Problem 39 - Precalculus : Building Concepts and Connections

Question

Find the square roots of each complex number. Round all numbers to three decimal places. Find the fifth roots of -1

Step-by-Step Solution

Answer

The fifth roots of -1 are \( \cos(\frac {\pi}{5}) + i \cdot \sin(\frac {\pi}{5}) \), \( \cos(\frac {3\pi}{5}) + i \cdot \sin(\frac {3\pi}{5}) \), -1, \( \cos(\frac {7\pi}{5}) + i \cdot \sin(\frac {7\pi}{5}) \) and \( \cos(\frac {9\pi}{5}) + i \cdot \sin(\frac {9\pi}{5}) \).

Step 1: Identify the number

We have -1 in the complex plane. Written in polar form, this becomes \( -1 = \cos(\pi) + i \cdot \sin(\pi) \)

Step 2: Apply De Moivre’s Theorem

Using De Moivre’s Theorem, we can compute the fifth roots of -1. The theorem tells us that if a complex number z is expressed in polar form as \( r(\cos \theta + i \sin \theta) \), its nth roots are given by \( r^{\frac{1}{n}}(\cos \frac{\theta +2 \pi k}{n} + i \sin \frac{\theta +2 \pi k}{n}) \), for k = 0, 1, 2, ..., n - 1. Here, r = 1, n = 5 and \(\theta = \pi\). Substituting these values shows us the five fifth roots of -1.

Step 3: Compute the roots

We calculate the above for k = 0, 1, 2, 3, 4. This would yield \( \cos(\frac {\pi}{5}) + i \cdot \sin(\frac {\pi}{5}) \), \( \cos(\frac {3\pi}{5}) + i \cdot \sin(\frac {3\pi}{5}) \), \( \cos(\pi) + i \cdot \sin(\pi) = -1 \), \( \cos(\frac {7\pi}{5}) + i \cdot \sin(\frac {7\pi}{5}) \) and \( \cos(\frac {9\pi}{5}) + i \cdot \sin(\frac {9\pi}{5}) \). These are the fifth roots of -1.

Key Concepts

De Moivre's Theorem

De Moivre's Theorem is a fundamental tool in complex number theory, named after the French mathematician Abraham de Moivre. It provides an elegant way to compute powers and roots of complex numbers when they are represented in their polar form.

De Moivre's Theorem states that for a complex number in polar form, expressed as \( r(\cos \theta + i \sin \theta) \), where \( r \) is the magnitude (or modulus) and \( \theta \) is the argument (or angle), the nth power of this complex number is given by \( r^n(\cos(n\theta) + i \sin(n\theta)) \).

This theorem not only simplifies the process of raising complex numbers to a power but also plays a vital role in finding the nth roots of a complex number. To find the root, we reverse the process and find such complex numbers that would give us the original number when raised to the nth power. Specifically, the nth roots of a complex number can be found using the formula:
\[ r^{\frac{1}{n}}\left(\cos\frac{\theta +2 \pi k}{n} + i \sin\frac{\theta +2 \pi k}{n}\right) \]
for \( k = 0, 1, 2, ..., n - 1 \). Here, each integer value of \( k \) yields a distinct nth root, illustrating the property that any non-zero complex number has exactly \( n \) distinct nth roots.

Polar Form of Complex Numbers

The polar form of a complex number is a different way of representing complex numbers, emphasizing their geometric interpretation. In place of the standard x and y coordinates you might be familiar with on the Cartesian plane, polar coordinates use an angle and a radius to locate a point.

A complex number \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit, is often rewritten in polar form as \( r(\cos \theta + i \sin \theta) \), where \( r \) is the magnitude of the complex number and \( \theta \) is the angle formed with the positive x-axis. The magnitude \( r \) is also known as the modulus, and it can be found using the Pythagorean theorem as \( \sqrt{a^2 + b^2} \), while the angle \( \theta \), called the argument, can be computed using the arctangent function as \( \arctan\left(\frac{b}{a}\right) \) for \( a > 0 \) and adjusted properly when \( a \) is less than or equal to zero.

Polar representation is particularly useful when performing multiplication, division, and finding powers and roots of complex numbers. This is because these operations can be performed by simply adjusting the magnitude and the angle of the complex number, without the need to deal with the complexities that arise in their algebraic (a + bi) form.

Root Computation of Complex Numbers

The computation of complex roots is made more intuitive through the use of the polar form of complex numbers paired with De Moivre's Theorem. When we talk about finding the nth root of a complex number, we are looking for all complex solutions that when raised to the nth power, will return our original number.

As visualized on the complex plane, each root corresponds to a point that is evenly distributed around a circle whose center is the origin (0), with each root forming an equal angle between them. This results from the periodic nature of sine and cosine functions.

To compute nth roots using the formula derived from De Moivre's Theorem, we divide the argument \( \theta \) by \( n \) and also take the n-th root of the magnitude \( r \). This is done for each \( k = 0 \) to \( n - 1 \) to find all distinct roots. The term \( 2 \pi k \) is important as it ensures that we account for the complete circle around the origin and thus find all the various roots.

Let's consider an example: to find the square roots of a complex number, we would set \( n = 2 \). If the number were -1, as in -1 written in polar form \( \cos(\pi) + i \cdot \sin(\pi) \), applying the formula would give us two values for \( k \): 0 and 1. These correspond to the two distinct square roots that when squared will return -1. The process is the same for any other nth root computation, with the number of roots equal to the value of \( n \). This method provides an organized and comprehensive solution to find all possible roots for any given complex number.