Problem 83 - Intermediate Algebra

Question

Divide. Write all answers in the form \(a+b i.\) $$ \frac{-4-\sqrt{-4}}{2+\sqrt{-1}} $$

Step-by-Step Solution

Answer

The simplified form is \(-\frac{6}{5}\).

Step 1: Simplify the Expression

First, recognize that within the expression \(-4 - \sqrt{-4}\), the square root of a negative number involves imaginary numbers. We know \(\sqrt{-4} = \sqrt{4} \cdot \sqrt{-1} = 2i\). So, rewrite \(-4 - \sqrt{-4}\) as \(-4 - 2i\). The denominator \(2 + \sqrt{-1}\) can be simplified to \(2 + i\) since \(\sqrt{-1} = i\). The expression becomes: \[\frac{-4 - 2i}{2 + i}\].

Step 2: Multiply by the Conjugate

To simplify the division of complex numbers, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of \(2 + i\) is \(2 - i\). Now multiply: \(\frac{(-4 - 2i)(2 - i)}{(2 + i)(2 - i)}\).

Step 3: Expand the Numerator and Denominator

Expand using the distributive property (also known as FOIL for binomials):- **Numerator**: \((-4)(2) + (-4)(-i) + (-2i)(2) + (-2i)(-i) = -8 + 4i - 4i - 2i^2\). Simplify using \(i^2 = -1\): \(-8 + 0 - 2(-1) = -8 + 2 = -6\).- **Denominator**: \((2)(2) + (2)(-i) + (i)(2) + (i)(-i) = 4 - 2i + 2i - i^2\). Simplify: \(4 + 1 = 5\) (since \(-i^2 = 1\)). The denominator simplifies to 5.

Step 4: Write Final Result

The expression simplifies to \(\frac{-6}{5} + \frac{0}{5}i\). This simplifies to \(-\frac{6}{5} + 0i\). Thus, the final answer is \(-\frac{6}{5}\).

Key Concepts

Imaginary Numbers

Imaginary numbers might sound a bit strange, but they're quite simple once you understand their purpose. They arise from taking the square root of negative numbers.
Normally, the square root of a positive number is straightforward; for example, \( \sqrt{9} = 3 \) because 3 squared is 9.
But what about \( \sqrt{-4} \)? It doesn't fit into the realm of regular numbers. That's where imaginary numbers come in. We define \( \sqrt{-1} \) as \( i \).
So:

• \( \sqrt{-4} = \sqrt{4} \times \sqrt{-1} = 2i \)
• Imagine replacing the \( \sqrt{-1} \) with \( i \)
• Suddenly, it all makes more sense!

Imaginary numbers are special and open new ways to solve problems beyond traditional mathematics. They are used to form complex numbers, which combine real numbers with imaginary numbers.

Division of Complex Numbers

Dividing complex numbers might seem tricky initially, but there's a reliable method to handle it.
We use something called the complex conjugate to simplify the division.
Imagine dividing \( (-4 - 2i) \/ (2 + i) \). Instead of dividing directly, we work around it by multiplying both the numerator and the denominator by the complex conjugate of the denominator.
Here’s what to do:

• Find the complex conjugate of the denominator. For \( 2 + i \), the conjugate is \( 2 - i \).
• Multiply the entire fraction by \( (2 - i) \/ (2 - i) \). Essentially, you multiply by 1, so you don't change the fraction's value.

The denominator, when multiplied by its conjugate, simplifies to a real number:
\( (2 + i)(2 - i) = 4 - i^2 = 4 + 1 = 5 \).
This clears away the imaginary part in the denominator, making it easier to work with.

Complex Conjugate

The complex conjugate is a key tool in managing complex numbers, especially in division.
It's almost like a mirror image of a complex number, with the only change happening to the sign of the imaginary part.
Consider a complex number \( a + bi \). Its complex conjugate is \( a - bi \).
This reflects the imaginary part across the real axis, resulting in:

• For example, the complex conjugate of \( 2 + i \) is \( 2 - i \).

Using the complex conjugate in operations like division transforms the denominator into a real number, simplifying calculations.
For example, when you have \( (2 + i) \), multiplying by its conjugate \( (2 - i) \) gives a real number:

• \( (2 + i)(2 - i) = 4 - i^2 = 4 + 1 = 5 \)

This simplification is essential for dividing complex numbers as it allows us to separate the real and imaginary parts cleanly.