Problem 60 - Precalculus Student Solutions Manual 5th

Question

Find each product. Write the answer in standard form. $$(6-4 i)(6+4 i)$$

Step-by-Step Solution

Answer

The product is 52.

Step 1: Identify the formula

Recognize that (6 - 4i)(6 + 4i) is a difference of squares formula, which can be written as (a - b)(a + b) = a^2 - b^2.

Step 2: Assign values to a and b

Assign the values a = 6 and b = 4i.

Step 3: Apply the formula

Substitute a = 6 and b = 4i in the difference of squares formula: (6 - 4i)(6 + 4i) = 6^2 - (4i)^2.

Step 4: Calculate the squares

Calculate the squares: 6^2 = 36 and (4i)^2 = 16i^2.

Step 5: Simplify i^2

Remember that i^2 = -1. So, 16i^2 = 16(-1) = -16.

Step 6: Perform the subtraction

Subtract the result: 36 - (-16) = 36 + 16 = 52.

Key Concepts

Difference of Squares

The 'difference of squares' is a useful algebraic identity. It states that for any two numbers 'a' and 'b', we have \[ (a - b)(a + b) = a^2 - b^2 \]. Here, 'a' and 'b' can be any values, including complex numbers. When you see an expression like \[ (6 - 4i)(6 + 4i) \], you can recognize it as a difference of squares format. This formula helps to simplify products without expanding the terms step by step.

Multiplication of Complex Numbers

When multiplying complex numbers, use the distributive property. Consider complex numbers in the form \[ (a + bi) \]. To multiply \[ (6-4i) \] by \[ (6+4i) \], you distribute each term: \[ 6(6) + 6(4i) - 4i(6) - 4i(4i) \]. Simplify to: \[ 36 + 24i - 24i - 16i^2 \]. Notice the terms \[ +24i \text{ and } -24i \] cancel out, leaving you with \[ 36 - 16i^2 \]. Next, replace \[ i^2 \] with \[ -1 \], because \[ i^2 = -1 \]. Thus, \[ -16i^2 \] becomes \[ 16 \]. So the result is: \[ 36 + 16 = 52 \]. Using the formula, you avoided complicated steps!

Imaginary Unit

The imaginary unit, denoted as \[ i \], is defined as \[ \sqrt{-1} \]. This number is crucial for dealing with complex numbers. A complex number combines a real part and an imaginary part and takes the form \[ a + bi \], where \[ 'i' \] represents the imaginary unit. One key property is \[ i^2 = -1 \]. Thus, for any power of \[ i \], you can simplify: \[ i^3 = i^2 \times i = -i \]. In problems, it's vital to know \[ i^2 = -1 \] to simplify complex operations. For example, in the expression \[ 16i^2 \], this becomes \[ 16(-1) = -16 \]. Grasping \[ i \]'s properties is essential for working with complex numbers.