Problem 33 - Algebra 2 and Trigonometry

Question

In \(19-34,\) write each sum or difference in terms of \(i\) $$ -\frac{1}{2}+\sqrt{-\frac{2}{3}}-\frac{1}{2}+\sqrt{-\frac{24}{9}} $$

Step-by-Step Solution

Answer

The sum or difference is \(-1 + i \left( \sqrt{\frac{2}{3}} + \frac{2 \sqrt{6}}{3} \right)\).

Step 1: Recognize Negative Radicals

Understand that when taking the square root of a negative number, we need to introduce the imaginary unit, denoted as \(i\), where \(i = \sqrt{-1}\).

Step 2: Simplify \( \sqrt{-\frac{2}{3}} \)

Writing \( \sqrt{-\frac{2}{3}} \) in terms of \(i\), we have: \( \sqrt{-\frac{2}{3}} = \sqrt{-1} \times \sqrt{\frac{2}{3}} = i \sqrt{\frac{2}{3}} \).

Step 3: Simplify \( \sqrt{-\frac{24}{9}} \)

First, note that \( \frac{24}{9} = \frac{8}{3} \). Therefore, \(\sqrt{-\frac{24}{9}} = \sqrt{-1} \times \sqrt{\frac{8}{3}} = i \sqrt{\frac{8}{3}} \), and further simplify \(\sqrt{\frac{8}{3}}\) into \(\frac{2\sqrt{6}}{3}\) as it can be factored to make simplification possible.

Step 4: Substitute Simplifications Back into Original Expression

Put the simplified forms back into the original expression: \(-\frac{1}{2} + i \sqrt{\frac{2}{3}} - \frac{1}{2} + i \cdot \frac{2 \sqrt{6}}{3}\).

Step 5: Combine Like Terms

Combine the real parts separately from the imaginary parts:- Real component: \(-\frac{1}{2} - \frac{1}{2} = -1\)- Imaginary component: \( i \sqrt{\frac{2}{3}} + i \cdot \frac{2 \sqrt{6}}{3} = i \left( \sqrt{\frac{2}{3}} + \frac{2 \sqrt{6}}{3} \right) \)

Key Concepts

Complex Numbers

Complex numbers are numbers that contain both a real part and an imaginary part. The standard form of a complex number is written as \( a + bi \), where \( a \) is the real part and \( bi \) is the imaginary part.
The imaginary unit \( i \) is crucial here as it allows us to work with the square roots of negative numbers.
In our exercise, the expression \(-\frac{1}{2} + i \sqrt{\frac{2}{3}} - \frac{1}{2} + i \cdot \frac{2 \sqrt{6}}{3}\) can be broken down into:

• Real part: \(-\frac{1}{2} - \frac{1}{2} = -1\)
• Imaginary part: \(i\left(\sqrt{\frac{2}{3}} + \frac{2 \sqrt{6}}{3}\right)\)

By performing operations separately on the real and imaginary parts, we see how complex numbers provide a comprehensive way to handle expressions in mathematics that involve roots of negative numbers.

Imaginary Unit

The imaginary unit, often denoted as \(i\), is a fundamental concept for dealing with the square roots of negative numbers.
It allows us to extend the real number system  into the complex plane. By definition, \(i\) is \(\sqrt{-1} \). Using \(i\), any negative square root can be reimagined in a manageable, and meaningful way. For instance:

• \(\sqrt{-1} = i\)
• \(\sqrt{-2} = i\sqrt{2}\)

In the given exercise, the terms \(\sqrt{-\frac{2}{3}}\) and \(\sqrt{-\frac{24}{9}}\) are rewritten as \(i\sqrt{\frac{2}{3}}\) and \(i\cdot \frac{2 \sqrt{6}}{3}\) respectively.
This simplification aids in easily combining or manipulating these terms when solving equations involving complex numbers.

Simplifying Radicals

Simplifying radicals is the process of making a radical expression as simple as possible. This includes factoring out perfect squares and rationalizing denominators.
Consider the example from the exercise, \(\sqrt{\frac{8}{3}}\), which simplifies to \(\frac{2\sqrt{6}}{3}\). This step involves:

• Breaking down \(8\) into \(4 \times 2\) (since \(4\) is a perfect square that simplifies to \(2\))
• The simplified form becomes \(\sqrt{4 \times 2/3} = \frac{2\sqrt{2}}{\sqrt{3}}\)
• Further rationalizing the denominator gives us \(\frac{2\sqrt{6}}{3}\)

Working with complex numbers often involves simplifying radicals. Hence, being comfortable with this method is vital for solving expressions that involve imaginary numbers.