Problem 26 - Algebra 2 and Trigonometry

Question

In \(26-37,\) find each product. $$ (1+5 i)(1+2 i) $$

Step-by-Step Solution

Answer

The product is \(-9 + 7i\).

Step 1: Apply the distributive property

Use the distributive property (also known as the FOIL method for binomials) to expand the expression \((1 + 5i)(1 + 2i)\) into four terms: \(1 imes 1, 1 \times 2i, 5i \times 1,\) and \(5i \times 2i\).

Step 2: Calculate each term

Perform the multiplication for each term: - \(1 \times 1 = 1\)- \(1 \times 2i = 2i\)- \(5i \times 1 = 5i\)- \(5i \times 2i = 10i^2\)

Step 3: Simplify \(10i^2\)

Recall that \(i^2 = -1\). Therefore, \(10i^2 = 10(-1) = -10\), simplifying the expression to \(1 + 2i + 5i - 10\).

Step 4: Combine like terms

Combine the real parts and the imaginary parts: \(1 - 10 = -9\) (real part) and \(2i + 5i = 7i\) (imaginary part).Thus, the expression simplifies to \(-9 + 7i\).

Key Concepts

Distributive Property

The distributive property is like a mathematical tool that helps us break down and handle algebraic expressions, especially when dealing with multiplication. Imagine you want to multiply something spread out in a bracket, such as \((a + b)\), with something else, say \((c + d)\). The distributive property lets us break this down like a champ, distributing each part of the first set to each part of the second, making sure nothing is left out.

• Take each term in the first binomial: \(a\) and \(b\).
• Multiply these with each term in the second binomial: \(c\) and \(d\).
• Bring it all together: \(a \times c + a \times d + b \times c + b \times d\).

In the context of complex numbers, the distributive property plays a crucial role in expanding expressions like \((1 + 5i)(1 + 2i)\). By treating \(1\) and \(5i\) as our \(a\) and \(b\), and in the second binomial \(1\) and \(2i\) as \(c\) and \(d\), the steps remain consistent:

• \(1 \times 1\)
• \(1 \times 2i\)
• \(5i \times 1\)
• \(5i \times 2i\)

This helps us break down and work through the problem systematically.

Foil Method

The FOIL method is a specific application of the distributive property tailored for multiplying two binomials. It's an acronym: **First, Outer, Inner, Last**. This process is super handy because it guides us to ensure every part of the binomials gets multiplied properly and nothing is left out.
Here's how it works on binomials:

• **First**: Multiply the first terms of each binomial \((1 \times 1)\).
• **Outer**: Multiply the outer terms \((1 \times 2i)\).
• **Inner**: Multiply the inner terms \((5i \times 1)\).
• **Last**: Multiply the last terms \((5i \times 2i)\).

In the example \((1 + 5i)(1 + 2i)\), applying the FOIL method will guide us through the process of breaking down the multiplication:

• First: \(1 \times 1 = 1\)
• Outer: \(1 \times 2i = 2i\)
• Inner: \(5i \times 1 = 5i\)
• Last: \(5i \times 2i = 10i^2\) (noting that \(i^2 = -1\) changes this term to \(-10\))

The FOIL method ensures every part of the multiplication is systematic and straightforward, turning complex expressions into something digestible.

Imaginary Unit

In the world of complex numbers, the imaginary unit is our esteemed guest, represented by the symbol \(i\). This special number is defined as the square root of \(-1\). Now, why do we care about \(i\)? Because it helps us extend our number system beyond real numbers.
Whenever you see \(i\), it means "real-number math just took a creative twist".

• \(i^2 = -1\) is a fundamental rule and the cornerstone of working with imaginary numbers.
• \(i^3 = i^2 \times i = -i\)
• \(i^4 = (i^2)^2 = 1\)

Understanding how \(i\) behaves is essential when simplifying expressions involving imaginary numbers, such as in our task \((1 + 5i)(1 + 2i)\). The key moment arises when you handle \(5i \times 2i\), resulting in \(10i^2\).
Remember, since \(i^2\) always equals \(-1\), this expression \(10i^2\) simplifies to \(-10\). That might seem like a small detail, but it's vital for keeping complex number calculations flowing smoothly.