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Chapter 10: Problem 89

Simplify. $$ i^{9} $$

Short Answer

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Step by step solution

01

Understand the powers of i

Recall that the imaginary unit, denoted as i, has cyclical powers: - \(i^1 = i\) - \(i^2 = -1\) - \(i^3 = -i\) - \(i^4 = 1\) This cycle repeats every four powers.
02

Divide the exponent by 4

To determine the simplified form of i raised to any power, divide the exponent by 4 and find the remainder. For \(i^9\): \(9 \div 4 = 2\) remainder \(1\)
03

Simplify using the remainder

Using the remainder from Step 2, which is 1, refer to the cycle of powers of i: \(i^1 = i\). Therefore: \(i^9 = i^1 = i\)

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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 \(i\) is unique because its powers follow a specific cycle. This cycle repeats every four powers. Here's how:
  • \(i^1 = i\)
  • \(i^2 = -1\)
  • \(i^3 = -i\)
  • \(i^4 = 1\)
This means that every power of \(i\) will fall into one of these four results. For instance, when you calculate higher powers like \(i^5\), it is effectively the same as starting the cycle over from \(i^1\).
Cycling through powers
Understanding the cycle for powers of \(i\) is crucial. This is because once you know the cycle, simplifying any power of \(i\) becomes much easier. Here's a quick way to see the cycle:
After \(i^4\) (which equals 1), the next power, \(i^5\), starts the cycle again, so \(i^5 = i\).
The pattern continues like this:
  • \(i^6 = -1\)
  • \(i^7 = -i\)
  • \(i^8 = 1\)
  • \(i^9 = i\)
Once you're familiar with this repeating sequence, you can quickly determine any power of \(i\) just by noting its position within this four-term cycle.
Remainder method
To simplify an expression like \(i^9\), use the remainder method. Here's how it works:
First, divide the exponent by 4 and find the remainder. For \(i^9\), you perform the division \(9 \div 4\). This gives a quotient of 2 and a remainder of 1. The remainder tells you which part of the cycle the power falls on.
For our example, a remainder of 1 indicates that \(i^9\) is the same as \(i^1\). Since \(i^1 = i\), it follows that \(i^9 = i\). The remainder method is a very efficient technique to simplify large exponents of \(i\).

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