Problem 48 - College Algebra with Corequisite Support

Question

For the following exercises, evaluate the expressions, writing the result as a simplified complex number. $$ \frac{1}{i^{11}}-\frac{1}{i^{21}} $$

Step-by-Step Solution

Answer

The simplified complex number is 0 + 0i.

Step 1: Simplify the powers of i

To begin, remember that the powers of the imaginary unit \( i \) repeat in a cycle: \( i^1 = i \), \( i^2 = -1 \), \( i^3 = -i \), and \( i^4 = 1 \). We will use this cycle to simplify \( i^{11} \) and \( i^{21} \).\( i^{11} = i^{(4 \times 2 + 3)} = (i^4)^2 \cdot i^3 = 1^2 \cdot (-i) = -i \)\( i^{21} = i^{(4 \times 5 + 1)} = (i^4)^5 \cdot i = 1^5 \cdot i = i \)

Step 2: Substitute the simplified powers into the expression

Replace \( i^{11} \) and \( i^{21} \) in the expression with their simplified forms:\[ \frac{1}{i^{11}} - \frac{1}{i^{21}} = \frac{1}{-i} - \frac{1}{i} \]

Step 3: Simplify the fractions

To simplify \( \frac{1}{-i} \) and \( \frac{1}{i} \), multiply the numerator and denominator by the complex conjugate:\( \frac{1}{-i} \cdot \frac{-i}{-i} = \frac{-i}{i^2} = \frac{-i}{-1} = i \)\( \frac{1}{i} \cdot \frac{-i}{-i} = \frac{-i}{i^2} = \frac{-i}{-1} = i \)

Step 4: Calculate the expression

Substitute the simplified fractions from Step 3 back into the expression:\[ i - i = 0 \]

Step 5: Write the result as a complex number

The result after subtraction is 0, which can be expressed as a complex number with no imaginary part: \[ 0 + 0i \]

Key Concepts

powers of i

Imaginary numbers revolve around the fundamental unit, denoted as \( i \), which is the square root of \(-1\). This property makes \( i \) special, giving it a cyclical pattern when raised to different powers. You can think of the powers of \( i \) as cycling through four distinct results:

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

Once you understand this cycle, simplifying expressions involving powers of \( i \) becomes much easier. For example, if you need to find \( i^{11} \), you simply divide 11 by 4 because the cycle repeats every 4 powers. The remainder will tell you which power in the cycle corresponds to \( i^{11} \). In this case, \( 11 \mod 4 = 3 \), so \( i^{11} = i^3 = -i \). By using this cycle, you simplify calculations and manage expressions involving \( i \) with ease.

simplifying expressions

Simplifying expressions with imaginary numbers often requires breaking down complex fractions or terms into more manageable pieces. Knowing how to handle powers of \( i \) is fundamental here, as you can reduce the problem into simpler steps. For the expression \( \frac{1}{i^{11}} - \frac{1}{i^{21}} \), replace each power of \( i \) with its simplified equivalent based on the cycle we previously discussed. After simplification, our expression becomes \( \frac{1}{-i} - \frac{1}{i} \).
To aptly simplify these fractions, leverage the properties of fraction arithmetic and the behavior of imaginary numbers. The objective is to multiply the numerator and the denominator by the complex conjugate of the denominator, which will help eliminate the imaginary unit from the denominator. This turns each fraction into its simplest form, making further arithmetic manageable.

complex conjugates

When working with complex numbers, especially in division, using complex conjugates is a powerful technique. The complex conjugate of a number \( a + bi \) is \( a - bi \). For any complex number, multiplying it by its complex conjugate results in a real number because the imaginary parts cancel each other out.

For fractions like \( \frac{1}{i} \) or \( \frac{1}{-i} \), the denominator is purely imaginary, i.e., just \( i \). By multiplying by the complex conjugate \( -i \), it transforms the denominator \( i^2 \) into \(-1\) since \( i^2 = -1 \). This effectively shifts the problem from one involving complex numbers to something more tangible with real numbers.

• For \( \frac{1}{-i} \), the simplification process goes as follows: \( \frac{1}{-i} \cdot \frac{-i}{-i} = \frac{-i}{-1} = i \).
• Similarly, for \( \frac{1}{i} \), perform: \( \frac{1}{i} \cdot \frac{-i}{-i} = \frac{-i}{-1} = i \).

By utilizing complex conjugates, the fractions are simplified efficiently, revealing a cleaner solution that is easy to handle and interpret in subsequent calculations.