Problem 67 - Precalculus Student Solutions Manual 5th

Question

Find each product. Write the answer in standard form. $$(2+i)(2-i)(4+3 i)$$

Step-by-Step Solution

Answer

The answer in standard form is \( 20 + 15i \).

Step 1: Multiply the first two complex numbers

First, we need to multiply \(2+i\) and \(2-i\). Use the formula \( (a+b)(a-b) = a^2 - b^2 \). Here \(a = 2\) and \(b = i\): \[ (2+i)(2-i) = 2^2 - i^2 = 4 - (-1) = 5 \]

Step 2: Multiply the result with the third complex number

Now, multiply the result from Step 1 (which is 5) by \(4+3i\). We use the distributive property: \[ 5(4+3i) = 5 \times 4 + 5 \times 3i = 20 + 15i \]

Key Concepts

complex conjugate

The complex conjugate of a complex number is formed by changing the sign of its imaginary part. If you have a complex number in the form of \(a + bi\), its conjugate will be \(a - bi\).

Here’s an example: the complex number \(3 + 4i\) has a conjugate \(3 - 4i\).

When we multiply a complex number by its conjugate, an interesting thing happens. Let's see:

• \((a + bi)(a - bi) = a^2 - (bi)^2\)


• Since \(i^2 = -1\), the equation becomes \(a^2 + b^2\).

This means that the result is always a real number, as the imaginary parts cancel out.

In the given exercise, \(2+i\) and \(2-i\) are conjugates. By multiplying them, we get:
\((2+i)(2-i) = 2^2 - i^2 = 4 + 1 = 5\).
Notice, it's a real number! This helps simplify further calculations.

product of complex numbers

Multiplying complex numbers involves using the distributive property (also known as the FOIL method in algebra). To multiply two complex numbers like \((a+bi)(c+di)\), follow these steps:

• Multiply the real parts: \(a \times c\).


• Multiply the real part by the imaginary part: \(a \times di\).


• Multiply the imaginary part by the real part: \(bi \times c\).


• Multiply the imaginary parts: \(bi \times di = bdi^2\).

Combine these results and use \(i^2 = -1\) to simplify the expression:
\(ac + adi + bci + bdi^2 = ac + adi + bci - bd\).

Let’s apply this to our problem. We initially multiplied \(2+i\) with \(2-i\) to get \(5\), then we multiplied \(5\) by \(4+3i\):

• Multiply the real parts: \(5 \times 4 = 20\).


• Multiply the real part by the imaginary part: \(5 \times 3i = 15i\).

Combine the results:
\(20 + 15i\).
This is our product, and we will put it in standard form in the next section!

standard form of complex numbers

The standard form of complex numbers is expressed as \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, which satisfies \(i^2 = -1\).

When dealing with complex numbers, it's crucial to ensure that your final answer is in this standard form. Here’s why:

• It's easier to identify the real part (\(a\)) and the imaginary part (\(b\)).


• It simplifies addition, subtraction, and comparison between complex numbers.

In our exercise, after multiplying \((2+i)(2-i)\) to get \(5\), and then multiplying \(5\) by \(4+3i\), we arrived at \(20 + 15i\).

This result is already in standard form, with \(20\) as the real part and \(15i\) as the imaginary part. Therefore, the final answer to the problem is
\(20 + 15i\).

Ensuring the result is in standard form helps you accurately describe and use complex numbers in further calculations and applications!