Chapter 1: Problem 41
Perform the indicated operation, and write each expression in the standard form \(a+\) bi. $$ 6 i^{3}-4 i^{5} $$
Short Answer
Expert verified
-10i
Step by step solution
01
Compute the value of i^3
Recall that \( i \) is the imaginary unit where \( i = \sqrt{-1} \). The powers of \( i \) follow a cyclical pattern: \( i^1 = i \), \( i^2 = -1 \), \( i^3 = -i \), \( i^4 = 1 \), and then it repeats. Therefore, \( i^3 = -i \).
02
Compute the value of i^5
Using the same cyclical pattern, note that \( i^5 \) can be rewritten as \( i^{4+1} = (i^4) \times i = 1 \times i = i \). Therefore, \( i^5 = i \).
03
Substitute the values back into the expression
Replace \( i^3 \) with \( -i \) and \( i^5 \) with \( i \) in the original expression. So, the expression becomes \[ 6(-i) - 4(i) \].
04
Simplify the expression
Simplify the expression \[ 6(-i) - 4(i) \]. This gives \[ -6i - 4i \].
05
Combine like terms
Combine the terms involving \( i \). \[ -6i - 4i = -10i \].
06
Write the expression in standard form
The standard form \( a + bi \) means we identify the real part and the imaginary part. Here the real part is 0, and the imaginary part is \( -10i \). Therefore, the expression in standard form is \[ 0 + (-10i) \] or simply \( -10i \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
imaginary unit
To understand this exercise, you need a firm grasp of the imaginary unit, often represented by the symbol \( i \). The imaginary unit is defined as \( i = \sqrt{-1} \). This might seem odd since you can’t take the square root of a negative number in the realm of real numbers. However, in the imaginary realm, we define this new number, \( i \), to make such mathematics possible. This means that when you square \( i \), you get \( -1 \); mathematically, \( i^2 = -1 \).
Without defining \( i \), much of advanced mathematics, particularly in fields like engineering and physics, wouldn’t be achievable. It’s a crucial concept that extends the number system into a new dimension, giving us the complex numbers.
Without defining \( i \), much of advanced mathematics, particularly in fields like engineering and physics, wouldn’t be achievable. It’s a crucial concept that extends the number system into a new dimension, giving us the complex numbers.
cyclical pattern of i
The powers of the imaginary unit \( i \) follow a repeating, cyclical pattern. This is essential for solving problems involving higher powers of \( i \). Here's the pattern you see repeatedly:
Knowing this pattern allows you to simplify any power of \( i \) by converting it into one of these four fundamental cases. For example, in the given exercise:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
Knowing this pattern allows you to simplify any power of \( i \) by converting it into one of these four fundamental cases. For example, in the given exercise:
- \( i^3 = -i \)
- \( i^5 = i \)
standard form of complex numbers
The standard form of a complex number is written as \(a + bi\), where:
For instance, in the solution, after computing the values of \(i^3\) and \(i^5\), we substituted them back into the original expression, simplified it, and combined like terms involving \(i\). The result was \(-10i\). Because there is no real part, this is equivalent to saying the real part \(a\) is 0, and the imaginary part \(bi\) is \(-10i\). Thus, the standard form is \(0 + (-10i)\) or just \(-10i\).
Remember, writing complex numbers in their standard form helps to clearly separate the real and imaginary components, making mathematical operations more straightforward.
- \(a\) is the real part
- \(bi\) is the imaginary part, and \(b\) is a real number
For instance, in the solution, after computing the values of \(i^3\) and \(i^5\), we substituted them back into the original expression, simplified it, and combined like terms involving \(i\). The result was \(-10i\). Because there is no real part, this is equivalent to saying the real part \(a\) is 0, and the imaginary part \(bi\) is \(-10i\). Thus, the standard form is \(0 + (-10i)\) or just \(-10i\).
Remember, writing complex numbers in their standard form helps to clearly separate the real and imaginary components, making mathematical operations more straightforward.