Chapter 1: Problem 17
Write the expression in the form \(a+b i\), where \(a\) and \(b\) are real numbers. (a) \(i^{73}\) (b) \(i^{-46}\)
Short Answer
Expert verified
(a) 0 + 1i (b) -1 + 0i
Step by step solution
01
Understanding powers of i
The imaginary unit \(i\) is defined such that \(i^2 = -1\). It follows that \(i^3 = i^2 \cdot i = -i\), \(i^4 = i^2 \cdot i^2 = 1\), and \(i^5 = i^4 \cdot i = i\). Thus, the powers of \(i\) cycle every 4: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and then it repeats.
02
Simplifying i^73
Since the powers of \(i\) cycle every 4, we need to find the remainder of 73 when divided by 4 to determine which power corresponds to \(i^{73}\). Calculating the remainder: \(73 \div 4 = 18\) remainder \(1\). Therefore, \(i^{73} = i^1 = i\).
03
Writing i^73 in a+bi form
Since \(i^{73} = i\), the expression can be directly written in the form \(a + bi\) as \(0 + 1i\), where \(a = 0\) and \(b = 1\).
04
Simplifying i^{-46}
To simplify \(i^{-46}\), first determine \(i^{46}\) using the cycle. Calculate \(46 \div 4 = 11\) remainder \(2\). Therefore, \(i^{46} = i^2 = -1\). Using the property \(i^{-n} = \frac{1}{i^n}\), we have \(i^{-46} = \frac{1}{i^{46}} = \frac{1}{-1} = -1\).
05
Writing i^{-46} in a+bi form
Since \(i^{-46} = -1\), we can write it in the form \(a + bi\) as \(-1 + 0i\), where \(a = -1\) and \(b = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Powers of Imaginary Unit
The imaginary unit, represented as \(i\), is a fundamental concept in mathematics, especially when dealing with complex numbers. It is defined such that \(i^2 = -1\). From this basic definition, we can derive the powers of \(i\). Let's take a look:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = i^2 \cdot i = -i\)
- \(i^4 = i^2 \cdot i^2 = 1\)
- \(i^5 = i^4 \cdot i = i\)
Cyclic Nature of Powers of i
One interesting property of the imaginary unit \(i\) is its cyclic nature.
- The powers of \(i\) cycle every four steps: \(i, -1, -i, 1\).
- This means that when handling a power of \(i\), you can find the equivalent smaller one by noting the remainder when dividing by 4.
- Divide 73 by 4 and get the remainder, which is 1.
- This remainder tells us that \(i^{73} = i^1 = i\).
Imaginary Unit in a+bi Form
Complex numbers are generally expressed in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. Simplifying powers of \(i\) can allow us to express results in this form. Let's reflect on our exercise:
- Taking \(i^{73}\), we found it simplifies to \(i\), which is directly written as \(0 + 1i\).
- Similarly, for \(i^{-46}\), recognizing the cycle, we determine it simplifies to \(-1\), expressible as \(-1 + 0i\).