Chapter 9: Problem 19
Find the four fourth roots of \(-16 .\)
Short Answer
Expert verified
The four fourth roots of \(-16\) are: \(\sqrt{2} + i\sqrt{2}\), \(-\sqrt{2} + i\sqrt{2}\), \(-\sqrt{2} - i\sqrt{2}\), and \(\sqrt{2} - i\sqrt{2}\).
Step by step solution
01
Understanding the Problem
We need to find the fourth roots of \(-16\), which means solving the equation \(z^4 = -16\) for the complex number \(z\). Complex numbers are often expressed in the form \(re^{i\theta}\), where \(r\) is the magnitude and \(\theta\) is the angle in radians.
02
Convert to Polar Form
To find the roots efficiently, we convert \(-16\) into its polar form. The magnitude \(r\) is \(16\), and the angle \(\theta\) is \(\pi\) radians (since \(-16\) lies on the negative real axis). So, \(-16\) can be expressed as \(16e^{i\pi}\).
03
Apply De Moivre's Theorem
To find the fourth roots, apply De Moivre's Theorem: if \(z = r^{1/n} \cdot e^{i(\theta + 2k\pi)/n}\), where \(n = 4\) and \(k = 0, 1, 2, 3\). Since \(r = 16\), we find \(r^{1/4} = 16^{1/4} = 2\).
04
Calculate the Roots
For each \(k\), calculate \(z_k = 2e^{i(\pi + 2k\pi)/4}\): - When \(k=0\), \(z_0 = 2e^{i\pi/4} = 2\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right) = \sqrt{2} + i\sqrt{2}\).- When \(k=1\), \(z_1 = 2e^{i3\pi/4} = 2\left(\cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4}\right) = -\sqrt{2} + i\sqrt{2}\).- When \(k=2\), \(z_2 = 2e^{i5\pi/4} = 2\left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right) = -\sqrt{2} - i\sqrt{2}\).- When \(k=3\), \(z_3 = 2e^{i7\pi/4} = 2\left(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4}\right) = \sqrt{2} - i\sqrt{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Form
When dealing with complex numbers, the polar form provides a way to express them using a magnitude and an angle instead of the usual real and imaginary parts. A complex number, typically represented as \( z = a + bi \), can be converted to polar form as \( z = r(\cos \theta + i\sin \theta) \). Here, \( r \) is the magnitude of the complex number, calculated as \( r = \sqrt{a^2 + b^2} \), and \( \theta \) is the argument or angle, found using \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).The polar form is particularly useful for simplifying mathematical operations like multiplication and division. In our exercise of finding the fourth roots of \(-16\), we first convert \(-16\) to polar form. The magnitude is \(16\) (distance from the origin on the complex plane), and since \(-16\) lies on the negative real axis, the angle \(\theta\) is \(\pi\) radians. Thus, \(-16\) can be represented as \(16e^{i\pi} \) in polar form.
De Moivre's Theorem
De Moivre's Theorem is an essential tool for dealing with powers and roots of complex numbers expressed in polar form. It states that for a complex number in polar form \( r(\cos \theta + i \sin \theta) \), its \( n \)-th power can be calculated as \( r^n (\cos(n\theta) + i \sin(n\theta)) \).This theorem can also be reversed to find roots. If we need to find an \( n \)-th root of a complex number, we modify \( \theta \) and \( r \) as follows:
- Compute the \( n \)-th root of the magnitude: \( r^{1/n} \).
- Divide the angle by \( n \) and add \( \frac{2k\pi}{n} \) for each root \( k = 0, 1, 2, ..., n-1 \).
Fourth Roots
To find the fourth roots of a complex number, you essentially divide the angle into four equal parts and find the corresponding magnitudes. The process involves:
- Finding the fourth root of the magnitude.
- Calculating the angles by evenly distributing them in the circle's 360 degrees, represented in radians.