Problem 45 - College Algebra

Question

Perform the indicated operation(s) and write the result in standard form. $$ (2-3 i)(1-i)-(3-i)(3+i) $$

Step-by-Step Solution

Answer

The result of the operation is -5 + i

Step 1: Multiply Complex Numbers - Part 1

Multiply first pair of complex numbers using the distributive property: (2-3 i)(1-i) = 2(1) - 2(i) - 3(i)(1) + 3(i)(-i)

Step 2: Multiply Complex Numbers - Part 2

Multiply second pair of complex numbers using the distributive property: (3-i)(3+i) = 3(3) - 3(i) - 1(i)(3) + 1(i)(i)

Step 3: Apply \(i^2=-1\) and Simplify

Simplify the above expressions using the rule \(i^2=-1\): (2-3 i)(1-i) = 2 - 2i - 3i + 3 = 5 - 5i and (3-i)(3+i) = 9 - 3i - 3i + 1=10 - 6i.

Step 4: Subtracting the Complex Numbers

Subtract the result of the second multiplication from the result of the first: (5 - 5i) - (10 - 6i) = (5-10) + (-5i - (-6i) )= -5 - (-1i) = -5 + i

Step 5: Final Result in Standard Form

The final result in standard form is -5 + i

Key Concepts

Distributive Property

The distributive property is a fundamental principle in algebra that allows us to multiply a single term across terms within parentheses. This property applies not only to real numbers but also to complex numbers, which are numbers that include both real and imaginary parts.

In complex numbers, using the distributive property involves multiplying each part of one complex number by each part of another. For instance, when you have the expression

• \((2-3i)(1-i)\),

you distribute each part as follows:

• First, multiply \(2 \times 1\), \(2 \times -i\),
• \(-3i \times 1\), and \(-3i \times -i\).

By carefully distributing each component, you ensure that no part of the expression is missed, allowing you to accurately simplify and solve the problem. The same principle is applied when multiplying the second pair,

• \((3-i)(3+i)\).

Standard Form

Standard form for a complex number is written as \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. This format helps in simplifying complex numbers for calculation and comparison.

In our exercise, once we perform the operations and simplify using properties like the distributive property and the rules involving the imaginary unit \(i\), we aim to express our final result neatly in this standard form.

For example, after performing the necessary operations and simplifications in our problem, the result is observed as

• \(-5 + i\)

where

• \(-5\)

is the real part, and

• \(i\)

is the imaginary part.

Subtracting Complex Numbers

To subtract complex numbers, combine the real parts and the imaginary parts separately. This means you treat it much like subtracting polynomials or vectors.

When you have two complex numbers, such as

• \((5-5i)\)

and

• \((10-6i)\),

you subtract

• \(10\)

from

• \(5\) to get \(-5\),
• \(-5i\)

from

• \(-6i\)

by using the operation

• (-5i - (-6i))

which simplifies to

• \(+i\).

Thus, the subtraction of these two complex numbers results in \(-5 + i\). This method ensures each part of the number is handled separately and correctly.

Imaginary Unit i

The imaginary unit \(i\) is a fundamental concept in the realm of complex numbers. It is defined as \(i = \sqrt{-1}\), and its square always equals \(-1\):

• \(i^2 = -1\).

This property helps when simplifying expressions involving complex numbers.

In our exercise, when expanding complex products, an \(i^2\) term will almost always appear. Recognizing that \(i^2 = -1\) allows one to substitute \(-1\) for each \(i^2\) term and proceed with additional simplifications.

For example, after distributing and simplifying

• \((2-3i)(1-i)\)

and

• \((3-i)(3+i)\),

the interpretation of \(3i(-i)\) or \(1(i^2)\) as \(-3\) (after reduction) illustrates the practical use of \(i^2 = -1\) for reaching the final standard form of the given complex number expression.