Problem 64 - A Graphical Approach to Precalculus with Limits

Question

Multiply as indicated. Write each product in standand form. $$(6-4 i)(6+4 i)$$

Step-by-Step Solution

Answer

The product in standard form is 52.

Step 1: Recall the formula for the difference of squares

The expression \((a-b)(a+b)\) is a difference of squares and can be expanded as \(a^2 - b^2\). Note that \(i^2 = -1\).

Step 2: Identify variables

In the given expression \((6-4i)(6+4i)\), identify \(a = 6\) and \(b = 4i\).

Step 3: Apply the difference of squares formula

Using the formula from Step 1, compute \(a^2 - b^2\) where \(a = 6\) and \(b = 4i\). Substitute the values: \[a^2 = 6^2 = 36\,b^2 = (4i)^2 = 16i^2 = 16(-1) = -16\,a^2 - b^2 = 36 - (-16) = 36 + 16 = 52\].

Step 4: Write the result in standard form

The product of \((6-4i)(6+4i)\) using the difference of squares is \(52\). This is already in standard form as there is no imaginary part.

Key Concepts

Difference of Squares

The difference of squares is a mathematical expression used for factoring specific types of expressions. It can be recognized in the pattern

• \((a-b)(a+b)\)

This pattern is known as the difference of squares because it can be expanded as

• \(a^2 - b^2\)

The reason behind this name is that you're subtracting the square of one term from another. Let’s break it down: if you have two terms, \(a\) and \(b\), multiplying their sum and their difference results in the difference between their squares. This concept is not just theoretical; it has applications in simplifying expressions, especially when dealing with polynomials and complex numbers.

Imaginary Unit

When dealing with complex numbers, the imaginary unit, represented by \(i\), plays a crucial role. The imaginary unit is defined by its unique property:

• \(i^2 = -1\)

This might seem strange at first because squaring a number usually yields a positive result. However, \(i\) is specifically designed to make sense of square roots of negative numbers, which don’t exist in the set of real numbers.
In complex numbers, the value of \(b\) in a difference of squares can be imaginary, such as \(4i\) in our original problem. This means when squared, it gives a negative result thanks to the property of \(i\):

• \((4i)^2 = 16i^2 = 16(-1) = -16\)

Understanding \(i\) helps in managing calculations involving complex numbers.

Standard Form

The standard form for complex numbers is expressed as

• \(a + bi\)

with \(a\) representing the real part, and \(b\) representing the imaginary part. When a complex number is written in standard form, it clearly shows the balance between the real and imaginary parts.
However, in the solution of the expression

• \((6-4i)(6+4i)\)

we derived a result

• \(52\)

which is purely a real number. This is already in standard form because the imaginary part is zero. When simplifying or solving expressions involving complex numbers, reaching or maintaining the standard form is often a key goal, ensuring interpretations are consistent and clear.