Problem 49 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The result is \(-2\).

Step 1: Simplify the square roots of negative numbers

First, express the square roots of negative numbers as complex numbers. We know that \( \sqrt{-8} = \sqrt{8} \, \cdot \, i = 2 \sqrt{2} \, i\) and \( \sqrt{-2} = \sqrt{2} \, i \). Substitute these into the expression, yielding \( \frac{2 + 2 \sqrt{2} \, i}{1 + \sqrt{2} \, i} \).

Step 2: Rationalize the denominator

To simplify, multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \( 1 + \sqrt{2} \, i \) is \( 1 - \sqrt{2} \, i \). Multiply both the top and bottom by the conjugate: \[ \frac{(2 + 2 \sqrt{2} \, i)(1 - \sqrt{2} \, i)}{(1 + \sqrt{2} \, i)(1 - \sqrt{2} \, i)} \].

Step 3: Simplify the denominator

The denominator becomes \((1)^2 - (\sqrt{2} \, i)^2 = 1 - (-2) = 1 + 2 = 3\).

Step 4: Simplify the numerator using the distributive property

Expand the numerator: \( (2 + 2 \sqrt{2} \, i)(1 - \sqrt{2} \, i) = 2 - 2\sqrt{2} \, i + 2\sqrt{2} \, i + 4 \cdot 2\cdot(-1) = 2 - 0 + 8 = 10\).

Step 5: Combine real and imaginary parts

Combine the terms from the expansion to find the real and imaginary parts. The real part is \( -6 \), and there is no imaginary part as the imaginary components cancelled out. So the numerator simplifies to just \(-6 \).

Step 6: Write the expression in standard form

Combine and simplify the fraction: \( \frac{-6}{3} = -2 \). Therefore, the expression in the form \( a + bi \) is \( -2 + 0i \), which is simply \( -2 \).

Key Concepts

Simplify Expressions

When dealing with complex numbers, the primary step often involves simplifying expressions. This means rewriting the expression in a form that's easier to work with or understand. In our exercise, the expression is \( \frac{2 + \sqrt{-8}}{1 + \sqrt{-2}} \). We start by recognizing that the square roots of negative numbers necessitate the use of imaginary units. The imaginary unit, denoted as \( i \), is defined such that \( i^2 = -1 \). To simplify, convert \( \sqrt{-8} \) and \( \sqrt{-2} \) to complex numbers as they involve square roots of negative numbers:

• \( \sqrt{-8} = \sqrt{8} \cdot i = 2 \sqrt{2} \cdot i \)
• \( \sqrt{-2} = \sqrt{2} \cdot i \)

By substituting these back, our expression morphs into \( \frac{2 + 2 \sqrt{2} \, i}{1 + \sqrt{2} \, i} \). This is a crucial step in dealing with complex numbers as it lays the groundwork for further simplification through arithmetic operations.

Complex Conjugate

The concept of a complex conjugate is essential when working with complex numbers, especially for simplifying expressions involving division. Given a complex number \( a + bi \), its complex conjugate is \( a - bi \). It 'flips' the sign of the imaginary part, and using it is instrumental in removing complex numbers from denominators. In our problem, to simplify \( \frac{2 + 2 \sqrt{2} \, i}{1 + \sqrt{2} \, i} \), we employ the complex conjugate of the denominator, \( 1 + \sqrt{2} \, i \). Thus, the conjugate is \( 1 - \sqrt{2} \, i \). When we multiply both the numerator and the denominator by this conjugate, it helps transform the denominator into a real number. The multiplication process leverages the property that discarding complex terms is possible through \( (a+bi)(a-bi) = a^2 - (bi)^2 \). This results in a new denominator \( 1 + 2 = 3 \), free from any imaginary parts.

Rationalizing the Denominator

Rationalizing the denominator is a method used to eliminate the imaginary part from the denominator of a fraction with complex numbers. This process involves multiplying both the numerator and the denominator by the conjugate of the denominator, turning the denominator into a real number.By multiplying \( (2 + 2\sqrt{2} \, i) \) by \( (1 - \sqrt{2} \, i) \), we simplify the fraction. The denominator becomes a real number, \( 3 \), as previously found. Focusing on the numerator for a moment, we expand as follows:

• \( 2 \cdot 1 = 2 \)
• \( 2 \cdot (-\sqrt{2} \, i) = -2\sqrt{2} \, i \)
• \( 2\sqrt{2} \, i \cdot 1 = 2\sqrt{2} \, i \)
• \( 2\sqrt{2} \, i \cdot (-\sqrt{2} \, i) = -4 \)

Once you combine these results, the imaginary terms cancel, and you are left with a real part, \( 2 - 4 = -6 \). Thus, the expression simplifies to \( \frac{-6}{3} = -2 \). Through rationalization, we neatly conclude our problem without any complex numbers in the denominator.