Problem 77 - Algebra A Combined Function

Question

Find each power of \(i\). $$ i^{-6} $$

Step-by-Step Solution

Answer

The value of \( i^{-6} \) is \(-1\).

Step 1: Understand the properties of 'i'

The imaginary unit ‘i’ is defined as the square root of -1. It has the following cyclical properties: - \( i^1 = i \)- \( i^2 = -1 \) - \( i^3 = -i \) - \( i^4 = 1 \)These properties repeat every four exponentiations, meaning for any integer n, \( i^{n} \) will cycle through \( i, -1, -i, \) and \( 1 \).

Step 2: Calculate the exponent modulo 4

Since the powers of \(i\) cycle every 4, we need to find \( -6 \mod 4 \) to determine the equivalent positive power. Calculate:\(-6 \div 4 = -1 \, .5\)This gives a remainder of 2, thus \( -6 \equiv 2 \mod 4 \).

Step 3: Apply the properties of 'i'

Using the results of \( -6 \equiv 2 \mod 4 \), we find that \( i^{-6} \equiv i^2 \). From the cyclical properties of \( i \), we know that:\( i^2 = -1 \)

Step 4: Conclude the solution

Since \( i^{-6} \equiv i^2 \) and \( i^2 = -1 \), we conclude that:\[ i^{-6} = -1 \]

Key Concepts

Imaginary Unit

The imaginary unit, denoted as 'i', is a fundamental concept in mathematics, particularly in the study of complex numbers. The core idea behind the imaginary unit is that it's the solution to the equation:\[x^2 = -1\]This means that the imaginary unit 'i' is defined as \( \sqrt{-1} \). This definition allows mathematicians to extend the number system beyond real numbers to include numbers that have both real and imaginary parts.

The concept of 'i' is critical because it enables the representation of complex numbers, which are of the form \( a + bi \), where 'a' and 'b' are real numbers. Here, 'a' is the real part and 'bi' is the imaginary part. What makes 'i' truly fascinating is how it behaves under multiplication. Multiplying 'i' by itself begins a cycle that is fundamental to understanding its properties:

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

This cycle repeats every four powers, which is a key element in solving equations involving powers of 'i'.

Cyclical Properties

The cyclical nature of the imaginary unit 'i' significantly simplifies computations involving powers of 'i'. Because the powers of 'i' repeat every four exponents, you only need to remember the outcomes for the first four powers. Here's the cyclical sequence once again:

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

After \(i^4\), the cycle repeats, meaning:

• \( i^5 = i \)
• \( i^6 = -1 \)
• \( i^7 = -i \)
• \( i^8 = 1 \)

Since the powers cycle, when dealing with larger exponents, you can reduce the exponent by using modulo 4. This allows you to determine what power of 'i' the number corresponds to within the four-step cycle. For example, if asked to compute \( i^{13} \), divide 13 by 4 to find the remainder, which is 1. Therefore, \( i^{13} = i \).

Exponentiation Modulo

Exponentiation modulo is a crucial technique to simplify computations involving powers of numbers like the imaginary unit 'i'. In simpler terms, the modulo operation finds the remainder when one number is divided by another. When dealing with powers of 'i', since we know the cyclical nature, modulo 4 becomes particularly useful.

Given an exponent 'n' for \( i^n \), you perform the operation:\[n \mod 4\]The result will suggest which of the four outcomes \( i, -1, -i, \) or \( 1 \) it will equal. This is exceptionally handy with negative exponents, too. For example, to find \( i^{-6} \):

• Compute \( -6 \mod 4 \). First, determine: dividing \(-6\) by 4 gives a quotient of \(-1.5\), which leaves a remainder of 2. Hence, \(-6 \equiv 2 \mod 4\).
• Since \( -6 \equiv 2 \mod 4 \), we find that \( i^{-6} = i^2 \).
• From cyclical properties, \( i^2 = -1 \).

Therefore, as a final answer, \( i^{-6} = -1 \). By always reducing the exponent using modulo 4, the workload is minimized, and calculations become rapidly solvable without manual iteration through all sequential powers.