Problem 47 - Algebra A Combined Function

Question

Multiply. Write your answers in the form \(a+b i\). $$ (6-2 i)(3+i) $$

Step-by-Step Solution

Answer

The product is 20.

Step 1: Expand the Expression

To multiply the two complex numbers, use the distributive property to expand \( (6 - 2i)(3 + i) \). This means applying the formula \( (a+b)(c+d) = ac + ad + bc + bd \).

Step 2: Multiply Each Term

First, multiply \(6\) by \(3\): \( 6 \times 3 = 18 \).

Step 3: Multiply and Combine Real and Imaginary Parts

Next, multiply \(6\) by \(i\): \( 6 \times i = 6i \). Then multiply \(-2i\) by \(3\): \(-2i \times 3 = -6i \). Finally, multiply \(-2i\) by \(i\): \(-2i \times i = -2i^2 \).

Step 4: Simplify Using \(i^2 = -1\)

Replace \(i^2\) with \(-1\): \(-2i^2 = 2 \), because \(i^2 = -1\). Now the expression becomes: \(18 + 6i - 6i + 2 \).

Step 5: Combine Like Terms

Combine the real numbers and imaginary numbers: \(18 + 2 = 20\) and \(6i - 6i = 0i\). Thus, the answer is \(20 + 0i\), which simplifies to \(20\).

Key Concepts

Distributive Property

The distributive property is a fundamental algebraic principle that helps us to simplify expressions and is particularly useful when multiplying complex numbers. When we see an expression like \[(6-2i)(3+i)\],we apply the distributive property by multiplying each term in the first complex number \((6 - 2i)\) by each term in the second complex number \((3 + i)\).
Here’s the breakdown:

• Multiply 6 by 3, which gives us 18.
• Multiply 6 by i, resulting in 6i.
• Multiply -2i by 3, giving -6i.
• Finally, multiply -2i by i, which results in \( -2i^2 \).

The key to using the distributive property is organizing each multiplication carefully to ensure all parts of one term are multiplied by all parts of the other. This helps to maintain the structural integrity of the expression, ultimately allowing us to simplify the overall equation.

Imaginary Unit i

The imaginary unit, denoted as \( i \),is pivotal in understanding and working with complex numbers. By definition, \( i \) is the square root of \( -1 \).Therefore, \( i^2 \) equals \( -1 \).This property of \( i \)z brings a unique twist when performing arithmetic operations, like multiplication, involving complex numbers.
In our example, after applying the distributive property, we encounter a term \(-2i^2\).Substituting \( i^2 = -1 \), we can convert \(-2i^2\) into 2: \(-2 \times (-1) = 2\).
Understanding how to manage the imaginary unit is crucial in accurately simplifying complex expressions, as failing to replace \(i^2\) with \(-1\) would lead to incorrect solutions.

Combining Like Terms

Once we have finished expanding the expression using the distributive property and managing the imaginary parts involving \( i \), the next step in solving a complex number multiplication problem is combining like terms.
In our example, after substitution, we are left with \(18 + 6i - 6i + 2\).
Combining like terms involves:

• Grouping and adding the real parts together: \(18 + 2 = 20\).
• Grouping the imaginary parts: \(6i - 6i\), which result in \(0i\). This effectively ''cancels out,'' meaning the imaginary terms sum to zero.

The expression simplifies to \(20 + 0i\), which we typically write simply as \(20\).
Understanding how to combine like terms is essential for simplifying and expressing complex numbers in their standard form \(a + bi\), where both real and imaginary parts are clearly identified.