Problem 86 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

The simplified expression is \(i\).

Step 1: Simplify the Power of i

Begin by recognizing that the powers of the imaginary unit \( i \) repeat every four steps: \( i^1 = i, \ i^2 = -1, \ i^3 = -i, \ \text{and} \ i^4 = 1 \). Therefore, to simplify \( i^{15} \), divide the exponent 15 by 4 and find the remainder.\[15 \div 4 = 3.75\]The remainder is 3, which means \( i^{15} = i^3 \). From the cycle, \( i^3 = -i \).

Step 2: Replace and Simplify the Expression

Now replace \( i^{15} \) with \( -i \) in the expression \( \frac{-1}{-i^{15}} \). This gives: \(\frac{-1}{-(-i)} = \frac{-1}{i}\)

Step 3: Rationalize the Denominator

To simplify \( \frac{-1}{i} \), multiply both the numerator and the denominator by \( i \) to eliminate the imaginary unit from the denominator: \(\frac{-1}{i} \times \frac{i}{i} = \frac{-1 \cdot i}{i^2} = \frac{-i}{-1}\). Since \( i^2 = -1 \), this further simplifies to \( \frac{-i}{-1} = i \).

Key Concepts

Imaginary Unit

To understand complex numbers, it's crucial to grasp the concept of the imaginary unit, denoted by the symbol \( i \). The imaginary unit is defined by the equation \( i^2 = -1 \). This definition forms the basis of all operations involving complex numbers.

The imaginary unit \( i \) is a unique number that, when squared, gives a negative result (-1). This property is fundamental because it allows us to extend the number line into a new dimension. While real numbers lie on a one-dimensional line, complex numbers occupy a two-dimensional plane called the complex plane.

The power of \( i \) follows a cyclical pattern that repeats every four powers. It progresses through:

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

Recognizing this cyclical sequence simplifies computations involving higher powers of \( i \). For example, if you need to find \( i^{15} \), you simply identify the position of 15 in the cycle by performing a division and checking the remainder.

Rationalizing Denominator

When dealing with fractions that have complex numbers in the denominator, it's a good practice to rationalize the denominator. This means eliminating the imaginary unit \( i \) from the denominator.

Here's the step-by-step procedure to rationalize the denominator. Let's consider the fraction \( \frac{-1}{i} \). To get rid of \( i \), multiply both the numerator and the denominator by \( i \), as shown below:

• Multiply: \( \frac{-1}{i} \times \frac{i}{i} \)
• This gives \( \frac{-1 \times i}{i^2} \)
• Since \( i^2 = -1 \), the expression simplifies further to \( \frac{-i}{-1} = i \)

The key here is that multiplying by \( i \) in both the numerator and denominator does not change the value of the fraction. It only changes the form by removing the imaginary unit from the bottom of the fraction. This process keeps calculations neatly in terms of real numbers wherever possible.

Exponents of i

Calculating the powers of the imaginary unit \( i \) can become very straightforward once you understand the cyclical nature of its exponents. This repetition is pivotal in simplifying expressions involving \( i \).

The cycle repeats every four powers: \( i^1 = i \), \( i^2 = -1 \), \( i^3 = -i \), and \( i^4 = 1 \). Anytime you encounter a higher power, such as \( i^{15} \), you can reduce it by dividing the exponent by 4 and examining the remainder:

• Perform: \( 15 \div 4 = 3.75 \)
• The remainder is 3, so \( i^{15} = i^3 = -i \)

Thus, knowing how to work with exponents of \( i \) efficiently allows you to simplify complex expressions and solve problems with ease. This cyclic repetition is exceptionally helpful in avoiding lengthy calculations, thereby making it a powerful tool in algebra involving complex numbers.