Problem 32 - Advanced Engineering Mathematics

Question

In Problems 27-32, use (10) to compute all roots. Sketch these roots on an appropriate circle centered at the origin. $$ (-1-\sqrt{3} i)^{1 / 4} $$

Step-by-Step Solution

Answer

The fourth roots are: \(\sqrt[4]{2}(\frac{\sqrt{3}}{2} + i\frac{1}{2})\), \(\sqrt[4]{2}(-\frac{\sqrt{3}}{2} + i\frac{1}{2})\), \(\sqrt[4]{2}(0-i)\), and \(\sqrt[4]{2}(\frac{\sqrt{3}}{2} - i\frac{1}{2})\).

Step 1: Convert the Complex Number to Polar Form

First, identify the complex number we're dealing with: \(-1 - \sqrt{3} i\). This can be expressed as a complex number in the form \(a + bi\) where \(a = -1\) and \(b = -\sqrt{3}\).Find the modulus (magnitude) \(r\) of the complex number using: \[ r = \sqrt{a^2 + b^2} = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2 \]Next, find the argument \(\theta\) using: \[ \theta = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{-\sqrt{3}}{-1}\right) = \tan^{-1}(\sqrt{3}) \]The angle \(\theta\) is \(\pi - \frac{\pi}{3} = \frac{2\pi}{3}\) because the complex number is in the second quadrant.Thus, the polar form is: \[ 2 (\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}) \]

Step 2: Use De Moivre’s Theorem to Find the Fourth Roots

For a complex number in polar form \( r (\cos \theta + i \sin \theta) \), the \(n\)-th root can be found as: \[ \sqrt[n]{r} \left( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} \right), \text{ for } k = 0, 1, \ldots, n-1 \]Here, \(n = 4\) and \(r = 2\), and \(\theta = \frac{2\pi}{3}\). Calculate the fourth roots:- For \( k = 0 \):\[ \sqrt[4]{2} \left( \cos \frac{\frac{2\pi}{3} + 0}{4} + i \sin \frac{\frac{2\pi}{3} + 0}{4} \right) = \sqrt[4]{2} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \]- For \( k = 1 \):\[ \sqrt[4]{2} \left( \cos \frac{\frac{2\pi}{3} + 2\pi}{4} + i \sin \frac{\frac{2\pi}{3} + 2\pi}{4} \right) = \sqrt[4]{2} \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right) \]- For \( k = 2 \):\[ \sqrt[4]{2} \left( \cos \frac{\frac{2\pi}{3} + 4\pi}{4} + i \sin \frac{\frac{2\pi}{3} + 4\pi}{4} \right) = \sqrt[4]{2} \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right) \]- For \( k = 3 \):\[ \sqrt[4]{2} \left( \cos \frac{\frac{2\pi}{3} + 6\pi}{4} + i \sin \frac{\frac{2\pi}{3} + 6\pi}{4} \right) = \sqrt[4]{2} \left( \cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6} \right) \]

Step 3: Simplify and Evaluate the Roots

Evaluate the trigonometric functions for the angles calculated:- For \( k = 0 \): \[ \sqrt[4]{2} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) = \sqrt[4]{2} \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \]- For \( k = 1 \): \[ \sqrt[4]{2} \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right) = \sqrt[4]{2} \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \]- For \( k = 2 \): \[ \sqrt[4]{2} \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right) = \sqrt[4]{2} (0 - i) \]- For \( k = 3 \): \[ \sqrt[4]{2} \left( \cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6} \right) = \sqrt[4]{2} \left( \frac{\sqrt{3}}{2} - i \frac{1}{2} \right) \]

Step 4: Sketch the Roots on the Complex Plane

Sketch the calculated roots on the complex plane, using a circle of radius \(\sqrt[4]{2}\) centered at the origin. Each root corresponds to a point equally spaced on the circle at the angles calculated: \(\frac{\pi}{6}\), \(\frac{5\pi}{6}\), \(\frac{3\pi}{2}\), and \(\frac{11\pi}{6}\). The points will form a square since all fourth roots of a given modulus will lie at equal angular distances on the circle.

Key Concepts

Polar Form

Polar form is a way to express complex numbers using magnitude and angle rather than real and imaginary parts. If you have a complex number in the form \(a + bi\), you can convert it to polar form by finding two key components:

• The modulus \(r\), which represents the distance from the origin to the point on the complex plane. This is calculated as \( r = \sqrt{a^2 + b^2} \).
• The argument \(\theta\), which is the angle formed with the positive x-axis, often found using \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).

Once these are determined, the polar form can be written as \( r(\cos \theta + i \sin \theta) \). This form is particularly useful for multiplying and dividing complex numbers, as well as finding powers and roots.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool for dealing with complex numbers in polar form. It simplifies the process of raising complex numbers to powers and finding roots. According to this theorem, for a complex number \( r(\cos \theta + i \sin \theta) \) raised to a power \( n \), the expression becomes:

\[ r^n (\cos(n\theta) + i \sin(n\theta)) \]

This formula allows you to easily compute powers of complex numbers by simply multiplying the angle \(\theta\) by the power \(n\), and raising the modulus \(r\) to the same power. Moreover, De Moivre's Theorem is essential for finding roots of complex numbers, as it provides a systematic way to find all possible solutions.

Roots of Complex Numbers

Finding roots of complex numbers can seem challenging, but using their polar form makes it manageable. When tasked with finding the \(n\)-th root of a complex number \( r(\cos \theta + i \sin \theta) \), De Moivre's Theorem assists with the following approach:

• Determine the \(n\)-th root of the magnitude, which is \( \sqrt[n]{r} \).
• Calculate the angles recursively by dividing \(\theta\) and each additional angle increment, \(2k\pi\), by \(n\). The general formula for the \(k\)-th root is: \( \sqrt[n]{r} (\cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n}) \), where \(k = 0, 1, 2, ..., n-1\).

The result gives \(n\) distinct roots, each of which is a point evenly spaced on a circle with radius \( \sqrt[n]{r} \). This property ensures that the roots are symmetrically distributed in the complex plane, which is a beautiful geometric result of complex numbers.