Problem 118 - Elementary and Intermediate Algebra

Question

Look Alikes ... a. \((3+i)+(2+4 i)\) b. \((3+i)(2+4 i)\)

Step-by-Step Solution

Answer

a: \(5+5i\); b: \(2+14i\).

Step 1: Identify the Problem Type

Determine if each expression is an addition or multiplication of complex numbers. The exercise involves both addition and multiplication of complex numbers.

Step 2: Solve Part (a): Addition of Complex Numbers

Identify the real and imaginary parts in each complex number: \((3+i)\) and \((2+4i)\). Add the real parts together and the imaginary parts together.Real parts: \(3 + 2 = 5\)Imaginary parts: \(1 + 4 = 5\)Thus, the sum is \(5 + 5i\).

Step 3: Solve Part (b): Multiplication of Complex Numbers

When multiplying complex numbers \((3+i)\) and \((2+4i)\), use the distributive property to expand.\((3+i)(2+4i) = 3 \cdot 2 + 3 \cdot 4i + i \cdot 2 + i \cdot 4i\)Calculate each term:\(3 \cdot 2 = 6\)\(3 \cdot 4i = 12i\)\(i \cdot 2 = 2i\)\(i \cdot 4i = 4i^2\), and recall that \(i^2 = -1\), so \(4i^2 = 4(-1) = -4\).Combine all terms:\(6 + 12i + 2i - 4 = 2 + 14i\).

Key Concepts

Addition of Complex Numbers

Adding complex numbers involves a simple process that is similar to adding two-digit numbers. Each complex number has a real part and an imaginary part. To perform the addition, we simply add the respective parts together. For instance, consider the complex numbers

• \((3+i)\)
• \((2+4i)\)

Here, the real parts are 3 and 2, and the imaginary parts are 1 (from \(i\)) and 4.
To add these numbers:

• Add the real parts: \(3 + 2 = 5\)
• Add the imaginary parts: \(1 + 4 = 5i\)

Thus, the sum of the complex numbers is \(5 + 5i\).
It's as straightforward as stacking like terms together, making sure to keep real and imaginary parts separate during addition.

Multiplication of Complex Numbers

Multiplication of complex numbers requires a bit more attention to detail compared to addition. When multiplying complex numbers, such as \((3+i)\) and \((2+4i)\), we use the distributive property to simplify the expression.
Let's break it down step-by-step:

• Apply the distributive property: \((3+i)(2+4i) = 3 \cdot 2 + 3 \cdot 4i + i \cdot 2 + i \cdot 4i\)
• Calculate each multiplication:
  • \(3 \cdot 2 = 6\)
  • \(3 \cdot 4i = 12i\)
  • \(i \cdot 2 = 2i\)
  • \(i \cdot 4i = 4i^2\)
• Recall that \(i^2 = -1\), so \(4i^2 = 4(-1) = -4\)
• Combine all terms: \(6 + 12i + 2i - 4\)
• Simplify by adding the real parts and the like imaginary parts together: \(2 + 14i\)

By methodically applying these steps, we handle each part separately, making the multiplication process manageable and the result accurate.

Distributive Property in Algebra

The distributive property is fundamental in algebra and helps simplify multiplication, especially in the context of complex numbers.
The property states that for any numbers \(a, b,\) and \(c\), the following holds true: \(a(b+c) = ab + ac\).
This property is at the heart of multiplying complex numbers. When working with \((3+i)(2+4i)\), we apply the distributive property twice:

• First, by distributing \(3\) over \((2+4i)\)
• Second, by distributing \(i\) over \((2+4i)\)

The distributive property helps break down complex multiplication into smaller, more manageable calculations, which are then combined to give the final result.
Understanding this concept deeply can significantly improve your ability to tackle more complex algebraic operations, not just with complex numbers but with algebra in general.