Problem 75 - A Graphical Approach to College Algebra

Question

Simplify each power of i to \(i, 1,-i,\) or \(-1\). $$i^{15}$$

Step-by-Step Solution

Answer

-i

Step 1: Identify the Pattern of Powers of i

The imaginary unit \(i\) has a repeating pattern every four powers: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\). This pattern repeats: \(i^5 = i\), \(i^6 = -1\), and so on.

Step 2: Determine the Remainder

To simplify \(i^{15}\), find the remainder when 15 is divided by 4. This is because the powers of \(i\) repeat every 4. 15 divided by 4 gives a quotient of 3 and a remainder of 3. So, the equivalent power of \(i^{15}\) is \(i^3\).

Step 3: Simplify Using the Pattern

From the pattern found in Step 1, \(i^3 = -i\). Therefore, \(i^{15} = -i\).

Key Concepts

Understanding the Imaginary Unit

The concept of the imaginary unit is crucial in understanding complex numbers. The imaginary unit is denoted by the symbol \(i\), which is defined as the square root of -1. In mathematical terms, it is expressed as \(i^2 = -1\). This definition opens up a new dimension for numbers that don't have a real square root.
For example, if we try to calculate the square root of -1 in the real numbers, it's impossible. This is where \(i\) comes into play. It allows us to deal with equations and problems that involve the square root of negative numbers by stepping outside the realm of real numbers.

• \(i\) is the imaginary unit
• \(i^2 = -1\)
• Helps solve equations with negative roots

By understanding \(i\), you open the door to solving complex number problems and grasping further mathematical concepts.

Powers of Imaginary Numbers and their Cyclical Nature

A fascinating aspect of imaginary numbers, particularly the powers of \(i\), is their cyclical pattern. This cyclical pattern repeats every four powers, making calculations more predictable and manageable.
The powers of \(i\) follow a distinct sequence:

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

This sequence repeats itself, so for any power of \(i\), you can determine its value by finding the remainder of the exponent divided by 4. This remainder will correspond to one of the four results above.
For example, when calculating \(i^{15}\), you divide 15 by 4, which leaves a remainder of 3. Referring back to our cycle, \(i^3\) simplifies to \(-i\). This approach streamlines the process, especially for large exponents.

Simplification of Powers Using Cyclic Pattern

The simplification of powers of i is a straightforward task once you understand the repeating pattern of its powers. This method helps convert higher powers into one of the base forms: \(i, -1, -i,\) and \(1\).
Here's how simplification works:1. **Identify the exponent**: Look at the power of \(i\) that needs simplification.2. **Divide by 4**: Divide this power by 4 and note the remainder.3. **Match the Remainder**: Use the remainder to match it with the corresponding power in the cyclical pattern:

• Remainder 1: \(i\)
• Remainder 2: \(-1\)
• Remainder 3: \(-i\)
• Remainder 0: \(1\)

Once you match the remainder with the pattern, you can quickly write down the simplified form of \(i^{ ext{{power}}}\).
By using this straightforward technique, you can efficiently deal with complex calculations of powers of \(i\), such as \(i^{15}\), simplifying it to \(-i\). This not only enhances problem-solving speed but also reduces computation errors.