Problem 71 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify Complex Conjugates

Notice that the expression \((2+i)(2-i)\) consists of a pair of complex conjugates. When two complex conjugates are multiplied, the result is a real number. Therefore, we should start by multiplying these two factors. This simplifies the expression significantly.

Step 2: Multiply the Complex Conjugates

Multiply \((2+i)(2-i)\). When multiplying complex conjugates, use the identity \((a+b)(a-b) = a^2 - b^2\). Set \(a = 2\) and \(b = i\): \((2+i)(2-i) = 2^2 - i^2 = 4 - (-1) = 5\). Therefore, the product of the complex conjugates is 5.

Step 3: Simplify the Expression

Substitute the result from Step 2 to simplify the expression to \(5(4+3i)\). This expression is now in the form of a real number multiplied by a complex number.

Step 4: Distribute to Find the Product

Distribute 5 across \(4+3i\): \[(5)(4) + (5)(3i) = 20 + 15i.\] This results in a new expression, which is already in the standard form of a complex number.

Step 5: Write the Final Expression in Standard Form

In the standard form of a complex number, the expression is \(a + bi\). From Step 4, the multiplication results in \(20 + 15i\), which is precisely in this form.

Key Concepts

Complex Conjugates

Complex conjugates are pairs of complex numbers that have the same real part but opposite imaginary parts. For example, the numbers \(2+i\) and \(2-i\) are complex conjugates. When you multiply these conjugates, something magical happens—they simplify into a real number!Here’s how it works:

• Write the conjugates as \((a+b)\) and \((a-b)\).
• Use the identity \((a+b)(a-b) = a^2-b^2\).
• In our case, \(a = 2\) and \(b = i\).
• Substitute these values into the identity to get \(2^2 - i^2 = 4 - (-1)\).

This evaluates to 5. The result of their multiplication is always a real, non-imaginary number. So, complex conjugates are super handy for simplifying complex expressions that might otherwise be scary! This trick makes complicated problems much more approachable.

Standard Form

The standard form of a complex number is simply written as \(a + bi\). Here, \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, defined as \(i^2 = -1\).Why is it called the "standard" form? It's like having a uniform for complex numbers—everyone knows what to expect!

• \(a\) is the real part of the complex number.
• \(bi\) is the imaginary part.

So when you see an expression like \(20 + 15i\), it's already in standard form, with 20 being the real part and 15i being the imaginary part. Writing numbers in this form ensures clarity and makes mathematical operations more predictable.Whether adding, subtracting, or multiplying complex numbers, presenting your final answer in standard form helps maintain consistency and helps others understand your results quickly.

Multiplication of Complex Numbers

When multiplying complex numbers, you often use the distributive property, just like with real numbers. Let's break it down using an example like \((2+i)(4+3i)\) to understand how these multiplications work.Here's the process:

• First, multiply the real parts: \((2)(4) = 8\).
• Next, multiply the real part by the imaginary part: \((2)(3i) = 6i\).
• Then, multiply the imaginary part by the real part: \((i)(4) = 4i\).
• Finally, multiply the imaginary parts: \((i)(3i) = 3i^2 = 3(-1) = -3\).

Combine all these results:

• Add the real numbers: \(8 - 3 = 5\).
• Add the imaginary numbers: \(6i + 4i = 10i\).

The resulting complex number is \(5 + 10i\). Multiplication becomes straightforward when broken down into these simple steps.In other cases, like in our original problem, simplifying within the expression using complex conjugates first, reduces the complexity and helps in accurate, faster multiplication.