Problem 17 - A Graphical Approach to Precalculus with Limits

Question

Find each power. Write the answer in rectangular form. Do not use a calculator. $$(-2-2 i)^{5}$$

Step-by-Step Solution

Answer

The rectangular form of \((-2 - 2i)^{5}\) is \(32 - 32i\).

Step 1: Identify the Complex Number

The complex number given in the problem is \((-2 - 2i)\). It is expressed in the form \(a + bi\) where \(a = -2\) and \(b = -2\).

Step 2: Convert to Polar Form

First, calculate the magnitude \(r\) of the complex number using the formula \(r = \sqrt{a^2 + b^2}\). For \((-2 - 2i)\), we have:\[ r = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\]Next, find the argument \(\theta\) using \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\). Here, \(\theta = \tan^{-1}\left(\frac{-2}{-2}\right) = \tan^{-1}\left(1\right) = \frac{3\pi}{4}\), because the number is in the third quadrant.

Step 3: Apply De Moivre's Theorem

De Moivre's Theorem states that for a complex number in polar form \(r(\cos \theta + i\sin \theta)\), its \(n\)th power is found by raising the magnitude to the power and multiplying the angle by \(n\). Thus,\[(-2 - 2i)^5 = (2\sqrt{2})^5 \left[\cos(5 \cdot \frac{3\pi}{4}) + i\sin(5 \cdot \frac{3\pi}{4})\right]\]Calculate the new magnitude, \[(2\sqrt{2})^5 = 32\sqrt{2}\]And new angle, \(5 \cdot \frac{3\pi}{4} = \frac{15\pi}{4} = \frac{7\pi}{4} \ ext{(modulo}\ 2\pi \text{)}\).

Step 4: Convert Back to Rectangular Form

Compute the rectangular form using the cosine and sine of the new angle.\[\cos(\frac{7\pi}{4}) = \cos(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2},\quad \sin(\frac{7\pi}{4}) = \sin(-\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\]Hence,\[(-2 - 2i)^5 = 32\sqrt{2} \left(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) = 32(1 - i) = 32 - 32i\]

Step 5: Final Answer

The rectangular form of \((-2 - 2i)^{5}\) is \(32 - 32i\).

Key Concepts

Polar Form

Polar form is a useful way to represent complex numbers, particularly when multiplying and raising them to powers. It combines a magnitude (or modulus) and an angle (or argument) for a compact representation. Given a complex number, like \(a + bi\), you can transform it into polar form using the magnitude \(r\) and the angle \theta\. \[\text{Magnitude: } r = \sqrt{a^2 + b^2}\]For our example \((-2 - 2i)\), \(a\) and \(b\) are both \(-2\), so the magnitude \(r = 2\sqrt{2}\). \[\text{Argument: } \theta = \tan^{-1}\left(\frac{b}{a}\right)\] When calculating the angle, remember that the arctangent function, \(\tan^{-1}\), gives values that need adjustment depending on the quadrant. For \((-2 - 2i)\), it resides in the third quadrant resulting in an angle \( heta = \frac{3\pi}{4}\). The polar form becomes:

• Magnitude: \(2\sqrt{2}\)
• Argument: \(\frac{3\pi}{4}\)

Thus, the complex number is represented in polar form as \(2\sqrt{2} \left( \cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4}) \right)\). This form makes it straightforward to apply the operations such as using De Moivre's Theorem.

De Moivre's Theorem

De Moivre's Theorem is incredibly handy when working with powers and roots of complex numbers in polar form. The theorem states that a complex number \(r(\cos \theta + i\sin \theta)\) raised to any power \(n\) becomes:\[r^n \left( \cos (n\theta) + i\sin (n\theta) \right)\]This means you can take the magnitude to the power of \(n\) and simply multiply the angle by \(n\) too. Let's apply it to our example \((-2 - 2i)^5\).

• The magnitude \(2\sqrt{2}\) becomes \( (2\sqrt{2})^5 = 32\sqrt{2}\).
• The angle \(\frac{3\pi}{4}\) is multiplied by \(5\), changing it to \(\frac{15\pi}{4}\).

This results in a new angle \(\frac{15\pi}{4}\), modulo \(2\pi\), simplifying to \(\frac{7\pi}{4}\) which effectively rotates the angle to within a single rotation.By simplifying the impact of powers in the angle and magnitude, De Moivre's Theorem makes calculations involving complex numbers very manageable.

Rectangular Form

Rectangular form, or sometimes called the standard form, is another common way to express complex numbers \(a + bi\). Here \(a\) is the real part, and \(b\) is the imaginary part. After using De Moivre's Theorem, it's common to convert the result back to this form to easily interpret the positive and negative components of the solution.When converting back from polar form, you use the new magnitude and angle to find the cosine and sine values which replace the real and imaginary parts.

• Use the cosine of the transformed angle for the real part: \(\cos(\frac{7\pi}{4}) = \frac{\sqrt{2}}{2}\).
• Use the sine for the imaginary part: \(\sin(\frac{7\pi}{4}) = -\frac{\sqrt{2}}{2}\).

Finally, multiply these with the magnitude:\[32\sqrt{2} \left(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) = 32 - 32i\]Thus, the rectangular form of \((-2 - 2i)^5\) is \32 - 32i\. This form makes it clear what the real and imaginary components are, enhancing comprehension of the complex number's behavior.