Question

Find the values of the following expressions: a. \((2+3 i)+(6-2 i)\) b. \((2-i)(1+3 i)\) c. \(i-(2+3 i)\)

Short answer

\(8 + i, 5 + 5i, -2 - 2i\)

Step-by-step solution

Part (a) - Adding Complex Numbers

To add two complex numbers, we separately add the real parts and the imaginary parts. Given expression: \((2+3 i)+(6-2 i)\) Step 1: Add the real parts together: \( 2 + 6 = 8 \) Step 2: Add the imaginary parts together: \( 3i - 2i = i \) Step 3: Combine the results: \( 8 + i \) Therefore, the value of the given expression is \( 8 + i \).

Part (b) - Multiplying Complex Numbers

To multiply complex numbers, we use the distributive property. Given expression: \((2-i)(1+3 i)\) Step 1: Distribute the first term in the first parenthesis: \( 2(1) + 2(3i) = 2 + 6i \) Step 2: Distribute the second term in the first parenthesis: \( -i(1) - i(3i) = -i - 3i^2 \) Step 3: Combine the results and simplify: \( (2 + 6i) + (-i - 3i^2) = 2 + 6i - i - 3(-1) \) \( = 2 + 5i + 3 = 5 + 5i \) Therefore, the value of the given expression is \( 5 + 5i \).

Part (c) - Subtracting Complex Numbers

To subtract complex numbers, we separately subtract the real parts and the imaginary parts. Given expression: \(i-(2+3 i)\) Step 1: Distribute the negative sign across the second complex number: \( i - 2 - 3i \) Step 2: Combine the real parts: \( -2 \) Step 3: Combine the imaginary parts: \( i - 3i = -2i \) Step 4: Combine the results: \( -2 - 2i \) Therefore, the value of the given expression is \( -2 - 2i \).

Key concepts

Adding Complex Numbers

When we talk about adding complex numbers, we are referring to combining two numbers of the form \( a + bi \) where \( a \) is the real part and \( bi \) is the imaginary part. To successfully add them, you perform addition for each corresponding part separately. For example, consider the sum \( (2+3i)+(6-2i) \).

To add these, you will:

• Add the real parts: \( 2 + 6 = 8 \).
• Add the imaginary parts: \( 3i - 2i = i \).

Then, you combine these results to get the sum \( 8 + i \). This process is intuitive and similar to adding two-dimensional vectors, where real numbers are treated like horizontal displacements, and imaginary numbers are like vertical displacements on a plane.

Multiplying Complex Numbers

Multiplying complex numbers incorporates the use of the distributive property (also known as the FOIL method in algebra). Let's dissect the multiplication of the complex numbers \( (2-i)(1+3i) \).

You perform the multiplication in the following steps:

• Multiply the real part of the first number by both parts of the second number: \( 2 \times 1 + 2 \times 3i = 2 + 6i \).
• Multiply the imaginary part of the first number by both parts of the second number: \( -i \times 1 - i \times 3i = -i + 3 \), since \( i^2 = -1 \).

Then, add the results to get \( 5 + 5i \). Remember that \( i^2 = -1 \), so anytime you see \( i^2 \) when multiplying complex numbers, you replace it with \( -1 \), simplifying the expression further.

Subtracting Complex Numbers

Subtracting complex numbers is very much like the addition, but instead, you subtract the corresponding parts. The subtraction of \( i-(2+3i) \) can be a bit tricker because of the subtraction sign. Follow these steps:

• Distribute the negative sign to both the real and imaginary parts of the second complex number: \( i - (2 + 3i) = i - 2 - 3i \).
• Subtract the real parts: Since there's no explicit real part attached to \( i \) in the first number, it's like subtracting \( 2 \) from \( 0 \), which is \( -2 \).
• Subtract the imaginary parts: \( i - 3i = -2i \).

Combine them to get your result: \( -2 - 2i \). It's essential to be careful with signs during subtraction to avoid errors.