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Chapter 5: Problem 68

Simplify. \(i^{14}\)

Short Answer

Expert verified
\(i^{14} = -1\).

Step by step solution

01

Understand the Properties of Imaginary Unit

The imaginary unit is represented by \(i\) and has the property \(i^2 = -1\). Understanding how \(i\) behaves under powers is crucial. Here are the first few powers of \(i\): \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\). Notice the cycle repeats every four powers: \(i^5 = i\), \(i^6 = -1\), \(i^7 = -i\), \(i^8 = 1\), and so forth.
02

Find the Remainder of the Exponent when Divided by 4

The cycle of powers of \(i\) repeats every 4 terms. To simplify \(i^{14}\), find the remainder when 14 is divided by 4. Performing the division, \(14 \div 4 = 3\) with a remainder of 2. Therefore, \(i^{14} = i^2\).
03

Simplify Using the Remainder

Using the property found in Step 2, since the remainder is 2, we simplify \(i^{14}\) to \(i^2\). Recall that from Step 1, \(i^2 = -1\). Therefore, \(i^{14} = -1\).

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Key Concepts

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

Imaginary Unit
The imaginary unit is a fundamental concept in the field of complex numbers. It's denoted by the symbol \(i\) and represents the square root of \(-1\). This is a special number because no real number squared gives a negative result. Here's why \(i\) is important:
  • It allows mathematicians to solve equations that don't have solutions in the set of real numbers. For instance, the equation \(x^2 + 1 = 0\) has solutions in complex numbers as \(x = i\) and \(x = -i\).
  • The introduction of \(i\) extends the number system from real numbers to complex numbers, where a complex number takes the form \(a + bi\), with \(a\) and \(b\) being real numbers.
Without \(i\), we wouldn't be able to explore these fascinating and critical areas of mathematics.
Powers of i
Understanding the powers of \(i\) is essential for simplifying expressions involving complex numbers. When \(i\) is raised to different powers, it cycles through a predictable pattern:
  • \(i^1 = i\)
  • \(i^2 = -1\)
  • \(i^3 = -i\)
  • \(i^4 = 1\)
After \(i^4\), the pattern repeats. This cycle continues with four-term increments:
  • \(i^5 = i\)
  • \(i^6 = -1\)
  • \(i^7 = -i\)
  • \(i^8 = 1\)
You can use this repeating cycle to simplify any power of \(i\) by dividing the exponent by 4 and finding the remainder. The remainder then determines which term of the cycle \(i^1\), \(i^2\), \(i^3\), or \(i^4\) to use.
Simplifying Expressions
Simplifying expressions involving powers of \(i\) is practical with the knowledge of its cyclical nature. Here's how you can simplify a specific case, such as \(i^{14}\):
  • First, divide the exponent 14 by 4, which gives a quotient of 3 and a remainder of 2, indicating that \(i^{14} = i^2\).
  • We already know from the cycle that \(i^2 = -1\).
  • Therefore, \(i^{14}\) simplifies to \(-1\).
This method is not just limited to \(i^{14}\) but can be applied to any power of \(i\). Simply find the remainder of the exponent divided by 4 to determine the equivalent power from the cycle, and simplify accordingly. This makes working with complex numbers efficient and straightforward.

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Most popular questions from this chapter

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