Problem 49 - Algebra A Combined Function

Question

Multiply. Write your answers in the form \(a+b i\). $$ (1-i)(1+i) $$

Step-by-Step Solution

Answer

2

Step 1: Apply the distributive property

Use the distributive property (also known as FOIL for binomials) to expand the expression \((1-i)(1+i)\). This means you multiply each term in the first binomial by each term in the second binomial:\( (1-i)(1+i) = 1 \cdot 1 + 1 \cdot i - i \cdot 1 - i \cdot i \).

Step 2: Perform the multiplications

Calculate each part of the expanded expression: 1. \(1 \cdot 1 = 1\)2. \(1 \cdot i = i\)3. \(-i \cdot 1 = -i\)4. \(-i \cdot i = -i^2\). Since \(i^2 = -1\), this becomes \(-(-1) = 1\).

Step 3: Combine like terms

Add the results from the multiplications: \(1 + i - i + 1\). Combine the like terms:- Combine the real parts: \(1 + 1 = 2\)- Combine the imaginary parts: \(i - i = 0\)

Step 4: Write the final answer in standard form

Consolidate the results into the standard form of a complex number, \(a + bi\): The real part is \(2\) and the imaginary part is \(0\), so the final answer is \(2 + 0i\), which simplifies to \(2\).

Key Concepts

Distributive Property

The distributive property is a fundamental property of multiplication used extensively in algebra. It allows us to simplify expressions by distributing the multiplication over addition or subtraction inside parentheses. When dealing with expressions like

• this means you multiply the terms outside of parentheses with each term inside.

For the expression \[(1-i)(1+i)\],applying the distributive property involves

• multiplying the first term in the first binomial (1-i)by each term in the second binomial(1+i). Similarly, you multiply -iby each term in (1+i),resulting in:
• • \(1 \cdot 1\),
  • \(1 \cdot i\),
  • \(-i \cdot 1\),
  • and \(-i \cdot i\).

Applying the distributive property this way ensures that each term in one binomial is multiplied with each term in the other binomial, ensuring no term is left out. This method is especially helpful for expanding and simplifying expressions with algebraic terms.

Binomials

A binomial is a polynomial with exactly two terms. It usually takes the form \(a + b\) where \(a\)and \(b\) are separate terms which could be constants, variables, or a mix of both. In our original exercise, the expression \((1-i)(1+i)\) consists of two binomials, \((1-i)\) and \((1+i)\).When you multiply binomials, like in our problem, you often use the FOIL method. This stands for

• First,
• Outer,
• Inner, and
• Last

The FOIL method ensures that you multiply each component of the first binomial by each component of the second binomial. For our example:

• **First**: Multiply the first terms: 1 \cdot 1 = 1
• **Outer**: Multiply the outer terms: 1 \cdot i = i
• **Inner**: Multiply the inner terms: -i \cdot 1 = -i
• **Last**: Multiply the last terms: -i \cdot i = -i^2

Using this technique facilitates handling similar binomial expressions, especially when they involve complex numbers.

Imaginary Unit

The concept of the imaginary unit is fundamental when dealing with complex numbers. The imaginary unit, denoted as \(i\),is defined such that \(i^2 = -1\).Imaginary numbers are multiples of \(i\)and are recognized when expressed in the form \(bi\),where \(b\)is a real number.In the exercise, when multiplying expressions like \(-i \cdot i\),the imaginary unit's defining property is highly relevant. Knowing that \(i^2 = -1\),we replace \(-i^2\)with \(-(-1)\),which simplifies to \(1\).This manipulation is essential in ensuring the resulting expression is in the standard form of a complex number, \(a + bi\),where aand bare real numbers. Mastering the use of \(i\)makes it easy to simplify and solve equations involving complex numbers.