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Chapter 12: Problem 43

Use the fact that \(i^{4}=1\) to simplify each expression (as in Example \(5(b)]\). $$i^{26}$$

Short Answer

Expert verified
The simplified form of \(i^{26}\) is \(-1\).

Step by step solution

01

Understand the Power Cycle of i

The imaginary unit \(i\) has cyclical powers: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and then it repeats. This means every power of \(i\) can be reduced by recognizing patterns every four powers.
02

Determine the Remainder of the Exponent When Divided by 4

To simplify \(i^{26}\), we need to find the exponent modulo 4. The remainder of 26 divided by 4 determines the simplified form of \(i^{26}\). Calculate \(26 \div 4 = 6\) remainder 2. Therefore, \(26 \equiv 2 \mod 4\).
03

Simplify the Expression Using the Remainder

Since \(26 \equiv 2 \mod 4\), we have \(i^{26} = i^2\). Therefore, replace the expression with \(i^2\).
04

Apply the Simplification of i^2

From the powers of \(i\), we know that \(i^2 = -1\). Substitute \(-1\) in place of \(i^2\) based on known values.

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Key Concepts

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

Powers of i
The powers of the imaginary unit, denoted as \(i\), follow a unique cycle. Understanding this cycle becomes the key to simplifying expressions involving powers of \(i\). As you explore the power sequence of \(i\), you'll notice the repetition every four exponents:
  • \(i^1 = i\)
  • \(i^2 = -1\)
  • \(i^3 = -i\)
  • \(i^4 = 1\)
  • And then, it repeats: \(i^5 = i, i^6 = -1, \) etc.
The critical takeaway is that after every four exponents, the sequence starts anew. With this understanding, any power of \(i\) can be simplified by recognizing which part of the cycle it aligns with. For instance, simplifying \(i^{26}\) entails determining its position in the sequence, which connects to modular arithmetic as outlined in later sections.
Complex Numbers
Complex numbers expand upon the concept of real numbers by including imaginary numbers. A complex number takes the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit. This form allows for a rich set of mathematical properties and operations.
  • **Real Part**: The real component \(a\), which lies along the horizontal axis in complex plane representation.
  • **Imaginary Part**: The imaginary component \(bi\), represented along the vertical axis.
  • **Conjugate**: A conjugate of a complex number \(a + bi\) is \(a - bi\), which reflects over the real axis.
Complex numbers provide solutions to equations that real numbers alone can't solve, such as \(x^2 + 1 = 0\). They enable calculations and transformations used across fields, including electrical engineering and quantum physics.
Modular Arithmetic
In simplifying powers of \(i\), modular arithmetic plays a pivotal role. This system of arithmetic considers numbers differently by working with their remainders when divided by a modulus. In our example, the modulus is 4 due to the cyclical nature of \(i\)'s powers.
  • **Concept**: When you divide a number by another (the modulus), the outcome focuses on the remainder rather than the quotient.
  • **Application**: For \(i^{26}\), dividing 26 by 4 gives a remainder of 2, represented by \(26 \equiv 2 \mod 4\).
  • **Result**: This calculation means \(i^{26}\) falls in the same category as \(i^2\) within the power cycle.
Utilizing modular arithmetic simplifies otherwise complex problems into manageable calculations using a straightforward remainder technique. This technique not only applies to powers of \(i\) but also extends to various fields like cryptography and computer science.

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Most popular questions from this chapter

(a) factor the denominator of the given \(\mathrm{ra}\) tional expression; (b) determine the form of the partial fraction decomposition for the given rational expression; and (c) determine the values of the constants in the partial fraction decomposition that you gave in part (b). To help you in spotting errors, use the fact that in part (c), each of the required constants turns out to be an integer. $$\frac{16 x^{2}+9 x-2}{3 x^{3}+x^{2}-2 x}$$

Determine the partial fraction decomposition for each of the given rational expressions. Hint: In Exercises \(17,18,\) and \(26,\) use the rational roots theorem to factor the denominator. $$\frac{x^{3}+2}{x^{4}+8 x^{2}+16}$$

(a) Let \(\alpha=\left(\frac{\sqrt{a^{2}+b^{2}}+a}{2}\right)^{1 / 2}\) and let \(\beta=\left(\frac{\sqrt{a^{2}+b^{2}}-a}{2}\right)^{1 / 2}\) Show that the square of the complex number \(\alpha+\beta i\) is \(a+b i\) (b) Use the result in part (a) to find a complex number \(z\) such that \(z^{2}=i\) (c) Use the result in part (a) to find a complex number \(z\) such that \(z^{2}=-7+24 i\)

Express the polynomial \(f(x)\) in the form \(a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\). (a) Find a third-degree polynomial function that has zeros \(-5,2,\) and 3 and a graph that passes through the point (0,1) (b) Use a graphing utility to check that your answer in part (a) appears to be correct.

You need to know that a prime number is a positive integer greater than 1 with no factors other than itself and 1. Thus the first seven prime numbers are 2,3,5,7,11,13 and 17. Find all prime numbers \(p\) for which the equation \(x^{2}+x-p=0\) has a rational root.
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