Problem 113 - Precalculus with Limits

Question

Show that \(\frac{1}{2}(1 - \sqrt{3}i)\) is a ninth root of \(-1\).

Step-by-Step Solution

Answer

The complex number \(\frac{1}{2}(1 - \sqrt{3}i)\) is indeed a ninth root of -1.

Step 1: Write down the original complex number

The complex number is \(\frac{1}{2}(1 - \sqrt{3}i)\).

Step 2: Raise to the power of 9

Calculate the ninth power of the complex number. This can be done by multiplying the complex number by itself 8 times or more easily using de Moivre's theorem, which states that if \(z = r(\cos \theta + i \sin \theta)\), then \(z^n = r^n(\cos n\theta + i \sin n\theta)\). First, put the number in the polar form \(r(\cos \theta + i \sin \theta)\). To do this, you should:1. Calculate the modulus \(r\) of the complex number: \(r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = 1\).2. Calculate the argument \(\theta\) (often denoted arg(\(z\)) of the complex number: \(\theta = arctan \frac{\sqrt{3}}{2}\). As the complex number lies in the third quadrant, the argument is \(\theta = \arctan\left(\frac{\sqrt{3}}{1}\right) - \pi = \frac{2\pi}{3}\).So, our polar form is \(1(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})\).Next, we can use De Moivre's theorem:\(\left[1\left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right)\right]^9 = 1^9\left(\cos \frac{18\pi}{3} + i \sin \frac{18\pi}{3}\right) = \cos {6\pi} + i \sin {6\pi} = -1\).

Step 3: Compare the result with -1

The operation brings out -1 which matches exactly with what we expected.

Key Concepts

De Moivre's Theorem

Understanding De Moivre's Theorem is crucial when working with complex numbers. This theorem is a formula that connects complex numbers with trigonometry. It says that for a complex number in polar form, given as \( z = r(\text{cos } \theta + i\text{ sin } \theta) \), the nth power of this number can be represented as \( z^n = r^n(\text{cos } n\theta + i\text{ sin } n\theta) \).

When using this theorem, we're essentially multiplying the modulus \( r \) of the complex number by itself \( n \) times and multiplying the angle \( \theta \) by \( n \) as well. This is particularly helpful since it allows for an easier computation of complex numbers raised to a power, as compared to multiplying the complex number by itself repeatedly. For instance, in our example of finding the ninth power, De Moivre's theorem simplifies the process, avoiding the need for time-consuming multiplications.

Ninth Root of a Complex Number

Finding the ninth root of a complex number may sound daunting, but using De Moivre's Theorem, it is quite manageable. To find the nth root, we essentially do the reverse of taking a power. Instead of multiplying the angle by \( n \), we divide it by \( n \). Moreover, we take the nth root of the modulus.

Given a complex number to the power of 9 that is equal to another complex number, we want to find a base complex number, such that when raised to the power of 9, we obtain the original number. To solve such problems, we can apply the formula inversely to root the modulus and divide the argument by 9.

Complex Number Modulus

The modulus of a complex number, usually denoted \( r \), is the distance from the origin to the point in the complex plane that represents the complex number. It is found using the formula \( r = \text{sqrt}(a^2+b^2) \), where \( a \) and \( b \) are the real and imaginary parts of the complex number, respectively.

In the example of \( \frac{1}{2}(1 - \text{sqrt}3i) \), we calculate the modulus to check the 'radius' of our number in the complex plane. It turns out to be 1, which means our number is located at a unit distance from the origin. Knowing the modulus is a key step in using De Moivre's Theorem to find powers and roots of complex numbers, as it's the base we raise to the nth power or nth root.

Complex Number Argument

The argument of a complex number is the angle formed by the line connecting the number to the origin with the positive x-axis. This angle is usually denoted \( \theta \) and is measured in radians. The argument can be found using inverse trigonometric functions, specifically the arctangent of the ratio of the imaginary part to the real part.

However, we have to be careful to adjust the angle depending on which quadrant the complex number lies in, as arctangent only gives us the angle for the first or fourth quadrant. In our example, the initial \( \text{arctan} \) calculation gave us an angle for the first quadrant, but since our complex number lies in the third quadrant, we adjusted by subtracting \( \text{pi} \). This gives us the correct argument, \( \text{theta} = \text{2pi/3} \), setting us up for application of De Moivre’s theorem.