Problem 10 - College Algebra Essentials

Question

In Exercises \(9-20,\) find each product and write the result in standard form. $$-8 i(2 i-7)$$

Step-by-Step Solution

Answer

The product is \(16 + 56i\).

Step 1: Distribute Real Part

First, distribute the real part \(-8i\) to each term of the binomial \(2i-7\). We find: \(-8i * 2i + -8i * -7\). This results in \(-16i^2 + 56i\).

Step 2: Simplify

Next, simplify the equation remembering the definition of \(i\), where \(i^2 = -1\). We substitute this in and obtain \(-16*(-1) + 56i\). This simplifies to \(16 + 56i\).

Step 3: Write in Standard Form

Finally, we must make sure the result is in standard form, that is \(a + bi\). The solution from step 2 is already in this format.

Key Concepts

Multiplication of Complex Numbers

Understanding the multiplication of complex numbers is akin to grasping the crossing of two electric currents—it might seem complex, but with the right formula, it makes perfect sense! Complex multiplication hinges on handling both the real and imaginary parts appropriately. Let's take the expression \( -8i(2i-7) \) as our teaching moment.

When approaching this, we use the distributive property to multiply \( -8i \) with each component in the binomial \( 2i \) and \( -7 \) separately. This gives us \( -16i^2 + 56i \) before simplification. Now, here's a neat trick: knowing that \( i^2 = -1 \) helps us turn \( -16i^2 \) into \( 16 \) because \( i^2 \) essentially flips the sign of the number preceding it. Add this to \( 56i \) and voila, you have \( 16 + 56i \), a product in standard form, where standard form means a real number plus an imaginary number \( a + bi \).

Imaginary Unit

The imaginary unit, denoted as \( i \), is the square root of \( -1 \) and is the foundation of all complex numbers. It's a bit like the secret ingredient in a magical recipe—it makes the impossible possible! Without \( i \) we wouldn't venture beyond the real number line. In simplifying the product of complex numbers, always remember that \( i^2 \) equals \( -1 \). This peculiar property is what allows us to convert parts of our expression into real numbers. For example, taking \( -16i^2 \) from our earlier problem, we replace \( i^2 \) with \( -1 \) and the equation transforms, leading to positive \( 16 \) as a result.

It's crucial for students to recall this property since it is the cornerstone when handling any operations involving imaginary numbers.

Distributive Property

The distributive property is a handy tool that lets us expand expressions like \( a(b + c) \) into \( ab + ac \). Imagine it as a multiplier stretching across each term within the brackets. In our complex numbers example, \( -8i \) disperses like a fragrance, touching \( 2i \) and \( -7 \) individually to give \( -16i^2 \) and \( 56i \) respectively.

By mastering the distributive property, students can deftly manage not only complex number multiplication but also simplify expressions in algebra, making it an essential mathematical maneuver. Always ensure to multiply each term thoroughly and consider any signs that could affect the outcome—negative times negative gives a positive, an easy-to-miss detail that swings the direction of your solution.