Problem 74 - Precalculus Mathematics for Calculus

Question

Find the indicated power using De Moivre’s Theorem. $$(\sqrt{3}-i)^{-10}$$

Step-by-Step Solution

Answer

\( \frac{1}{2048} + i \frac{\sqrt{3}}{2048} \)

Step 1: Convert to Polar Form

To use De Moivre’s Theorem, we first convert the complex number into polar form. The complex number given is \( \sqrt{3} - i \). To find its polar form, we calculate the modulus \( r \) and the argument \( \theta \). The modulus is found as \( r = \sqrt{ (\sqrt{3})^2 + (-1)^2 } = \sqrt{3 + 1} = 2 \).Next, find \( \theta \) using the formula \( \tan^{-1} \left( \frac{-1}{\sqrt{3}} \right) \). Since the point is in the fourth quadrant, \( \theta = -\frac{\pi}{6} \).So in polar form: \( z = 2 \text{cis} \left(-\frac{\pi}{6}\right) \), where \( \text{cis} \theta = \cos \theta + i \sin \theta \).

Step 2: Apply De Moivre’s Theorem

De Moivre’s Theorem states \((r \text{cis} \theta)^n = r^n \text{cis}(n\theta)\). To find \((\sqrt{3} - i)^{-10}\), we substitute \( r = 2 \), \( \theta = -\frac{\pi}{6} \), and \( n = -10 \).This results in: \[ z^{-10} = (2 \text{cis}(-\frac{\pi}{6}))^{-10} = 2^{-10} \text{cis}(-10 \times -\frac{\pi}{6}) = 2^{-10} \text{cis} \left(\frac{10\pi}{6}\right) \] Simplify \( \frac{10\pi}{6} \) to \( \frac{5\pi}{3} \) and adjust to be within \( [0, 2\pi) \), giving \( \text{cis}(\frac{\pi}{3}) \).

Step 3: Calculate the Result

Next, we calculate \( z^{-10} = 2^{-10} \text{cis} \left( \frac{\pi}{3} \right) \). First, find \( 2^{-10} \): \[ 2^{-10} = \frac{1}{1024} \] Then, compute \( \text{cis} \left( \frac{\pi}{3} \right) = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i \frac{\sqrt{3}}{2} \).Thus, \[ z^{-10} = \frac{1}{1024} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = \frac{1}{2048} + i \frac{\sqrt{3}}{2048} \]

Key Concepts

Complex Numbers

Complex numbers are numbers that include a real part and an imaginary part. They are usually written in the form \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part, with \( i \) being the imaginary unit. For example, in the given exercise, \( \sqrt{3} - i \) is a complex number where \( a = \sqrt{3} \) and \( b = -1 \).
Complex numbers extend the real numbers and are pivotal in fields such as engineering and physics. They allow for the solution of equations that wouldn’t be solvable within the set of real numbers.
Imaginary numbers are numbers that yield a negative result when squared. The imaginary unit \( i \) is defined by \( i^2 = -1 \). This extension to the number system enables us to handle problems involving square roots of negative numbers.

• Real part, \( a \): The horizontal component on the complex plane.
• Imaginary part, \( b \): The vertical component on the complex plane.

Understanding the interplay between the real and the imaginary parts is key to grasping the nature of complex numbers.

Polar Form

Polar form is a way to express complex numbers using a radius and an angle, instead of a real and an imaginary component. It is particularly useful for complex multiplication and powers, as illustrated in De Moivre's Theorem.
In polar form, a complex number is shown as \( z = r \text{cis} \theta \), where \( r \) is the modulus or absolute value of the complex number, and \( \theta \) is the argument or angle of the complex number with the positive x-axis on the complex plane.
To convert a complex number \( a + bi \) to polar form:

• Calculate the modulus \( r = \sqrt{a^2 + b^2} \).
• Find the argument \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \), taking into consideration the quadrant where the complex number is located.

In our example, \( \sqrt{3} - i \) is transformed to polar form as \( 2 \text{cis} \left(-\frac{\pi}{6}\right) \). This switch to polar representation simplifies operations like powering and multiplication.

Complex Conjugates

Complex conjugates are pairs of complex numbers that have the same real part but opposite imaginary parts. Given a complex number \( a + bi \), its conjugate is \( a - bi \).
Using complex conjugates is valuable for simplifying the division of complex numbers and finding their roots. The product of a complex number and its conjugate gives a real number.

• If \( z = a + bi \), then \( \overline{z} = a - bi \).
• The magnitude of a complex number, \( |z| \), remains unchanged by conjugation.
• Product: \( z \cdot \overline{z} = a^2 + b^2 \), always a non-negative real number.

In calculations involving De Moivre’s Theorem, understanding the role of conjugates can help reduce errors and simplify expressions. However, in this specific exercise, the main focus is on powering a complex number, so conjugates are used primarily in refining calculations where necessary.