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Expand and simplify the given expressions by use of the binomial formula. $$(1-j)^{6}(j=\sqrt{-1})$$

Short Answer

Expert verified
The expanded and simplified expression is \(2 + 8j\).

Step by step solution

01

Identify the Binomial Formula

The binomial theorem can be expressed as \((a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k\). This allows us to expand expressions of the form \((a+b)^n\). We will use this to expand \((1-j)^6\).
02

Apply the Binomial Formula Terms

For \((1-j)^6\), set \(a = 1\) and \(b = -j\), and \(n = 6\). This gives us:\[(1-j)^6 = \sum_{k=0}^{6}\binom{6}{k}1^{6-k}(-j)^k = \sum_{k=0}^{6}\binom{6}{k}(-j)^k\]
03

Calculate Binomial Coefficients and Powers

Compute each term of the series: - For \(k = 0\), the term is \(\binom{6}{0}(-j)^0 = 1\)- For \(k = 1\), the term is \(\binom{6}{1}(-j)^1 = -6j\)- For \(k = 2\), the term is \(\binom{6}{2}(-j)^2 = -15\)- For \(k = 3\), the term is \(\binom{6}{3}(-j)^3 = 20j\)- For \(k = 4\), the term is \(\binom{6}{4}(-j)^4 = 15\)- For \(k = 5\), the term is \(\binom{6}{5}(-j)^5 = -6j\)- For \(k = 6\), the term is \(\binom{6}{6}(-j)^6 = 1\)
04

Sum Up the Terms

Combine all the terms from the binomial expansion:\[1 - 6j - 15 + 20j + 15 - 6j + 1 = (1 - 15 + 15 + 1) + (-6j + 20j - 6j)\]This simplifies to:\[2 + 8j\]
05

Final Simplification

After consolidating like terms, the final simplified expression is:\[2 + 8j\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Numbers
Complex numbers are numbers that have a real part and an imaginary part. The imaginary unit is represented by \( j \) or \( i \), where \( j = \sqrt{-1} \). This means \( j^2 = -1 \). In the expression \((1-j)^6\), \( j \) is used in the expansion process as part of the binomial formula. Understanding complex numbers is crucial in mathematics because they allow the calculation of square roots of negative numbers, which is impossible with real numbers alone.
  • The real part of a complex number is the number without the imaginary unit \( j \).
  • The imaginary part is the term with \( j \).
  • In our problem, \(1\) is the real part and \(-j\) is the imaginary part.
When performing operations like the binomial expansion, each power of \( j \) follows a specific cycle:
  • \( j^1 = j \)
  • \( j^2 = -1 \)
  • \( j^3 = -j \)
  • \( j^4 = 1 \)
These rules are applied during the expansion process to simplify complex terms consistently.
Polynomials
A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. Polynomials are a crucial element of algebra because they generalize the notion of arithmetic sums to include combinations of variables and powers.

In the context of the binomial theorem, consider the expression \((1-j)^6\) as a polynomial where each term's degree decreases with the value of \( k \) from 6 to 0. The coefficients in each term come from the binomial coefficient \( \binom{n}{k} \), which determines how many ways we can choose \( k \) elements from \( n \) elements without regard to order.
  • Standard polynomial terms look like \(ax^n + bx^{n-1} + \ldots + c\).
  • Here, \((1-j)^6\) expands into multiple polynomial terms.
  • The degree of the polynomial is represented by the largest exponent of the variable, which is 6 in this case.
Polynomials provide a structured way to handle large calculations involving powers, making them an indispensable part of algebra and calculus.
Mathematical Expansion
Mathematical expansion refers to expressing a complicated expression as a sum of simpler terms. The Binomial Theorem allows us to expand expressions like \((a+b)^n\) to a sum of terms involving powers of \(a\) and \(b\). This is particularly useful because it lets us handle complex expressions more easily.

By applying the binomial theorem, we can break down \((1-j)^6\) into individual terms, each calculated using:
  • \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) which gives us the binomial coefficients.
  • Powers like \((-j)^k\) calculated through known powers of complex number \(j\).
  • Summing all the terms to revert to simpler constant and imaginary terms.
This expansion method is not only applicable for solving algebraic problems, but it's also used in statistics, calculus, and more to model and solve real-world problems with multiple interacting variables.

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