Problem 16 - Precalculus: Functions and Graphs

Question

Write the expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$\text { (a) } i^{92} \quad \text { (b) } i^{-33}$$

Step-by-Step Solution

Answer

(a) 1 + 0i, (b) 0 - i

Step 1: Understand Powers of i

Recall that the imaginary unit \(i\) satisfies \(i^2 = -1\). Powers of \(i\) cycle every four: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and then it repeats.

Step 2: Simplify i^{92}

To simplify \(i^{92}\), find the remainder of 92 divided by 4, since the powers of \(i\) repeat every 4. 92 divided by 4 is 23 with a remainder of 0, so \(i^{92} = (i^4)^{23} = 1^{23} = 1\).

Step 3: Simplify i^{-33} using Positive Exponents

Convert negative exponent to a positive one: \(i^{-33} = \frac{1}{i^{33}}\).Now find 33 mod 4. 33 divided by 4 is 8 with a remainder of 1, so \(i^{33} = i\). Therefore, \(i^{-33} = \frac{1}{i} = -i\) (since dividing by \(i\) is equivalent to multiplying by \(-i\)).

Step 4: Express Results in a + bi Form

For part (a), \(i^{92} = 1\), which is already in the form \(a + bi\) with \(a = 1\) and \(b = 0\).For part (b), \(i^{-33} = -i\), which is in the form \(a + bi\) with \(a = 0\) and \(b = -1\).

Key Concepts

Powers of Imaginary Unit

The imaginary unit, denoted as \(i\), is a fascinating concept in mathematics used to solve equations that don't have real solutions. It is defined by the property \(i^2 = -1\). This characteristic allows \(i\) to become a backbone for complex numbers. When it comes to the powers of \(i\), they follow a repeating pattern every four exponents. Here's how it works:

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

The cycle then repeats: \(i^5 = i\), \(i^6 = -1\), and so on. This cyclic nature makes it straightforward to simplify expressions involving powers of \(i\). By understanding and leveraging this pattern, even very large or negative exponents can be managed easily.

Exponents in Mathematics

Exponents are a fundamental tool in mathematics that signify repeated multiplication of a number by itself. For instance, \(a^n\) means \(a\) multiplied by itself \(n\) times. They're not just limited to positive integers; exponents can be zero, negative, or even fractions, leading to interesting results.

Zero exponents, as in \(a^0\), result in 1 as long as \(a\) is not zero. This is because dividing any number by itself leaves us with 1. Negative exponents, such as \(a^{-n}\), represent the reciprocal of the positive exponent, giving us \(\frac{1}{a^n}\). Fractional exponents point to roots, where \(a^{1/n}\) is the \(n\)-th root of \(a\).

When simplifying expressions involving exponents, it's crucial to follow these rules and convert expressions between different forms to uncover simpler equivalents. This helps when dealing with complex numbers powered by exponentials, especially handling imaginary units like \(i\).

Simplifying Expressions

Simplifying mathematical expressions is about reducing them to their simplest form while maintaining equivalence. This process involves consolidating like terms, converting into known forms, and applying arithmetic operations cleanly.

When simplifying expressions with imaginary numbers, it often involves considering the cyclic behavior of \(i\) and its exponents. For instance, when faced with \(i^{92}\), recognizing that powers of \(i\) repeat every four allows us to directly state the result without multiplying \(i\) repeatedly. Simply find the remainder of 92 divided by 4 to see where it falls within the cycle, giving \(i^{92} = 1\).

Additionally, managing negative exponents involves using reciprocal operations. Thus, \(i^{-33}\) becomes \(\frac{1}{i}\), and since \(i^3 = -i\), the expression simplifies to \(-i\). These simplifications allow us to rewrite expressions efficiently in the form \(a + bi\), essential for clarity and further mathematical operations.