Problem 74 - Precalculus with Limits

Question

In Exercises 73 - 78, use the Binomial Theorem to expand the complex number. Simplify your result. \( \left(2 - i\right)^5 \)

Step-by-Step Solution

Answer

-58 - 121i

Step 1: Understand the Binomial Theorem

The Binomial Theorem states that: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\] where \( \binom{n}{k} = \frac{n!}{k!(n-k)!}\), \(a\) and \(b\) are any numbers, and \(n\) is a natural number.

Step 2: Identify a and b

In the given problem, \(a = 2\) and \(b = -i\).

Step 3: Apply the Binomial Theorem

Expand \((2 - i)^5\) using the Binomial Theorem: \[ (2 - i)^5 = \sum_{k=0}^{5} \binom{5}{k} 2^{5-k} (-i)^k \]

Step 4: Simplify the expression

Expand using Binomial Theorem and simplify each term separately: \[ (2 - i)^5 = \binom{5}{0} 2^{5} (-i)^{0} + \binom{5}{1} 2^{4} (-i)^{1} + \binom{5}{2} 2^{3} (-i)^{2} + \binom{5}{3} 2^{2} (-i)^{3} + \binom{5}{4} 2^{1} (-i)^{4}+\binom{5}{5} 2^{0} (-i)^{5}\] Simplify terms, \[= 32 - 80i + 80i^2 - 40i^3 + 10i^4 - i^5 \] As \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\) and \(i^5 = i\), substitute these values, \[= 32 - 80i - 80 - 40i - 10 - i \]

Step 5: Write the final answer

Combine like terms to simplify: \[= -58 - 121i \] This is the simplified form of \((2 - i)^5\), using the Binomial Theorem.

Key Concepts

Expansion of Complex Numbers

The expansion of complex numbers using the Binomial Theorem can at first seem daunting, but it becomes manageable with a systematic approach. By using this theorem, you can expand expressions of the form \((a + b)^n\) where \(a\) and \(b\) might include complex numbers. For example, in the problem \((2 - i)^5\), we can set \(a = 2\) and \(b = -i\).
You start by applying the formula:

• Identify the parameters \(a\), \(b\), and \(n\).
• Use the Binomial Theorem: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
• This generates terms for each value of \(k\) between 0 and \(n\).

Once the terms are generated, complex powers of \(i\) (such as \(i^2, i^3\), etc.) must be simplified.
This method allows you to fully expand and obtain the simplified form of any complex binomial raised to a power, using known identities of complex numbers.

Combinatorics

Combinatorics involves counting and arrangement principles, which is crucial in the application of the Binomial Theorem. When dealing with binomials like \((a + b)^n\), knowing how to calculate the coefficients, \( \binom{n}{k} \), is essential.
This notation, \( \binom{n}{k} \), represents the number of combinations. It tells you how many ways you can choose \(k\) elements from \(n\), and is calculated as

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
• The factorial \(n!\) means multiplying \(n\) by all integers less than the number down to 1 (e.g., \(4! = 4 \times 3 \times 2 \times 1\)).

This computation forms the backbone of properly expanding any binomial, including when complex numbers are in play.
It gives each term in the expansion the correct coefficient, ensuring the result reflects the power \((a + b)^n\) accurately.

Complex Numbers

Complex numbers are numbers that have both a real part and an imaginary part, expressed as \(a + bi\), where \(i\) is the imaginary unit with the property that \(i^2 = -1\). They extend the concept of one-dimensional numbers to two dimensions.
The crucial thing to remember when working with complex numbers, especially in expansions, is how their powers cycle:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\), and repeats thereafter.

Understanding this pattern is key to simplifying terms in a binomial expansion involving complex numbers. Each time you apply a power of \(i\), you cycle through these values.
Recognizing this not only makes calculations more straightforward, but it's also a time saver. Complex numbers, while initially challenging, start uncovering a new layer of math that brings simplicity with practice.