Problem 16 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-34: Write the expression in the form \(a+b i\), where \(a\) and \(b\) are real numbers. (a) \(i^{92}\) (b) \(i^{-33}\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Powers of i

Recall that the imaginary unit \(i\) has a cyclical pattern for its powers: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\). This means every fourth power of \(i\) returns to 1, and the cycle repeats.

Step 2: Simplify i^92

Divide 92 by 4, since the powers of \(i\) repeat every four terms: \(92 \div 4 = 23\) with a remainder of 0. Thus, \(i^{92} = (i^4)^{23} = 1^{23} = 1\). So, \(a = 1\) and \(b = 0\), resulting in \(1 + 0i\).

Step 3: Simplify i^{-33}

Convert the negative exponent to a positive one by using the property \(i^{-n} = 1/i^n\). Calculate \(33 \div 4\) which gives a remainder of 1, meaning \(i^{33} = i\). Consequently, \(i^{-33} = 1/i = -i\) because the reciprocal of \(i\) is \(-i\). Therefore, \(a = 0\) and \(b = -1\), resulting in \(0 - i\).

Key Concepts

Powers of i

When dealing with complex numbers and specifically the imaginary unit, the powers of the imaginary unit, denoted as \( i \), follow a predictable cycle. This cycle repeats every four steps. The pattern is as follows:

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

Understanding this cycle helps in simplifying complex expressions, as any integer power of \( i \) can be reduced to one of these four outcomes. For instance, if you are asked to simplify \( i^{92} \), you would divide 92 by 4 since the cycle repeats every four terms, yielding a remainder of 0. Therefore, \( i^{92} = (i^4)^{23} = 1^{23} = 1 \). Identifying these cycles makes it much easier to work with large exponents of \( i \), reducing them to a much simpler form, like \( a + bi \).
In the above example, \( i^{92} = 1 \), which becomes \( 1 + 0i \) since there is no imaginary component.

Imaginary Unit

The imaginary unit \( i \) is a fundamental building block in the world of complex numbers. It is defined by the equation \( i^2 = -1 \). This definition allows for the extension of the real number system into complex numbers, providing solutions to equations that have no real solutions.
For example, any time you encounter the square root of a negative number, it can be expressed as a multiple of \( i \). The expression \( \sqrt{-1} \) is represented as \( i \), making it possible to work with negative square roots. Complex numbers take the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit. This form is particularly useful in various fields such as engineering, physics, and computer science, where they often represent quantities like electrical currents and signal waveforms.

Negative Exponents

Negative exponents in mathematics indicate the reciprocal of the base raised to the positive of that exponent. This rule is consistent even when dealing with the imaginary unit \( i \). For instance, \( i^{-n} = \frac{1}{i^n} \). This property is very helpful, especially when simplifying expressions with negative exponents.
Consider the expression \( i^{-33} \). To simplify, first convert it to \( \frac{1}{i^{33}} \). Then, you calculate \( i^{33} \) by understanding the cycle of powers of \( i \). As \( 33 \div 4 \) gives a remainder of 1, \( i^{33} = i \). Hence, \( i^{-33} = \frac{1}{i} = -i \), since the reciprocal of \( i \) is \(-i \) as derived from multiplying the numerator and the denominator by the conjugate, \( i \).
Understanding negative exponents helps in solving complex number problems, particularly when needing to express answers in the form \( a + bi \). For \( i^{-33} \), it simplifies to \( 0 - i \), which is \( a = 0 \) and \( b = -1 \).