Problem 23 - Precalculus

Question

Find the result of each expression using De Moivre's theorem. Write the answer in rectangular form. $$(-\sqrt{3}+i)^{6}$$

Step-by-Step Solution

Answer

The result of the expression \((-\sqrt{3} + i)^{6}\) is \(-64\) in rectangular form.

Step 1: Convert to Polar Form

To apply De Moivre's theorem, we need to express the complex number in polar form. A complex number \( a + bi \) can be written in polar form as \( r(\cos \theta + i \sin \theta) \), where \( r = \sqrt{a^2 + b^2} \) and \( \theta = \tan^{-1}(b/a) \). For the given number \( -\sqrt{3} + i \), we calculate: \( r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2 \). The angle \( \theta = \tan^{-1}(1 / -\sqrt{3}) = -\pi/6 \), adjusting for the correct quadrant gives: \( \theta = 5\pi/6 \). Thus, the polar form is \( 2(\cos 5\pi/6 + i \sin 5\pi/6) \).

Step 2: Apply De Moivre's Theorem

De Moivre's theorem states that \( [r(\cos \theta + i \sin \theta)]^n = r^n (\cos(n\theta) + i \sin(n\theta)) \). Here, \( r = 2 \), \( \theta = 5\pi/6 \), and \( n = 6 \). So we have: \[ [2(\cos 5\pi/6 + i \sin 5\pi/6)]^6 = 2^6 (\cos(6 \times 5\pi/6) + i \sin(6 \times 5\pi/6)) \].

Step 3: Calculate Power and Angles

First, calculate \( 2^6 = 64 \). Next, compute the angle: \( 6 \times 5\pi/6 = 5\pi \). Now, simplify \( \cos 5\pi \) and \( \sin 5\pi \): Since \( \cos(5\pi) = -1 \) and \( \sin(5\pi) = 0 \), we substitute these values into the polar form expression.

Step 4: Convert Back to Rectangular Form

Substituting the cosine and sine values into the equation, we have: \( 64(-1 + 0i) = -64 \). Thus, the final rectangular form of the expression \((-\sqrt{3} + i)^{6}\) is \(-64\).

Key Concepts

Polar Form

The polar form of a complex number is an alternative way to represent complex numbers. While the rectangular (or Cartesian) form uses coordinates on a plane as \(a + bi\), the polar form expresses a complex number through its magnitude and angle. This is particularly useful for multiplication and exponentiation, as seen when using De Moivre's Theorem.

In polar form, a complex number is written as \(r(\cos \theta + i \sin \theta)\), where:

• \(r\) represents the magnitude, calculated as \(\sqrt{a^2 + b^2}\).
• \(\theta\) is the phase angle, found using \(\tan^{-1}(b/a)\).

For example, if you have the complex number \(-\sqrt{3} + i\):

• The magnitude \(r\) is calculated as \(\sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{4} = 2\).
• The angle \(\theta\) is initially \(\tan^{-1}(1 / -\sqrt{3})\), which needs adjusting to match the number's position in its correct quadrant, resulting in \(5\pi/6\).

Rectangular Form

The rectangular form of a complex number is the standard format most people are familiar with, expressed as \(a + bi\), where \(a\) is the real part, and \(b\) is the imaginary part. This form is intuitive for addition and subtraction, thanks to its straightforward representation on a Cartesian plane.

While it offers simplicity in visualization and basic operations, the rectangular form is less convenient for certain calculations, like powers and roots, where the polar form excels. Converting from polar to rectangular form helps when the problem requires a final answer in rectangular format, as seen in the exercise's solution ending with a result of \(-64\) for \((-\sqrt{3} + i)^6\).

In practice, once calculations are performed in polar form, you take your final result and translate it back into rectangular form by calculating \(r \cos(\theta)\) for the real part and \(r \sin(\theta)\) for the imaginary part. If, for instance, your final calculations yield no imaginary component, it simplifies to a real number directly, as demonstrated by the exercise.

Complex Numbers

Complex numbers extend the real numbers by introducing an imaginary unit \(i\), defined as \(i^2 = -1\). This revolutionary idea broadens the numerical landscape to solve equations that real numbers alone cannot. A complex number contains a real part \(a\) and an imaginary part \(bi\), denoted as \(a + bi\).

Key aspects of complex numbers include:

• Addition and Subtraction: Occurs component-wise, \((a + bi) + (c + di) = (a+c) + (b+d)i\).
• Multiplication: Follows distributive law with combined terms adjusted for \(i\), e.g., \((a + bi)(c + di) = (ac - bd) + (ad + bc)i\).
• Conjugates: The conjugate of \(a + bi\) is \(a - bi\), useful for rationalizing denominators.

The polar form is particularly beneficial in simplifying how we handle powers and roots of complex numbers, courtesy of De Moivre's Theorem.

Consider our example \((-\sqrt{3} + i)^6\). By recognizing it as a complex number, we utilize both its rectangular and polar forms to efficiently perform exponentiation, ultimately translating complex operations into a more manageable format.