Chapter 1: Problem 16
Write the expression in the form \(a+b i\), where \(a\) and \(b\) are real numbers. (a) \(i^{92}\) (b) \(i^{-33}\)
Short Answer
Expert verified
(a) \(1 + 0i\); (b) \(0 - i\)
Step by step solution
01
Understanding Powers of i
The imaginary unit 'i' is defined as \(i = \sqrt{-1}\). Therefore, the powers of 'i' follow a cyclical pattern:- \(i^1 = i\)- \(i^2 = -1\)- \(i^3 = -i\)- \(i^4 = 1\)- \(i^5 = i\) (and the pattern repeats with a cycle of 4)...
02
Determine the Remainder for \(i^{92}\)
To find \(i^{92}\), we need the remainder of 92 divided by 4 because the powers of 'i' cycle every 4. Performing the division, we get 92 divided by 4 equals 23 with a remainder of 0. Thus, \(i^{92} = i^0 = 1\).
03
Determine the Remainder for \(i^{-33}\)
For negative powers, the same cycle rules apply. Start by converting \(i^{-33}\) into a positive power by utilizing the identity \(i^{-n} = 1 / i^n\). The remainder of 33 divided by 4 is 1 (since 33 = 32 + 1, where 32 is a multiple of 4). Therefore, \(i^{-33} = 1/i^1 = 1/i = -i\) as \(1/i = -i\) by the property \(i \cdot i = -1\).
04
Write Each Expression in Standard Form
Now express both answers in the form \(a + bi\):- For \(i^{92}\), we have the result as 1. This means \(a = 1\) and \(b = 0\), so it's written as \(1 + 0i\).- For \(i^{-33}\), we conclude \(-i\), thus \(a = 0\) and \(b = -1\), resulting in \(0 - i\) or \(0 - 1i\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Imaginary Unit
The imaginary unit, often represented by the symbol "i," is a fundamental concept in mathematics. This special number is defined by the equation \(i = \sqrt{-1}\). The introduction of the imaginary unit allowed mathematicians to solve equations that do not have solutions in the set of real numbers.
Here's why the imaginary unit is essential:
Understanding \(i\) is crucial for grasping more complex algebraic structures. Keep in mind, when you're exploring equations involving \(i\), you're operating in a broader spectrum beyond the real number line.
Here's why the imaginary unit is essential:
- It enables the calculation of square roots of negative numbers, which are otherwise impossible in the realm of real numbers.
- By working with the imaginary unit, we can gain a deeper understanding and representation of complex phenomena.
Understanding \(i\) is crucial for grasping more complex algebraic structures. Keep in mind, when you're exploring equations involving \(i\), you're operating in a broader spectrum beyond the real number line.
Powers of i
Powers of the imaginary unit "i" cycle through a predictable pattern. This pattern repeats every 4 powers, making it crucial to remember when working with higher powers of \(i\).
Here is the basic cycle:
Here is the basic cycle:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)
- The result of \(i^n\) depends on the remainder when \(n\) is divided by 4.
- If the remainder is 0, then \(i^n = 1\).
- If the remainder is 1, then \(i^n = i\).
- If the remainder is 2, then \(i^n = -1\).
- If the remainder is 3, then \(i^n = -i\).
Complex Number Form
Complex numbers bring together real and imaginary numbers in a neat package. They are typically written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit.
The structure of a complex number allows for both real and imaginary parts:
For example, in the expression \(i^{92}\) which simplifies to \(1\), you convert it to complex form as \(1 + 0i\). In the case of \(i^{-33}\) simplifying to \(-i\), the expression becomes \(0 - 1i\) in complex number form.
This form can also facilitate operations such as addition, subtraction, and multiplication of complex numbers, making it a versatile and powerful tool in mathematics.
The structure of a complex number allows for both real and imaginary parts:
- The "real part" is represented by \(a\).
- The "imaginary part" is represented by \(b\), the coefficient of \(i\).
For example, in the expression \(i^{92}\) which simplifies to \(1\), you convert it to complex form as \(1 + 0i\). In the case of \(i^{-33}\) simplifying to \(-i\), the expression becomes \(0 - 1i\) in complex number form.
This form can also facilitate operations such as addition, subtraction, and multiplication of complex numbers, making it a versatile and powerful tool in mathematics.