Simplify Each Of The Following Powers Of I

Simplifying Powers of i: A Comprehensive Guide

The imaginary unit, denoted by i, is a fundamental concept in complex numbers. Understanding how to simplify powers of i is crucial for anyone working with complex numbers, whether in algebra, calculus, or more advanced mathematical fields. This comprehensive guide will delve into the intricacies of simplifying powers of i, providing you with a solid understanding and practical techniques to tackle any problem.

Understanding the Imaginary Unit, i

Before we embark on simplifying powers of i, let's establish a solid foundation. The imaginary unit, i, is defined as the square root of -1:

i = √(-1)

This seemingly simple definition opens up a whole new world of mathematical possibilities. Because no real number multiplied by itself can result in a negative number, i is defined as an imaginary number. This doesn't mean it's unreal or unimportant; rather, it extends the number system beyond the real numbers to encompass complex numbers.

The Cyclic Nature of Powers of i

The beauty of simplifying powers of i lies in its cyclical nature. Let's explore the first few powers:

• i¹ = i (This is the base definition)
• i² = -1 (The square of i is -1, by definition)
• i³ = i² * i = -1 * i = -i
• i⁴ = i² * i² = (-1) * (-1) = 1

Notice something significant? After i⁴, the pattern repeats. Let's continue:

• i⁵ = i⁴ * i = 1 * i = i
• i⁶ = i⁴ * i² = 1 * (-1) = -1
• i⁷ = i⁴ * i³ = 1 * (-i) = -i
• i⁸ = i⁴ * i⁴ = 1 * 1 = 1

This cyclical pattern of i, -1, -i, 1 repeats indefinitely. This cyclic property is the key to efficiently simplifying any power of i.

Methods for Simplifying Powers of i

There are two primary methods for simplifying powers of i:

Method 1: Using the Cyclical Pattern

This method leverages the repeating cycle of i, -1, -i, 1. To use this method, we divide the exponent by 4 and examine the remainder.

• Step 1: Divide the exponent by 4. Find the quotient and the remainder.
• Step 2: Use the remainder to determine the simplified form. The remainder will correspond to a position in the cycle:
  • Remainder of 0: i⁴ = 1
  • Remainder of 1: i¹ = i
  • Remainder of 2: i² = -1
  • Remainder of 3: i³ = -i

Example 1: Simplify i¹⁷

1. Divide 17 by 4: 17 ÷ 4 = 4 with a remainder of 1.
2. The remainder is 1, so i¹⁷ simplifies to i.

Example 2: Simplify i²⁶

1. Divide 26 by 4: 26 ÷ 4 = 6 with a remainder of 2.
2. The remainder is 2, so i²⁶ simplifies to -1.

Example 3: Simplify i⁻⁵

1. First, we rewrite i⁻⁵ as 1/i⁵.
2. Then, we simplify i⁵ using the method above. 5 divided by 4 gives a remainder of 1, so i⁵ = i.
3. Therefore, i⁻⁵ = 1/i. To express this without a denominator containing i, we multiply the numerator and the denominator by the conjugate, which is -i: (1 * -i) / (i * -i) = -i / 1 = -i

Method 2: Using the Property i⁴ = 1

This method utilizes the fact that i⁴ = 1. We rewrite the power of i as a product involving i⁴ and another power of i. Since i⁴ = 1, this simplifies the expression considerably.

Example 4: Simplify i¹⁹

1. Rewrite i¹⁹ as (i⁴)⁴ * i³.
2. Since i⁴ = 1, this simplifies to 1⁴ * i³ = 1 * i³ = i³ = -i.

Example 5: Simplify i⁻¹²

1. Rewrite i⁻¹² as 1/i¹².
2. Rewrite i¹² as (i⁴)³ = 1³ = 1.
3. Therefore, i⁻¹² = 1/1 = 1.

Advanced Applications and Problem Solving Strategies

While the basic methods are sufficient for many problems, let’s consider some more advanced scenarios:

Scenario 1: Dealing with large exponents

When the exponent is exceptionally large, using a calculator to find the remainder after division by 4 is highly recommended. This greatly accelerates the process.

Scenario 2: Expressions involving both real and imaginary parts

Simplifying expressions involving both real and imaginary parts might require a combination of simplifying powers of i along with other algebraic manipulations.

Example 6: Simplify 3i⁷ + 2i¹⁰ - 5i¹³

1. Simplify each power of i individually:

   • i⁷ = -i
   • i¹⁰ = (i⁴)² * i² = 1² * (-1) = -1
   • i¹³ = (i⁴)³ * i = 1³ * i = i

2. Substitute the simplified powers of i back into the original expression: 3(-i) + 2(-1) - 5(i) = -3i - 2 - 5i = -2 - 8i

Scenario 3: Equations involving powers of i

Sometimes, you might encounter equations where you need to solve for a variable that involves powers of i. Understanding the cyclical nature and simplifying the powers of i is essential here.

Example 7: Solve for x: x² = -9

Taking the square root of both sides:

x = ±√(-9) = ±√(9 * -1) = ±√9 * √(-1) = ±3i

Conclusion

Mastering the simplification of powers of i is a fundamental skill in the realm of complex numbers. By understanding the cyclic nature of powers of i and employing the efficient methods outlined above, you can confidently handle various problems, from simple simplifications to more complex equations involving real and imaginary components. Remember the core principle: the powers of i cycle through i, -1, -i, and 1. With practice, you’ll find this process becomes second nature. This will lay the foundation for tackling more advanced topics within complex numbers. This mastery will serve you well in various mathematical contexts, enhancing your problem-solving abilities and deepening your understanding of complex numbers.