Problem 41 - College Algebra

Question

For the following exercises, perform the indicated operation and express the result as a simplified complex number. $$ i^{22} $$

Step-by-Step Solution

Answer

\(-1\)

Step 1: Understand the Properties of i

The imaginary unit \( i \) has a fundamental property such that \( i^2 = -1 \). This forms a cyclical pattern: \( i^3 = i^2 \cdot i = -i \), and \( i^4 = i^3 \cdot i = 1 \). So, every power of \( i \) repeats every four numbers: \( i, -1, -i, 1 \).

Step 2: Determine the Remainder

To simplify \( i^{22} \), notice that the powers of \( i \) repeat every 4. Therefore, determine the remainder when 22 is divided by 4. The division gives 22 divided by 4 equals 5 with a remainder of 2. So we are really calculating \( i^2 \).

Step 3: Simplify Using the Remainder

Since \( i^2 = -1 \), and from Step 2, \( i^{22} \) simplifies to \( i^2 \), we have \( i^{22} = -1 \).

Key Concepts

Imaginary Unit

The imaginary unit, often denoted as \( i \), is a fundamental concept in complex numbers. Its primary definition is that it satisfies the equation \( i^2 = -1 \). This unique property allows us to work with and understand numbers that are not real, thus expanding the number system into the complex plane. Complex numbers are expressed in the form \( a + bi \), where \( a \) represents the real part and \( b \) the imaginary part with \( i \) being the imaginary unit.

• \( i \) is used to represent the square roots of negative numbers.
• It is not a real number but opens the door to calculations involving the square root of negative values.

Grasping the concept of \( i \) is essential because it forms the basis for understanding more complex operations and mathematical ideas involving complex numbers.

Cyclical Pattern

The cyclical pattern of the powers of \( i \) is an interesting aspect of the imaginary unit. Because of the defining property \( i^2 = -1 \), when you multiply \( i \) by itself several times, it leads to a repeating cycle. Here is how it works:

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

Then, the cycle starts again. This means every fourth power of \( i \) will result in the same outcome as one of the first four powers. When doing calculations such as \( i^{22} \), recognize that because of this pattern, the power can be reduced by taking the remainder of the exponent divided by 4. In this case, 22 divided by 4 leaves a remainder of 2, which corresponds to \( i^2 = -1 \). This cyclical nature is a powerful tool to simplify powers of \( i \).

Exponentiation of i

The exponentiation of \( i \) involves raising this imaginary unit to various powers and understanding its cyclical nature. We've seen the cycle of \( i, -1, -i, 1 \) every four powers previously. This is crucial because it allows us to simplify expressions involving \( i \) raised to high powers.To simplify powers of \( i \):1. Determine the exponent, for example, \( i^n \).2. Divide \( n \) by 4 to find the remainder, which helps identify which part of the cycle the exponent lands on.3. Simplify the expression based on the remainder: - Remainder 0 gives you \( i^4 = 1 \). - Remainder 1 gives you \( i^1 = i \). - Remainder 2 gives you \( i^2 = -1 \). - Remainder 3 gives you \( i^3 = -i \).This method allows getting the answer quickly without manual calculations for each power. In solving \( i^{22} \), the remainder was 2, leading directly to the result of \( i^2 = -1 \). This process makes handling complex number exponents much more manageable.