Problem 79 - Algebra 2

Question

If \(i^{2}=-1,\) then what is the value of \(i^{71} ?\) \(\mathbf{F}-1\) \(\mathbf{G} 0\) \(\mathbf{H}-\boldsymbol{i}\) \(\mathbf{J} i\)

Step-by-Step Solution

Answer

The value of \(i^{71}\) is \(-i\).

Step 1: Understand the Powers of i

The imaginary unit \( i \) has cyclical powers that repeat every 4 steps: \( i^1 = i \), \( i^2 = -1 \), \( i^3 = -i \), \( i^4 = 1 \), and then it repeats. This will help in simplifying any power of \( i \) by reducing the exponent modulo 4.

Step 2: Find 71 Modulo 4

Calculate \( 71 \mod 4 \) to reduce the exponent to one within the repeating cycle of the powers of \( i \). Perform the division: \( 71 \div 4 = 17 \) with a remainder of \( 3 \). Hence, \( 71 \equiv 3 \mod 4 \).

Step 3: Relate the Modulo Result to Powers of i

Since \( 71 \equiv 3 \mod 4 \), we have \( i^{71} = i^{3} \). From the known powers of \( i \), we know that \( i^{3} = -i \).

Step 4: Conclude the Simplification

Based on the reduced power and its corresponding value in the cycle, we conclude \( i^{71} = -i \), which corresponds to option \( \mathbf{H} \).

Key Concepts

Powers of Imaginary Unit

The concept of imaginary numbers revolves around the imaginary unit, denoted as \( i \), which is defined by the property \( i^2 = -1 \). This might seem counterintuitive because no real number squared gives a negative result. However, imaginary numbers are a masterstroke in mathematics, enabling solutions to equations that previously had none. When working with powers of \( i \), a straightforward cyclic pattern emerges. Specifically:

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

After \( i^4 \), the cycle repeats. Recognizing this sequence is crucial. By understanding these four basic powers, you can evaluate any power of \( i \). For example, for \( i^{71} \), instead of calculating every single multiplication, you can rely on this cyclical pattern to simplify it.

Modular Arithmetic

Modular arithmetic is an essential concept in simplifying powers of the imaginary unit, especially when dealing with large exponents. Think of it as a clever way of "wrapping around" numbers in a cycle, analogous to how clocks wrap around every 12 hours.To use modular arithmetic, divide the exponent by the cycle length, which for \( i \) is 4. The remainder tells you the equivalent lower power. For instance, with \( i^{71} \):

• Divide 71 by 4: \( 71 \div 4 = 17 \) remainder 3
• Thus, \( 71 \equiv 3 \mod 4 \)

This remainder, 3, indicates that \( i^{71} \) corresponds to \( i^3 \) in the pattern. Modular arithmetic isn't just for imaginary numbers; it's widely used in computer science, cryptography, and number theory due to its elegance and simplicity.

Cyclical Patterns of Powers of i

The cyclical pattern of powers of \( i \) is both a testament to the beauty and simplicity inherent in mathematics. This cycle is unchanging and repeats every four exponents, which means you only ever need to remember the first four powers of \( i \):

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

After \( i^4 \), it all starts again from \( i^1 \). When confronted with any power of \( i \), breaking it down using this cycle makes the problem far less daunting. Understanding this concept also reinforces how modular arithmetic and cyclical patterns are intertwined. Once you calculate the remainder using modular arithmetic for an exponent, you immediately know which stage of this cycle corresponds to that power of \( i \), as seen with \( i^{71} = i^3 = -i \). Embracing these patterns equips you with a powerful tool for dealing with complex numbers.