Problem 39 - Intermediate Algebra

Question

Multiply. Write the product in the form \(a+b i .\) See Example 4. $$ (4-2 i)^{2} $$

Step-by-Step Solution

Answer

The product is \(12 - 16i\).

Step 1: Understand the Expression

We're tasked with squaring the complex number \((4 - 2i)\). This involves multiplying the complex number by itself: \((4 - 2i) \times (4 - 2i)\).

Step 2: Apply the Distributive Property

Use the distributive property \((a-b)(c-b)=ac-ad-bc+bd\) on \((4 - 2i)(4 - 2i)\):\[(4 - 2i)(4 - 2i) = 4(4) + 4(-2i) - 2i(4) - 2i(-2i)\].

Step 3: Perform the Multiplications

Calculate each term: - \(4 \times 4 = 16\)- \(4 \times (-2i) = -8i\)- \(-2i \times 4 = -8i\)- \(-2i \times (-2i) = 4i^2\) (Note: \(i^2 = -1\)) resulting in \(4 \times -1 = -4\).

Step 4: Combine Like Terms

Now, combine all the results:\[16 - 8i - 8i - 4\].Simplify it to:\[16 - 4 - 16i = 12 - 16i\].

Step 5: Finalize the Expression in Standard Form

The product of the square of \((4 - 2i)\) in standard form \(a+bi\) is:\[12 - 16i\].

Key Concepts

Distributive Property

The distributive property is a fundamental principle in algebra that allows us to break down and simplify expressions. In the context of multiplying complex numbers, it helps us manage the distribution of every part of one binomial to every part of another.
For example, when multiplying \((4 - 2i)(4 - 2i)\), each term in the first parenthesis is multiplied by each term in the second. This can be visualized as:

• \(4 \times 4\)
• \(4 \times (-2i)\)
• \(-2i \times 4\)
• \(-2i \times (-2i)\)

By breaking down the multiplication like this, it becomes easier to manage complex numbers by simplifying their real and imaginary parts separately.

Imaginary Unit

The imaginary unit, denoted as \(i\), is a unique number defined by the property \(i^2 = -1\). It serves as the foundation of imaginary numbers, acting as a bridge between real numbers and complex numbers.
When dealing with expressions involving complex numbers, especially multiplication, remembering the property of the imaginary unit is crucial.
For instance, in our problem when multiplying \(-2i \times -2i\), once you compute it, you bring in the imaginary unit to get \(-4i^2\). Then, applying \(i^2 = -1\), this simplifies to \(4\), converting something that might initially seem complex into a real number.

Multiplication of Complex Numbers

When multiplying complex numbers, the goal is to apply the distributive property across all terms, just like in traditional algebra. Each term in the first complex number is multiplied by each term in the second.
Let's demystify the squaring of \((4 - 2i)\). You'll first distribute:

• \(4 \cdot 4 = 16\)
• \(4 \cdot (-2i) = -8i\)
• \(-2i \cdot 4 = -8i\)
• \(-2i \cdot (-2i) = 4i^2\)

Upon calculating these products, combine like terms and replace \(i^2\) with \(-1\), meticulously simplifying the complex expression into its real and imaginary components. The term \(4i^2\) translates to \(-4\), reducing the complexity of the expression.

Standard Form of Complex Numbers

The standard form of a complex number is expressed as \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit.
After performing operations on complex numbers, it's crucial to express the final result in standard form, ensuring clarity and correctness.
Let's consider \((4 - 2i)^2\). After distributing, calculating, and simplifying, we gather all real elements together, and do the same for the imaginary ones. This process results in \(12 - 16i\), where:

• The real part, \(a\), is \(12\).
• The imaginary part, \(b\), is \(-16\).

Putting everything back together, the complex number in standard form makes it straightforward to identify both its real and imaginary parts.