Problem 57 - A Graphical Approach to Precalculus with Limits

Question

Multiply as indicated. Write each product in standand form. $$(2+4 i)(-1+3 i)$$

Step-by-Step Solution

Answer

The product is \(-14 + 2i\).

Step 1: Apply the Distributive Property

Begin by distributing each term in the first binomial by every term in the second binomial. This is a process similar to the FOIL method. Multiply each term: \[ (2 + 4i)(-1 + 3i) = 2(-1) + 2(3i) + 4i(-1) + 4i(3i) \]

Step 2: Simplify Each Term

Calculate each term from the distribution:- \(2(-1) = -2\)- \(2(3i) = 6i\)- \(4i(-1) = -4i\)- \(4i(3i) = 12i^2\)Remember that \(i^2 = -1\), so \(12i^2 = 12(-1) = -12\).

Step 3: Combine Like Terms

Now, combine the real parts together and the imaginary parts together:- Real parts: \(-2 - 12 = -14\)- Imaginary parts: \(6i - 4i = 2i\).

Step 4: Write the Result in Standard Form

Combine the simplified results to get the final expression in standard form. The standard form for a complex number is \(a + bi\).So, \((2 + 4i)(-1 + 3i) = -14 + 2i\).

Key Concepts

Distributive Property

When it comes to multiplying complex numbers, the distributive property is your best friend! The distributive property is a rule in algebra that simplifies expressions with multiple terms. It allows you to "distribute" or multiply each term in one set with every term in another. In our problem, we are dealing with two binomials:

• First binomial: \(2 + 4i\)
• Second binomial: \(-1 + 3i\)

The key is to multiply each term in the first binomial by each term in the second one. This is similar to the FOIL method used when multiplying two binomials.
By applying this method, you ensure that no part of the original expressions is left unmultiplied. For example, in the step by step solution, the process is carried out like this:

• Multiply \(2\) with \(-1\) and \(3i\)
• Multiply \(4i\) with \(-1\) and \(3i\)

This ensures that every component contributes to the final result, which makes it so effective.

Imaginary Unit

The imaginary unit, denoted as \(i\), is a fundamental concept when working with complex numbers. By definition, \(i\) is the square root of \(-1\). This is quite unique because most numbers have positive square roots. However, \(i\) is special and allows mathematicians to work with numbers that have negative square roots.
Understanding how \(i\) behaves is crucial in solving problems involving complex numbers. One important property of \(i\) is that when you square it, you get \(-1\). Mathematically, this is expressed as:

• \(i^2 = -1\)

In the step-by-step solution of our problem, this property comes in handy when multiplying \(4i(3i)\). The \(i\) terms together give \(12i^2\), and since \(i^2 = -1\), the term becomes \(-12\).
This transformation also helps in converting expressions to their simplest form, especially when coupled with other real numbers. It's essential for fully understanding how complex numbers operate.

Standard Form of Complex Numbers

Complex numbers often need to be written in what we call \'standard form\'. This standard form is expressed as \(a + bi\), where \(a\) is the real part, and \(b\) is the imaginary part. Having a common format helps to better organize and simplify complex numbers mathematically.
In our exercise, after distributing and simplifying, the expression \((2 + 4i)(-1 + 3i)\) results in \(-14 + 2i\). This is the final expression in standard form, where:

• \(-14\) is the real part (\(a\))
• \(2i\) is the imaginary part (\(bi\))

Arranging complex numbers in this form makes it easier to add, subtract, multiply, or even compare them with other complex numbers. Moreover, identifying the real and imaginary parts allows for a simplified process when combining like terms. In our solution, once the separate terms are combined into real and imaginary groups, it's easy to write down the answer neatly and concisely in standard form!