Problem 17 - Precalculus: Functions and Graphs

Question

Write the expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$\text { (a) } i^{73} \quad(b) i^{-46}$$

Step-by-Step Solution

Answer

(a) \(0 + 1i\) (b) \(-1 + 0i\)

Step 1: Simplify Powers of i

Recall that the complex number \(i\) has a cyclical pattern: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\). After every four powers, the pattern repeats. To simplify \(i^{73}\), find the remainder when \(73\) is divided by \(4\) to determine where we are in the cycle: \(73 \mod 4 = 1\). This means \(i^{73} = i^1 = i\). So, \(i^{73}\) translates to \(0 + 1i\).

Step 2: Simplify the Negative Power of i

For \(i^{-46}\), first simplify as a positive power: \(i^{-46} = \frac{1}{i^{46}}\). From the cyclical nature of \(i\), \(i^{46} = i^{44+2} = (i^4)^{11} \cdot i^2 = 1^{11} \cdot -1 = -1\). So, \(i^{-46} = \frac{1}{-1} = -1\).Therefore, \(i^{-46}\) translates to \(-1 + 0i\).

Key Concepts

Powers of i

The imaginary unit, often denoted by the letter \(i\), is the cornerstone of all complex numbers. Understanding the powers of \(i\) is essential for working with complex numbers. At its core, \(i\) is defined as the square root of \(-1\). Consequently, we have the following powers of \(i\):

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

These four results form a repeating cycle. Each subsequent power of \(i\) can be determined by its position in this cycle. Therefore, to simplify complex expressions involving powers of \(i\), we find the remainder of the exponent division by 4 and use it to locate the corresponding power in the cycle. This cyclical pattern saves calculation time and reduces potential for error.

Simplifying Complex Expressions

Simplifying complex expressions often involves transforming expressions to their simplest form, particularly when raised to various powers. When asked to write a complex expression like \(i^{73}\) in the form \(a + bi\) , we leverage our understanding of the powers of \(i\) and its cyclical nature.
For example, simplifying \(i^{73}\) requires finding the remainder when 73 is divided by 4, which is 1. This tells us that \(i^{73} = i^1 = i\), resulting in the complex number \(0 + 1i\).
Similarly, with negative exponents such as \(i^{-46}\), the concept is similarly applied by first converting to positive exponents: \(i^{-46} = \frac{1}{i^{46}}\). Understanding that \(i^{46}\) simplifies to \(-1\), allows us to write \(i^{-46}\) in the form \(-1 + 0i\).
Mastery of simplifying these patterns allows for a deeper exploration of more complicated expressions.

Cyclical Pattern of i

A fundamental aspect of working with imaginary numbers is recognizing their cyclical nature due to the pattern exhibited by powers of \(i\). As previously mentioned, when computing \(i^n\), where \(n\) is any integer, it is efficient to realize that \(i\) follows a 4-step repeating cycle:

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

After reaching \(i^4\), the cycle restarts. Thus, understanding the cyclical pattern eliminates guesswork when simplifying expressions with high powers of \(i\).
This knowledge is practical because any power of \(i\) can be reduced by examining the exponent modulo 4, quickly determining the corresponding equivalent in the cycle.
This cyclical pattern not only aids in simplifying powers of \(i\) but also reinforces efficient calculation techniques useful throughout complex number operations.