Problem 72 - A Graphical Approach to Precalculus with Limits

Question

Multiply as indicated. Write each product in standand form. $$(3-i)(3+i)(2-6 i)$$

Step-by-Step Solution

Answer

The product is \(20 - 60i\) in standard form.

Step 1: Multiply the First Two Binomials

We start by multiplying the first two binomials \((3-i)(3+i)\). These are conjugates, and their product is given by the difference of squares: \( (3)^2 - (i)^2 = 9 - (-1) = 9 + 1 = 10\). The result of this multiplication is 10.

Step 2: Multiply the Result with the Third Binomial

Next, we multiply the result from Step 1, which is 10, with the last binomial \((2-6i)\). Apply the distributive property: \(10(2-6i) = 10 \times 2 - 10 \times 6i = 20 - 60i\). Both terms are real numbers.

Step 3: Write in Standard Form

The expression obtained in Step 2 is already in the standard form for complex numbers, which is \(a + bi\). In this case, our expression is \(20 - 60i\).

Key Concepts

Conjugates

Conjugates play a crucial role in complex numbers, simplifying calculations and helping to transform complex expressions into simpler forms. The conjugate of a complex number is found by changing the sign of the imaginary part. For a complex number expressed as \(a + bi\), the conjugate is \(a - bi\). This operation on complex numbers is very useful when we need to perform arithmetic operations like multiplication, especially in case of division.

When you multiply a complex number by its conjugate, you obtain a real number. This result comes from the property of conjugates resulting in the "difference of squares." For example, multiplying \((3-i)\) by its conjugate \((3+i)\), results in the product \(3^2 - (i)^2\).

• Real part squared: \(3^2 = 9\)
• Imaginary part squared: \(i^2 = -1\), replacing gives \(-(i^2) = +1\).

So, the conjugate multiplication becomes \(9 + 1 = 10\), a real number. This property of conjugates makes such multiplications easy and simplifies expressions substantially.

Difference of squares

The difference of squares is a powerful algebraic expression that can simplify many mathematical problems. It is written as \(a^2 - b^2\) and can be factored as \((a + b)(a - b)\). This identity proves extremely useful when dealing with complex numbers.

In the context of the original exercise, the expression \((3-i)(3+i)\) uses the difference of squares. Here, the number \(3\) represents \(a\) and \(i\) represents \(b\). Applying the formula, we see:

• \(a^2 = 3^2 = 9\)
• \(b^2 = i^2 = -1\), resulting in subtracting a negative, which adds 1.

This results in \(9 - (-1) = 9 + 1 = 10\). So, by using the difference of squares, you can simplify to a single real number, which becomes a crucial and simplified intermediate result in problems involving complex numbers.

Distributive property

The distributive property is a basic yet essential principle of algebra. It allows you to distribute multiplication over addition or subtraction, which helps simplify expressions and solve equations efficiently. The rule is expressed as \(a(b + c) = ab + ac\).

In complex numbers, this property is often used to multiply numbers where one of the factors is a real number. In our exercise, after finding the product of the first two binomials as 10, we use the distributive property to multiply this by the next binomial \((2 - 6i)\).

• Multiply 10 by the real component: \(10 \times 2 = 20\)
• Multiply 10 by the imaginary component: \(10 \times (-6i) = -60i\)

Combining these results, the term becomes \(20 - 60i\). This illustrates how the distributive property can break down what may appear complex into simpler arithmetic operations, providing an efficient pathway to a solution.