Problem 80 - Intermediate Algebra

Question

Perform each indicated operation. Write the result in the form \(a+b i\). $$ 4(2-i)^{2} $$

Step-by-Step Solution

Answer

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

Step 1: Expand the Squared Term

First, use the formula for squaring a binomial: \ \[(a-b)^2 = a^2 - 2ab + b^2\]. In this exercise, let \(a = 2\) and \(b = i\). So, we have: \ \[(2-i)^2 = (2)^2 - 2 \cdot 2 \cdot i + (i)^2\].\Now, calculate each term: \ 1. \((2)^2 = 4\). \ 2. \(-2 \cdot 2 \cdot i = -4i\). 3. \(i^2 = -1\). Combine all: \ \[(2-i)^2 = 4 - 4i - 1\]. \ Simplify: \ \[(2-i)^2 = 3 - 4i\].

Step 2: Multiply by the Coefficient

After obtaining the result of \((2-i)^2\) as \(3-4i\), multiply this result by 4: \ \[4(3 - 4i) = 4 \times 3 + 4 \times (-4i)\]. \ Calculate each term: \ 1. \(4 \times 3 = 12\). \ 2. \(4 \times (-4i) = -16i\). \ So, the product is: \ \[12 - 16i\].

Step 3: Express the Result in Standard Form

The result of the operation \(4(2-i)^2\) is already in the required form \(a + bi\), where \(a = 12\) and \(b = -16\). \Thus, the final expression is \(12 - 16i\).

Key Concepts

Binomial Expansion

When we talk about binomial expansion, we are referring to expanding expressions raised to a power. The classic formula for this is known as the binomial theorem. For a binomial expression, such as \((a + b)\), raised to the power of 2, the expansion can be written using the identity:

• \((a - b)^2 = a^2 - 2ab + b^2\)

This formula helps us to expand \((a - b)^2\) quickly without having to multiply the binomial by itself manually. In the provided example, we use \(a = 2\) and \(b = i\) to transform \((2-i)^2\) into a simple polynomial that can be easily evaluated.
This concept is not just useful for simple arithmetic but extends into higher-level mathematics and is valuable for solving algebraic equations, especially involving complex numbers, which bring us to the next section.

Imaginary Unit

The imaginary unit, represented as \(i\), is a mathematical concept used to extend the real number system into the complex number system. It is defined by the property that \(i^2 = -1\).

Unlike real numbers, imaginary and complex numbers open up a new realm of mathematical exploration.

Here are a few key points about the imaginary unit \(i\):

• It is used to represent the square root of negative numbers. For instance, \(\sqrt{-1} = i\).
• Combining real and imaginary components creates complex numbers, expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers.
• In calculations, always remember \(i^2 = -1\), helping you simplify expressions that include even higher powers of \(i\), like \(i^3 = -i\) and \(i^4 = 1\).

In the exercise solution, the step \(i^2 = -1\) is crucial as it simplifies the term involving \(i\). This step changes the nature of the calculation, turning an imaginary component into a real number, leading to its appropriate simplification in the context of complex numbers.

Complex Multiplication

Multiplying complex numbers might seem intimidating at first, but it follows straightforward rules based on distributing the multiplication over addition and remembering the property of \(i\).

Complex multiplication involves expanding each term, similar to multiplying binomials, but with an added twist due to the imaginary component. Here's what is typically involved:

• Given two complex numbers, \((a + bi)\) and \((c + di)\), you can find their product using the distributive property: \((a + bi)(c + di) = ac + adi + bci + bdi^2\).
• Substituting \(i^2 = -1\) allows you to combine the real and imaginary parts: \(ac - bd\) forms the new real part, and \((ad + bc)i\) forms the imaginary part.

In the example given, multiplying each component of \(3 - 4i\) by \(4\) after binomial expansion transforms the expression into standard form.
This operation emphasizes the importance of correct distribution and recognition of \(i^2\) to ensure accurate simplification and final outcomes.