Problem 15 - Precalculus: Functions and Graphs

Question

Write the expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$\begin{array}{ll} \text { (a) } i^{43} & \text { (b) } i^{-20} \end{array}$$

Step-by-Step Solution

Answer

(a) i^{43} = 0 - i; (b) i^{-20} = 1

Step 1: Recall the Exponent Cycle of i

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

Step 2: Determine i^{43} Using Modulo Operation

Since the cycle repeats every 4 powers, calculate \(43 \mod 4\). This equals 3. Hence, \(i^{43} = i^3 = -i\).

Step 3: Write i^{43} in the Form a + bi

Since \(i^{43} = -i\), the expression in the form \(a + bi\) is 0 + (-1)i, where \(a = 0\) and \(b = -1\).

Step 4: Determine i^{-20} Using Inverse Property

Negative exponents indicate the reciprocal, so \(i^{-20} = \frac{1}{i^{20}}\). Determine \(20 \mod 4 = 0\), so \(i^{20} = 1\). Thus, \(i^{-20} = \frac{1}{1} = 1\).

Step 5: Write i^{-20} in the Form a + bi

Since \(i^{-20} = 1\), the expression in the form \(a + bi\) is 1 + 0i, where \(a = 1\) and \(b = 0\).

Key Concepts

Imaginary Unit

The imaginary unit, denoted as \(i\), is a fundamental concept in complex numbers, where the square of \(i\) is equal to \(-1\). It's important to understand that \(i\) does not have a value on the real number line. Instead, it helps us expand our number system to handle equations that don't have real solutions. For example, the equation \(x^2 = -1\) does not have any solutions if we only consider real numbers, but it has solutions in the complex number system: \(x = i\) and \(x = -i\).

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

This sequence demonstrates the imaginary unit repeating in a cycle every four powers. This repetition allows us to simplify calculations significantly when dealing with powers of \(i\).

Exponent Cycle

The concept of an exponent cycle is closely tied to understanding powers of \(i\). Because \(i\)'s powers repeat every 4 steps, this natural cycle allows us to handle high powers or negative powers without endless calculation.For instance, if an exponent is very large, like \(i^{43}\), you can use the cycle to determine its simplification quickly. You can perform the calculation \(43 \mod 4\), which results in 3, so \(i^{43} = i^3 = -i\). This approach helps bypass tedious arithmetic while keeping calculations manageable and accurate.The cycle is:

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

Each cycle ends with \(i^4 = 1\), simplifying complex calculations by aligning higher powers with simpler, well-known results.

Negative Exponents

Negative exponents indicate a reciprocal operation. For complex numbers involving \(i\), the same rules apply. Instead of computing directly, you invert the base of the exponent, which is particularly useful when the base is a complex number.Consider \(i^{-20}\):

• First, rewrite it using positive exponents: \(i^{-20} = \frac{1}{i^{20}}\).
• Next, note from the exponent cycle that \(i^{20} = 1\), since \(20 \mod 4 = 0\).
• Thus, \(i^{-20} = \frac{1}{1} = 1\).

This manipulation leverages the properties of exponent cycles and reciprocals to solve what might initially seem a complex problem simply and efficiently.

Modulo Operation

The modulo operation is a mathematical operation that finds the remainder of division between two numbers. In the context of complex numbers and particularly the powers of \(i\), it assists in determining which cycle a high power lands within.For example, when finding \(i^{43}\), instead of calculating \(i\) raised to a high power directly, you compute \(43 \mod 4\), which is 3. Therefore, \(i^{43} = i^3 = -i\). Performing modulo operation allows you to:

• Identify the equivalent lower power within a known cycle.
• Simplify the expression using familiar results from the cycle.
• Avoid unnecessary lengthy calculations by reducing substantial powers to smaller, manageable numbers.

Using the modulo operation strategically in combinations with the exponential cycle of \(i\) can significantly streamline calculations involving complex numbers.