Problem 53 - Precalculus Mathematics for Calculus

Question

Evaluate the radical expression and express the result in the form \(a+b i\) $$\frac{2+\sqrt{-8}}{1+\sqrt{-2}}$$

Step-by-Step Solution

Answer

The expression simplifies to \( 2 + 0i \).

Step 1: Simplify the square roots of negative numbers

Rewrite \( \sqrt{-8} \) as \( \sqrt{8} i \) and \( \sqrt{-2} \) as \( \sqrt{2} i \). So the original expression becomes: \[ \frac{2 + \sqrt{8} i}{1 + \sqrt{2} i}. \]

Step 2: Simplify with rational numbers

Since \( \sqrt{8} \) can be simplified to \( 2\sqrt{2} \), rewrite the expression as: \[ \frac{2 + 2\sqrt{2} i}{1 + \sqrt{2} i}. \]

Step 3: Multiply by the conjugate of the denominator

To remove the imaginary unit \( i \) from the denominator, multiply the numerator and denominator by the conjugate of the denominator, \( 1 - \sqrt{2} i \). The expression becomes: \[ \frac{(2 + 2\sqrt{2} i)(1 - \sqrt{2} i)}{(1 + \sqrt{2} i)(1 - \sqrt{2} i)}. \]

Step 4: Calculate the denominator

Use the formula \((a + bi)(a - bi) = a^2 + b^2\) to calculate the denominator: \[ (1)^2 - (\sqrt{2} i)^2 = 1 - 2(-1) = 3. \]

Step 5: Expand the numerator

Calculate the product \((2 + 2\sqrt{2} i)(1 - \sqrt{2} i)\): \(2 \times 1 = 2\), \(2 \times -\sqrt{2} i = -2\sqrt{2} i\), \(2\sqrt{2} i \times 1 = 2\sqrt{2} i\), and \(2\sqrt{2} i \times -\sqrt{2} i = -4i^2 = 4\). This results in: \[ 2 - 2\sqrt{2} i + 2\sqrt{2} i + 4 = 6. \]

Step 6: Combine real and imaginary parts

Notice that the imaginary parts \(-2\sqrt{2} i\) and \(+2\sqrt{2} i\) cancel each other out, leaving just the real part. The numerator becomes 6. The expression is now \( \frac{6}{3} \).

Step 7: Simplify the fraction

The fraction simplifies to \( \frac{6}{3} = 2 \). Since there is no imaginary part, the expression in the form \( a + bi \) is \( 2 + 0i \).

Key Concepts

Imaginary Unit

Complex numbers are intriguing entities composed of a real part and an imaginary part. The heart of these numbers is the imaginary unit, denoted as \( i \). By definition, \( i \) is equal to the square root of -1, written mathematically as \( i = \sqrt{-1} \).
Imaginary numbers are extensions of the real number system, allowing us to take the square root of negative numbers—a concept not possible within the real numbers alone. Here are a few key points about the imaginary unit:

• \( i^2 = -1 \). This property is crucial and often used in operations involving complex numbers.
• Every imaginary number can be expressed in the form of a real number multiplied by \( i \), such as \( 3i \) or \( -2i \).
• The powers of \( i \) repeat in a cycle: \( i, -1, -i, \) and \( 1 \).

These properties allow us to simplify expressions like \( \sqrt{-8} \) into \( \sqrt{8} i \), turning seemingly complex radical expressions into manageable forms.

Radical Expressions

Radical expressions involve roots, such as square roots, and in the context of complex numbers, they can involve negative numbers under the radical. It's here that imaginary numbers become incredibly useful.
First, let's understand how to deal with square roots of negative numbers:

• When taking the square root of a negative number, such as \( \sqrt{-8} \), you break it down into \( \sqrt{8} \times \sqrt{-1} \).
• By recognizing \( \sqrt{-1} \) as \( i \), you can rewrite \( \sqrt{-8} \) as \( \sqrt{8} i \), simplifying further if possible.
• In our exercise, \( \sqrt{8} \) simplifies to \( 2\sqrt{2} \), thus \( \sqrt{-8} \) becomes \( 2\sqrt{2} i \).

This method allows us to simplify and express complex roots in a clearer form, leveraging the properties of imaginary and real numbers to find an understandable solution.

Complex Conjugate

The complex conjugate is a key concept in dealing with complex numbers, especially when simplifying expressions. Let's explore what it is and how it is used.
A complex conjugate of a complex number is formed by changing the sign of the imaginary part. If you have a complex number \( a + bi \), its complex conjugate is \( a - bi \).

• This transformation is particularly useful for eliminating the imaginary unit in the denominator of a fraction. By multiplying both the numerator and denominator by the conjugate of the denominator, the denominator becomes a real number.
• For example, with the denominator \( 1 + \sqrt{2} i \), its conjugate is \( 1 - \sqrt{2} i \).
• Using the identity \((a+bi)(a-bi)=a^2+b^2\), the product in the denominator results in a real number free of any \( i \)'s.

This technique simplifies complex expressions, transforming them into a form where real and imaginary components are easily distinguishable. Understanding and applying the complex conjugate is vital in performing operations with complex numbers efficiently.