Problem 10 - A Graphical Approach to Precalculus with Limits

Question

Find each power. Write the answer in rectangular form. Do not use a calculator. $$(2 \sqrt{2}-2 i \sqrt{2})^{6}$$

Step-by-Step Solution

Answer

The power \((2\sqrt{2} - 2i\sqrt{2})^6\) is \(4096i\) in rectangular form.

Step 1: Express in Polar Form

First, we express the complex number \(2\sqrt{2} - 2i\sqrt{2}\) in polar form. To do this, find the modulus \(r\) and the argument \(\theta\). The modulus is \(r = \sqrt{(2\sqrt{2})^2 + (-2\sqrt{2})^2} = \sqrt{8 + 8} = 4\). The argument is \(\theta = \arctan\left(\frac{-2\sqrt{2}}{2\sqrt{2}}\right) = \arctan(-1) = -\frac{\pi}{4}\). Thus, the polar form is \(4(\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}))\).

Step 2: Apply De Moivre's Theorem

Use De Moivre's Theorem to find \((r(\cos \theta + i\sin \theta))^n = r^n(\cos(n\theta) + i\sin(n\theta))\). For \(n = 6\), we have \(r^6 = 4^6 = 4096\) and \(n\theta = 6(-\frac{\pi}{4}) = -\frac{3\pi}{2}\).

Step 3: Simplify Trigonometric Functions

Evaluate \(\cos\left(-\frac{3\pi}{2}\right)\) and \(\sin\left(-\frac{3\pi}{2}\right)\). The angle \(-\frac{3\pi}{2}\) corresponds to one full rotation counterclockwise minus a \(\frac{\pi}{2}\) turn, landing on the negative y-axis. Therefore, \(\cos\left(-\frac{3\pi}{2}\right) = 0\) and \(\sin\left(-\frac{3\pi}{2}\right) = 1\).

Step 4: Compute Final Rectangular Form

Substitute the simplified trigonometric values into the polar form. This gives \(4096(0 + i\cdot 1) = 4096i\). Therefore, the rectangular form is \(4096i\).

Key Concepts

Rectangular Form

The rectangular form of a complex number is its representation using the real and imaginary parts. For a complex number like \( a + bi \), \( a \) is the real component while \( b \) represents the imaginary component. When solving problems involving complex numbers, the rectangular form is quite intuitive.
To understand it better:

• The complex number is positioned on a 2D plane where the x-axis denotes the real part and the y-axis denotes the imaginary part.
• The complex number \( 2\sqrt{2} - 2i\sqrt{2} \) has real part \( 2\sqrt{2} \) and imaginary part \( -2\sqrt{2} \).

The rectangular form is especially useful when performing operations such as addition and subtraction since both parts are directly accessible. Students may often switch to this form after deriving results in polar form to interpret results more straightforwardly in practical problems.

Polar Form

The polar form presents complex numbers based on their magnitude and direction. Instead of simply seeing a complex number as a sum of real and imaginary parts, polar form involves expressing the number as a distance from the origin and an angle from the positive real axis. This is denoted as \( r(\cos \theta + i\sin \theta) \), where \( r \) is the modulus and \( \theta \) is the argument.
To derive the polar form:

• Calculate the modulus \( r = \sqrt{a^2 + b^2} \). For \( a + bi = 2\sqrt{2} - 2i\sqrt{2} \), \( r \) becomes 4.
• Find the argument \( \theta = \arctan\left(\frac{b}{a}\right) \). Here, \( \theta = \arctan(-1) = -\frac{\pi}{4} \).

The polar form provides a clear understanding of the complex number's magnitude and its rotation on the plane. It's particularly advantageous when multiplying or dividing complex numbers, which can be cumbersome in rectangular form.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool used primarily to compute powers and roots of complex numbers. It states that for any real number \( \theta \) and integer \( n \), \((r(\cos \theta + i\sin \theta))^n = r^n(\cos(n\theta) + i\sin(n\theta))\). This effectively ties polar form to exponentiation and allows complex powers to be solved efficiently.
In practical applications:

• Identify the modulus \( r \) and argument \( \theta \) from the polar form.
• Compute the power of the modulus \( r^n \). For example, with \( r = 4 \) and \( n = 6 \), we get \( 4^6 = 4096 \).
• Multiply the argument by the power \( n\theta \). Here, \( n\theta = 6(-\frac{\pi}{4}) = -\frac{3\pi}{2} \).

De Moivre’s Theorem simplifies expressions and facilitates conversions between polar and rectangular forms by using trigonometric identities. By evaluating \( \cos(-\frac{3\pi}{2}) \) and \( \sin(-\frac{3\pi}{2}) \) straightforwardly as 0 and 1, we can revert back to the rectangular form efficiently, obtaining results like \( 4096i \) directly.