Problem 66 - Precalculus with Limits

Question

Finding a Power of a Complex Number In Exercises \(65-80\) , use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form. $$(2+2 i)^{6}$$

Step-by-Step Solution

Answer

The sixth power of (2+2i) is -64i

Step 1: Converting the Complex Number to Polar Form

A complex number can be written in polar form as \(r (\cos \theta + i \sin \theta)\), where \(r\) is the modulus of the complex number and \(\theta\) is the argument. Here, the modulus \(r = \sqrt{(2^2 + 2^2)} = 2\sqrt{2}\) and the argument \(\theta = \arctan(2/2) = \frac{\pi}{4}\). So the complex number in polar form is \(2\sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})\).

Step 2: Applying DeMoivre's Theorem

DeMoivre's Theorem states that for any complex number in polar form \(r (\cos \theta + i \sin \theta)\), its nth power is \(r^n (\cos n\theta + i \sin n\theta)\). Applying this to our complex number, we get \((2\sqrt{2})^6 (\cos \frac{6 \pi}{4} + i \sin \frac{6 \pi}{4}) = 64 (\cos \frac{3 \pi}{2} + i \sin \frac{3 \pi}{2})\).

Step 3: Transforming Back to Rectangular Form

Finally, we convert this back to rectangular form using the identities \(\cos(\frac{3 \pi}{2}) = 0\) and \(\sin(\frac{3 \pi}{2}) = -1\). So the final result is \(64 \cdot (0 - i) = -64i\).

Key Concepts

Complex Numbers

Complex numbers are numbers that have a real part and an imaginary part. They are usually expressed in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part with \(i\) representing the square root of -1. This means that complex numbers include all the familiar numbers we use in real life, plus numbers that can be thought of as being roots of negative numbers.

Here's a breakdown:

• The real part, \(a\), is just like any regular number you're used to. It's the part of the complex number that doesn't involve \(i\).
• The imaginary part, \(bi\), is where things get interesting. It includes \(i\), our notation for the imaginary unit, which has the property that \(i^2 = -1\).

Complex numbers are helpful in many areas of math and engineering, especially when dealing with oscillations and waves, where phases need to be taken into account.

Polar Form

Polar form is another way to express complex numbers, offering an alternative to the rectangular form. It's particularly useful when you are dealing with multiplication or division of complex numbers or finding powers and roots. In polar form, a complex number is represented as \(r (\cos \theta + i \sin \theta)\), where:

• \(r\) is the modulus of the complex number, which is the distance from the origin in the complex plane, calculated as \(r = \sqrt{a^2 + b^2}\).
• \(\theta\) is the argument, or angle with the positive real axis, found using \(\theta = \arctan(b/a)\).

In our exercise, the modulus is \(2\sqrt{2}\) and the argument is \(\frac{\pi}{4}\), corresponding to a point on the unit circle where the angle formed is exactly \(45^\circ\).

This form makes it straightforward to use DeMoivre’s Theorem, which elegantly expresses powers and roots by simply multiplying the modulus and angle.

Rectangular Form

Finally, returning to the rectangular form involves expressing the result of a complex operation back in the familiar \(a + bi\) notation. This is especially valuable for presenting your final answers in a clear, standard format that is easy to interpret.

To convert back from polar form to rectangular form, you use:

• The cosine (\(\cos(\theta)\)) to find the real part.
• The sine (\(\sin(\theta)\)) to determine the imaginary part.

Taking the complex number we processed earlier, we found that the cosine and sine of \(\frac{3\pi}{2}\) are 0 and -1, respectively. This simplifies the expression \(64 (0 - i)\), leading us directly to the final answer: \(-64i\).

Thus, rectangular form provides a neat and clear way to display complex numbers, especially after operations that involve powers or roots, ensuring that your complex number is easy to compute with and understand.