Problem 46 - Algebra and Trigonometry

Question

Evaluate the expression and write the result in the form a bi. $$ \frac{(1+2 i)(3-i)}{2+i} $$

Step-by-Step Solution

Answer

The result is \( 3 + i \).

Step 1: Expand the Numerator

First, let's distribute the expression in the numerator. We have \[ (1 + 2i)(3 - i) = 1 \cdot 3 + 1 \cdot (-i) + 2i \cdot 3 + 2i \cdot (-i) \]This simplifies to \[ 3 - i + 6i - 2i^2 \] Remember that \( i^2 = -1 \), so substitute \( -2i^2 \) with \( 2 \). Hence, it becomes \[ 3 - i + 6i + 2 = 5 + 5i \].

Step 2: Convert Denominator into Real Number

To eliminate the imaginary part from the denominator, multiply the numerator and the denominator by the conjugate of the denominator. The denominator is \( 2 + i \), and its conjugate is \( 2 - i \). Thus, multiply as follows:\[ \frac{(5 + 5i)}{2+i} \cdot \frac{(2-i)}{(2-i)} \] In the denominator, we have \[ (2+i)(2-i) = 4 - i^2 = 4 + 1 = 5 \].

Step 3: Expand the New Numerator

Now multiply the expanded numerator by the conjugate of the original denominator:\[ (5+5i)(2-i) = 5 \cdot 2 + 5 \cdot (-i) + 5i \cdot 2 + 5i \cdot (-i) \] This expands to:\[ 10 - 5i + 10i - 5i^2 \]Recalling that \( i^2 = -1 \), substitute \( -5i^2 \) with 5:\[ 10 - 5i + 10i + 5 = 15 + 5i \].

Step 4: Simplify the Expression

Now, divide the result from Step 3 by the real denominator obtained in Step 2:\[ \frac{15 + 5i}{5} \] This simplifies to \[ 3 + i \].

Step 5: Express in Standard Form a+bi

The simplified result \( 3 + i \) is already in the form \( a + bi \), where \( a = 3 \) and \( b = 1 \).

Key Concepts

Imaginary Unit

The imaginary unit, denoted as \( i \), is the cornerstone of complex numbers. It is defined as \( i = \sqrt{-1} \). This unit is fundamental because it allows us to expand our number system beyond real numbers to include complex numbers, which can represent solutions to equations that don't have real solutions.
When squared, the imaginary unit has a unique property:

• \( i^2 = -1 \)

This property is essential in simplifying expressions involving complex numbers. For instance, in the given problem, \( i^2 = -1 \) is used to convert terms like \(-2i^2\) into real numbers, which simplifies the computation process.
The imaginary unit opens doors to represent numbers of the form \( a + bi \), where \( a \) and \( b \) are real numbers. This is crucial for the manipulation of expressions and equations in the complex plane.

Complex Conjugate

The complex conjugate of a complex number is a concept that helps simplify the division of complex numbers. For any complex number \( a + bi \), its complex conjugate is \( a - bi \). This changes the sign of the imaginary part and is denoted as \( \overline{a+bi} = a - bi \).
Using the complex conjugate allows us to eliminate the imaginary part from the denominator of a fraction. In the exercise, we multiplied both the numerator and the denominator by the conjugate of the denominator \( (2+i) \), which is \( (2-i) \).
This operation relies on the formula:

• \( (a+bi)(a-bi) = a^2 + b^2 \)

Illustrated in the solution, multiplying \((2+i)(2-i) = 4 + 1 = 5\) confirms that the denominator becomes a real number. This technique simplifies complex fractions, making the arithmetic more straightforward. Understanding how to use the complex conjugate is crucial for working with problems that involve division of complex numbers.

Standard Form of Complex Numbers

The standard form of a complex number is expressed as \( a + bi \), where \( a \) represents the real part and \( b \) represents the imaginary part. This form is essential as it allows us to clearly separate the real and imaginary components, providing a simple way to perform arithmetic operations.
In the exercise solution, we ended with the expression \( 3 + i \). Here:

• \( a = 3 \) is the real part
• \( b = 1 \) is the imaginary part

This format makes it easy to add, subtract, or multiply complex numbers, and it also helps in visualizing them as points in the complex plane. Using the standard form ensures consistency in calculations and is a go-to representation for most problems involving complex numbers.