Problem 84 - A Graphical Approach to College Algebra

Question

Simplify each power of i to \(i, 1,-i,\) or \(-1\). $$\begin{aligned} &1\\\ &\frac{1}{i^{-46}} \end{aligned}$$

Step-by-Step Solution

Answer

The simplified expression of \(\frac{1}{i^{-46}}\) is \(-1\).

Step 1: Identify the Pattern of Powers of i

We know that the powers of the imaginary unit i repeat every four terms: \[i = i \] \[i^2 = -1 \] \[i^3 = -i \] \[i^4 = 1 \] After \(i^4\), the pattern repeats, so \(i^5 = i\), \(i^6 = -1\), etc.

Step 2: Simplify the Power using Modulo 4

Given the expression \(i^{-46}\), we first convert the negative power to a positive power by using the reciprocal property: \[i^{-46} = \frac{1}{i^{46}}\] To find \(i^{46}\), calculate the remainder of 46 divided by 4: \[46 \div 4 = 11 \text{ remainder } 2\].This means \(i^{46} = i^2 = -1\).

Step 3: Complete the Expression

Now replace \(i^{46}\) with its equivalent from the pattern: \[i^{-46} = \frac{1}{i^{46}} = \frac{1}{-1}\].

Step 4: Final Simplification

Simplify \(\frac{1}{-1}\) to \[-1\].

Key Concepts

Powers of i

Imaginary numbers are an exciting concept in mathematics, primarily used to extend our understanding of numbers beyond the real number line. The imaginary unit, denoted as \(i\), is defined as \(i = \sqrt{-1}\). One of the fascinating properties of \(i\) is its cyclical pattern when raised to integer powers. This cycle repeats every four terms, which makes determining powers of \(i\) relatively straightforward.

Let's explore this repeating pattern:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

After \(i^4\), the powers of \(i\) repeat in this same sequence. So, for any integer \(n\), the power \(i^n\) is found by observing the remainder when \(n\) is divided by 4. For example, if \(n\) is 5, then \(i^5 = i^{4+1} = 1 \times i = i\). This cyclical property simplifies our work significantly when dealing with complex numbers.

Simplifying Expressions

When dealing with powers of \(i\), simplifying expressions is key to solving problems efficiently. Using the cyclical pattern of \(i\), we can find its power by modulo operation, essentially finding how many times a power completes the cycle of four.

For an expression like \(i^{-46}\), negative exponents can be intimidating at first, but they become manageable by understanding them as reciprocals. We start by rewriting the negative power as a reciprocal: \(i^{-46} = \frac{1}{i^{46}}\). The next step is simplifying \(i^{46}\) by determining "46 modulo 4."

Here’s how it works:

• 46 divided by 4 gives a quotient of 11 and a remainder of 2. Thus, \(i^{46} = i^2\).
• From the cycle, \(i^2 = -1\).

By substituting \(-1\) into our rewritten expression, we have \(i^{-46} = \frac{1}{-1}\), which simplifies to \(-1\). This simplification approach turns complex-looking expressions into simpler ones that are easier to interpret.

Algebraic Manipulation

Algebraic manipulation involves the strategic rearrangement and simplification of equations and expressions. In the context of imaginary numbers, it helps clarify and resolve complex expressions involving powers of \(i\).

To demonstrate this, consider the expression \(i^{-46}\). Initially, this seems complex due to the negative exponent and large power of \(i\). However:

• First, transform the negative exponent by moving \(i^{-46}\) to the denominator, yielding \( \frac{1}{i^{46}} \).
• Next, we determine the equivalent power using the cycle of powers of \(i\). As discovered, \(i^{46} = i^2 = -1\).
• Finally, substitute to simplify: \( \frac{1}{-1} = -1\).

This method showcases how algebraic manipulation, combined with understanding the repetition of powers and handling negative exponents, can simplify seemingly daunting expressions efficiently.