/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Find the indicated powers of com... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 44

Find the indicated powers of complex numbers. $$i^{31}$$

Short Answer

Expert verified
The value of \( i^{31} \) is \(-i\).

Step by step solution

01

Understand the Power of i

The imaginary unit, denoted as \( i \), is defined such that \( i^2 = -1 \). Higher powers of \( i \) can be simplified by recognizing the pattern in the powers of \( i \). The powers of \( i \) cycle every four exponents as follows: \( i^1 = i \), \( i^2 = -1 \), \( i^3 = -i \), and \( i^4 = 1 \).
02

Use the Pattern to Simplify Higher Powers

Since the powers of \( i \) repeat every four exponents, any power of \( i \) can be reduced by finding the remainder when the exponent is divided by 4. For example, \( i^{31} \) can be simplified by calculating \( 31 \mod 4 \).
03

Divide the Exponent by 4

Divide 31 by 4 to find the remainder: \(31 \div 4 = 7 \text{ remainder } 3\). This means \( 31 \mod 4 = 3 \).
04

Relate the Result to the Power Pattern

Since the remainder is 3, \( i^{31} = i^3 \). From the pattern in Step 1, we know that \( i^3 = -i \).
05

State the Final Answer

Thus, \( i^{31} = -i \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Powers of i
The imaginary unit, represented as \( i \), follows an interesting pattern when raised to different powers. The key is understanding that these powers cycle every four exponents, allowing for simplification. The pattern is:
  • \( i^1 = i \)
  • \( i^2 = -1 \)
  • \( i^3 = -i \)
  • \( i^4 = 1 \)
After \( i^4 \), the sequence repeats itself, so:
  • \( i^5 = i^1 \)
  • \( i^6 = i^2 \)
  • \( i^7 = i^3 \)
  • \( i^8 = i^4 \) = 1
Understanding this repeating pattern makes it simpler to handle higher powers like \( i^{31} \).
Exponents of Imaginary Unit
When facing exponents of the imaginary unit \( i \), the key idea is reducing the exponent using the repeating pattern we've discussed. To simplify \( i^{31} \), you follow these steps:
  • Recognize the four-term pattern of \( i \): \( i, -1, -i, 1 \).
  • Calculate the remainder when the exponent is divided by 4, since the pattern repeats every four numbers.
In our example,, divide the exponent 31 by 4:
\[ 31 \div 4 = 7 \text{ remainder } 3 \] This tells us that \( 31 \mod 4 = 3 \), meaning \( i^{31} = i^3 \). Referring back to our pattern, we know that \( i^3 = -i \). So, \( i^{31} = -i \).
Modulo Operation
The modulo operation is crucial in simplifying higher powers of \( i \). This mathematical operation finds the remainder when one number is divided by another. Specifically, to simplify \( i^{31} \), we use modular arithmetic with a base of 4. Here鈥檚 why and how:
  • The pattern in the powers of \( i \) repeats every 4 cycles: \( i, -1, -i, 1 \).
  • By dividing the exponent by 4, we find which part of the cycle the exponent falls into.
In our calculation:
\[ 31 \div 4 = 7 \text{ remainder } 3 \] The remainder (3 in this case) indicates the position in the cycle: \( i^3 = -i \), simplifying \( i^{31} \) directly. This highlights the usefulness of modulo operations in handling exponents, especially when identifying repetitive patterns in sequences.

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Most popular questions from this chapter

Simplify. $$\frac{\sqrt{6}}{2} \cdot \frac{1}{\sqrt{3}}$$

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