Problem 82 - A Graphical Approach to College Algebra

Question

Simplify each power of i to \(i, 1,-i,\) or \(-1\). $$\frac{1}{i^{12}}$$

Step-by-Step Solution

Answer

The simplified expression is 1.

Step 1: Understand the powers of i

The imaginary unit \(i\) has a cyclical pattern with powers: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\). These powers repeat every four steps.

Step 2: Determine the power cycle position

Calculate the remainder when 12 is divided by 4 (since the cycle length is 4). This is done to figure out which power in the cycle \(i^1, i^2, i^3,\) or \(i^4\) corresponds to \(i^{12}\). 12 divided by 4 leaves a remainder of 0.

Step 3: Identify the equivalent power of i

According to the cycle, \(i^4 = 1\) (and thus \(i^8 = 1\), \(i^{12} = 1\), etc.). Since the remainder is 0, this means \(i^{12}\) is equivalent to \(i^0\), which is the same as \(i^4 = 1\).

Step 4: Solve the expression

Replace \(i^{12}\) with 1 in the original expression: \[\frac{1}{i^{12}} = \frac{1}{1} = 1\].

Key Concepts

Powers of i

Imaginary numbers often require dealing with the powers of the imaginary unit, denoted by the letter \(i\). In mathematics, the imaginary unit is foundational in extending the number system beyond real numbers. Essential properties of \(i\) involve its powers, which form a repeating cycle. Understanding this cyclical pattern is critical.

• \(i^1 = i\): This is just the imaginary unit itself.
• \(i^2 = -1\): Multiplying \(i\) by itself results in -1.
• \(i^3 = -i\): Here, \(i^3\) is \(i \times i^2\), resulting in \(-i\).
• \(i^4 = 1\): This power returns the product back to the real number 1.

Beyond these values, powers of \(i\) beyond \(i^4\) start repeating this pattern, which makes simplification tasks straightforward—key to the exercise is recognizing the cycle and applying it.

Imaginary Unit

The imaginary unit \(i\) is not just a mathematical curiosity, but a practical tool to handle equations that involve the square root of negative numbers. Within complex numbers, \(i\) is defined by its property that \(i^2 = -1\). Remember, without this designation, the square root of a negative number cannot be expressed within the realm of real numbers. With \(i\), however, it is possible.When combining real and imaginary numbers, we reach the realm of complex numbers, written in the form \(a + bi\), where \(a\) and \(b\) are real numbers. This makes the imaginary unit vital, as it assists in solving equations that have no solutions solely within the real number set, especially quadratic equations with negative discriminants.

Cyclic Patterns

Understanding cyclic patterns is crucial when simplifying powers of the imaginary unit \(i\). Since \(i\) exhibits a repeating cycle every four powers, this means that calculating high powers of \(i\) can be simplified by identifying which part of the cycle they correspond to. Every power of \(i\) can be seen as part of one of the four key values: \(i, -1, -i, \text{ or } 1\).To determine the appropriate power of \(i\) for simplification, you need to utilize division to find the remainder:

• When you divide the exponent by 4, the remainder tells you which cycle position to use.
• A remainder of 0 corresponds to \(i^4 = 1\).
• A remainder of 1 corresponds to \(i^1 = i\).
• A remainder of 2 corresponds to \(i^2 = -1\).
• A remainder of 3 corresponds to \(i^3 = -i\).

Let's take the example of \(i^{12}\): When we divide 12 by 4, the remainder is 0, indicating \(i^{12} = i^0\), which simplifies to 1. Cyclic patterns, thus, offer a simplified framework to tackle complex arithmetic without unnecessary computational overload.