, where \( r \) is the magnitude and \( \theta \) is the angle with the positive real axis.
- The angle, \( \theta \), is measured in radians, and \( r \) is always non-negative.
- In our exercise, the number 16 can be represented as \( 16 \times e^{i \cdot 0} \). Here, \( r = 16 \) and \( \theta = 0 \), meaning the number is purely real and non-negative, lying on the positive real axis.
Polar form is particularly helpful in finding roots because it distributes angles evenly, making calculations systematic and structured.
Rectangular Form
Rectangular form expresses a complex number as a sum of its real and imaginary components. It closely resembles the way we use coordinates to pinpoint locations on a plane.
- In rectangular form, a complex number \( z \) is written as \( z = a + bi \) where \( a \) is the real part and \( b \) is the imaginary part.
- Conversion from polar to rectangular uses two key relations: \( a = r \cos(\theta) \) and \( b = r \sin(\theta) \). This converts the polar angle \( \theta \) into a coordinate-based format.
- From the exercise, converting the fourth root \( 2 \times e^{i \cdot 0} \) gives \( 2 + 0i \), and similarly for others: \( 0 + 2i \), \(-2 + 0i \), and \(0 - 2i \).
Each conversion uses the unit circle's symmetrical properties, transforming a rotation perspective into a straightforward coordinate, making operations tangible and intuitive.
- In rectangular form, a complex number \( z \) is written as \( z = a + bi \) where \( a \) is the real part and \( b \) is the imaginary part.
- Conversion from polar to rectangular uses two key relations: \( a = r \cos(\theta) \) and \( b = r \sin(\theta) \). This converts the polar angle \( \theta \) into a coordinate-based format.
- From the exercise, converting the fourth root \( 2 \times e^{i \cdot 0} \) gives \( 2 + 0i \), and similarly for others: \( 0 + 2i \), \(-2 + 0i \), and \(0 - 2i \).
Each conversion uses the unit circle's symmetrical properties, transforming a rotation perspective into a straightforward coordinate, making operations tangible and intuitive.
Fourth Roots
Finding the fourth roots of a complex number retrieves four distinct numbers (or roots), equally spaced around the origin in the complex plane. This is due to the fundamental property of complex roots having symmetric distribution.
- The formula to find an \( n \)th root, \( z^{1/n} = r^{1/n} \times e^{i (\theta + 2k\pi)/n} \) where \( k \) runs from 0 to \( n-1 \).
- The solution uses \( n = 4 \) (since we want fourth roots), so \( k \) takes values 0, 1, 2, and 3 to give each root. Each root angle is found by altering \( k \), yielding symmetry across the circle.
- For our example, \( r^{1/4} = 16^{1/4} = 2 \). The roots derived give distinctly spread out points: \( 2 \angle 0 \), \( 2 \angle \frac{\pi}{2} \), \( 2 \angle \pi \), and \( 2 \angle \frac{3\pi}{2} \).
These computations demonstrate a balance, positioned perfectly around the unit circle, visually illustrating rotations and solving elegantly.
- The formula to find an \( n \)th root, \( z^{1/n} = r^{1/n} \times e^{i (\theta + 2k\pi)/n} \) where \( k \) runs from 0 to \( n-1 \).
- The solution uses \( n = 4 \) (since we want fourth roots), so \( k \) takes values 0, 1, 2, and 3 to give each root. Each root angle is found by altering \( k \), yielding symmetry across the circle.
- For our example, \( r^{1/4} = 16^{1/4} = 2 \). The roots derived give distinctly spread out points: \( 2 \angle 0 \), \( 2 \angle \frac{\pi}{2} \), \( 2 \angle \pi \), and \( 2 \angle \frac{3\pi}{2} \).
These computations demonstrate a balance, positioned perfectly around the unit circle, visually illustrating rotations and solving elegantly.