Problem 10 - Algebra 2 and Trigonometry

Question

In \(3-10,\) write the expansion of each binomial. $$ (1-i)^{5} ; i=\sqrt{-1} $$

Step-by-Step Solution

Answer

The expansion of \((1 - i)^5\) is \(-4 + 6i\).

Step 1: Understand the Binomial Theorem

The binomial theorem states that \((a + b)^n\) can be expanded using the formula \(\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k\). Here, \(a = 1\), \(b = -i\), and \(n = 5\).

Step 2: Identify the Binomial Coefficients

The binomial coefficients \(\binom{5}{k}\) where \(k = 0, 1, 2, 3, 4, 5\) can be determined from Pascal's triangle or calculated as \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\). For this case, the coefficients are 1, 5, 10, 10, 5, 1.

Step 3: Expand the Binomial

Apply the coefficients and the powers of \(a\) and \(-i\):\[(1 - i)^5 = \binom{5}{0}(1)^5(-i)^0 + \binom{5}{1}(1)^4(-i)^1 + \binom{5}{2}(1)^3(-i)^2 + \binom{5}{3}(1)^2(-i)^3 + \binom{5}{4}(1)^1(-i)^4 + \binom{5}{5}(1)^0(-i)^5\]

Step 4: Simplify Each Term

Calculate each term separately:- \(\binom{5}{0}(1)^5(-i)^0 = 1 \cdot 1 \cdot 1 = 1\)- \(\binom{5}{1}(1)^4(-i)^1 = 5 \cdot 1 \cdot (-i) = -5i\)- \(\binom{5}{2}(1)^3(-i)^2 = 10 \cdot 1 \cdot (-1) = -10\)- \(\binom{5}{3}(1)^2(-i)^3 = 10 \cdot 1 \cdot i = 10i\)- \(\binom{5}{4}(1)^1(-i)^4 = 5 \cdot 1 \cdot 1 = 5\)- \(\binom{5}{5}(1)^0(-i)^5 = 1 \cdot 1 \cdot i = i\)Note that \(i^2 = -1\) and \((-i)^2 = -1, (-i)^3 = -i, (-i)^4 = 1, (-i)^5 = i\).

Step 5: Combine the Terms

Combine all of the calculated terms: \[(1 - i)^5 = 1 - 5i - 10 + 10i + 5 + i\]Combine like terms separately for real and imaginary parts: - Real: \(1 - 10 + 5 = -4\)- Imaginary: \(-5i + 10i + i = 6i\)Thus, the expansion of \((1 - i)^5\) is \(-4 + 6i\).

Key Concepts

Complex Numbers

Complex numbers are numbers that have both a real part and an imaginary part. They are generally expressed in the form of \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. The imaginary part arises because of the presence of the imaginary unit \(i\), which is defined as \(i = \sqrt{-1}\).

Complex numbers are crucial in advanced mathematical contexts, including polynomial equations and engineering applications. They help with the solutions that cannot be solved using only real numbers. For instance, when asked to find the square root of a negative number, complex numbers provide a means to handle these situations.

One of the neat properties of complex numbers is how they can be added, subtracted, and even multiplied together. When you're performing operations with complex numbers, you'll handle the real and imaginary parts separately:

• **Addition/Subtraction**: Combine like terms by adding or subtracting the real parts together and the imaginary parts together (e.g., \((a + bi) + (c + di) = (a+c) + (b+d)i\)).
• **Multiplication**: Use distributive property to expand (e.g., \((a + bi)(c + di) = ac + adi + bci + bdi^2\)). Remember that \(i^2 = -1\), which simplifies the expression.

Complex numbers extend our understanding of numbers and open the door for broader and deeper connections in mathematics.

Imaginary Unit

The imaginary unit, denoted as \(i\), is an essential element in mathematics, particularly in complex numbers. It is defined by its unique property: \(i^2 = -1\). This definition allows for the square root of negative numbers and leads to complex numbers, which are essential in advanced mathematics.

Imagine trying to solve the equation \(x^2 + 1 = 0\). In the realm of real numbers, there is no solution since no real number satisfies \(x^2 = -1\). However, introducing \(i\) resolves this, as \(x = \pm i\) becomes a solution to the equation.

The imaginary unit follows specific properties that can be used in polynomial expansions and other calculations:

• **Powers of \(i\)**: The powers of \(i\) cycle every four steps:
• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)
• After \(i^4\), the cycle repeats.

Understanding the imaginary unit and its properties is key to mastering complex numbers, and it plays a critical role in binomial expansions involving complex terms.

Polynomial Expansion

Polynomial expansion using the binomial theorem allows us to raise expressions to any integer power. The binomial theorem provides a formula to break down expressions like \((a + b)^n\) into a series of terms, which are easily computed.

The formula for the binomial expansion is \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^{k} \]In this equation:

• **Binomial Coefficient**: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) determines how each term combines contributions from \(a\) and \(b\).
• **Terms**: Each term in the expansion is calculated by a specific combination of powers of \(a\) and \(b\), and multiplied by the corresponding binomial coefficient.
• **Simplification**: Any term where \(b\) is imaginary requires special attention since powers of \(i\) have unique effects on the terms as seen in binomial expansions with complex numbers.

By applying this theorem, you can systematically expand polynomials to better understand or simplify otherwise cumbersome expressions. Particularly with complex numbers, this theorem becomes a powerful tool to manage and interpret complex binomials.