Problem 80 - A Graphical Approach to Precalculus with Limits

Question

Simplify each expression to \(i, 1,-i,\) or \(-1\) $$i^{-15}$$

Step-by-Step Solution

Answer

\(i^{-15} = -i\).

Step 1: Understand the Power of i

The imaginary unit, denoted by \(i\), is defined such that \(i^2 = -1\). The powers of \(i\) cycle every four: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\). This cycle repeats, so \(i^5 = i\), \(i^6 = -1\), \(i^7 = -i\), and \(i^8 = 1\), and so on.

Step 2: Simplify the Exponent

To simplify \(i^{-15}\), first find \(15 \mod 4\) because the powers repeat every four steps. Divide 15 by 4 to get a remainder of 3. This is because \(15 \div 4 = 3.75\), which means 15 is 3 full cycles of 4 with a remainder of 3.

Step 3: Identify the Equivalent Power for the Remainder

Based on the remainder found in Step 2, \(i^{15}\) simplifies to \(i^3\) because 15 leaves a remainder of 3 when divided by 4. Therefore, \(i^{-15}\) simplifies to \(i^{-3}\) (or equivalently, the inverse of \(i^3\)).

Step 4: Calculate the Expression

Since \(i^3 = -i\), the inverse \(i^{-3}\) is also \(-i\). Thus, \(i^{-15} = -i\).

Key Concepts

Imaginary Unit

The imaginary unit, widely recognized as the letter \( i \), plays a crucial role in complex numbers. It is fundamental in extending our number system beyond real numbers. The key property of \( i \) is that its square equals \( -1 \), represented mathematically as:

• \( i^2 = -1 \)

This mathematical entity allows us to explore numbers in a two-dimensional plane, where the real numbers sit on one axis and the imaginary numbers lie on the other. Imaginary numbers provide solutions to equations that have no real solutions, such as squaring negative numbers. By incorporating \( i \) into calculations, we can achieve new dimensions in mathematics that extend the study of algebra into complex number theory.

Powers of i

Understanding the powers of \( i \) is essential for simplifying expressions involving complex numbers. The powers of \( i \) exhibit a cyclic pattern that repeats every fourth power. Let's break down the cycle:

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

Beyond this point, the pattern repeats: \( i^5 = i \), \( i^6 = -1 \), and so forth. This cyclic nature arises from the definition of \( i \), specifically that \( i^4 \) equals 1, resetting the powers to their initial state. Recognizing this pattern is pivotal when simplifying larger exponents of \( i \), since it dramatically reduces complex calculations to one of the four base values without extensive computation.

Simplifying Expressions

Simplifying expressions with \( i \) may seem challenging at first, but understanding the cyclic nature of \( i \) can significantly ease the process. Let's consider the case of simplifying \( i^{-15} \). The goal here is to express this power in terms of one of the basic values of the cycle: \( i \), \( -1 \), \( -i \), or \( 1 \).When you come across an exponent, say \(-15\), use the modulus operation to find where it sits in the \( i \) cycle. We calculate the exponent modulo 4 because the powers of \( i \) cycle every four terms:

• \( 15 \mod 4 = 3 \)

This result implies, for our simplified form, \( i^{15} \) equates to \( i^3 \), which equals \( -i \). Since we originally had \( i^{-15} \), the result is the reciprocal, or simpler the equivalent representation which is also \( -i \). Remember, recognizing and applying this cyclical pattern streamlines many complex number problems.