Problem 86 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The expression simplifies to \( i \).

Step 1: Analyze the expression

We are given the expression \( \frac{-1}{-i^{15}} \). The goal is to simplify it to one of the values \( i, 1, -i, \) or \( -1 \).

Step 2: Simplify the negative denominator

The expression \( \frac{-1}{-i^{15}} \) can be restructured by removing the negative sign from the denominator, resulting in \( \frac{1}{i^{15}} \), since two negatives cancel each other out.

Step 3: Determine the power of \(i\)

The imaginary unit \(i\) has a cyclic pattern in its powers:\[ i^1 = i, \]\[ i^2 = -1, \]\[ i^3 = -i, \]\[ i^4 = 1. \]These values repeat every four powers. To find \(i^{15}\), compute the remainder of 15 divided by 4.

Step 4: Calculate the remainder

When 15 is divided by 4, the remainder is 3. Therefore, \(i^{15} = i^3 = -i\).

Step 5: Simplify using the result

Substituting \(i^{15} = -i\) into the expression, we have \( \frac{1}{-i} \).

Step 6: Rationalize the denominator

To simplify \( \frac{1}{-i} \), multiply both the numerator and denominator by \(i\):\[ \frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2}. \]Since \(-i^2 = 1\), this simplifies to \( \frac{i}{1} = i \).

Key Concepts

Imaginary Unit

In mathematics, the imaginary unit, represented as \( i \), is a fundamental concept when working with complex numbers. Specifically, the imaginary unit is defined by its property \( i^2 = -1 \). This property forms the basis for complex numbers and allows calculations involving square roots of negative numbers. Complex numbers have the form \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) stands for the imaginary part. The imaginary unit is crucial as it facilitates many areas of advanced mathematics, including engineering and physics, where it represents solutions to various equations. By understanding the imaginary unit, students can simplify expressions that involve powers of \( i \) through recognizing its cyclical pattern.

Exponentiation

Exponentiation is a mathematical operation involving numbers raised to a power. When dealing with complex numbers, especially those involving the imaginary unit \( i \), it's important to understand the behavior of powers of \( i \).

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

These results demonstrate a repeating cycle every four powers. This cyclic nature greatly simplifies calculations involving powers of \( i \). For example, when tasked with simplifying \( i^{15} \), one can find the remainder when 15 is divided by 4. The remainder is 3, and hence, \( i^{15} = i^3 = -i \). Leveraging this cyclical pattern simplifies otherwise complex exponentiation problems.

Simplification

Simplification is the process of reducing an expression into its simplest form. When tackling expressions that involve complex numbers, simplification often involves rationalizing the denominator or combining like terms. In the given problem, we are tasked with simplifying \( \frac{1}{-i} \), which initially seems complex.To simplify, we apply a technique called rationalization. This involves multiplying the numerator and the denominator by the conjugate of the denominator—in this case, \( i \). Perform the multiplication:\[ \frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2} \]Since \( -i^2 = 1 \), the expression simplifies to:\[ \frac{i}{1} = i \]This step-by-step simplification is key to converting complex number expressions into manageable forms. By mastering techniques like rationalization, students can easily simplify complex expressions and reach elegant solutions.