Detailed Explanations:

Calculating the Square Root of Negative Numbers

In this example, we will calculate the square root of -4. Understanding the square root of negative numbers is a fundamental concept in complex numbers and imaginary numbers.

Step 1: Break Down the Number

Recognize that -4 can be written as -1 times 4:

\(-4 = -1 \times 4\)

Step 2: Apply the Property of Square Roots

Apply the property of square roots that says the square root of a product is the product of the square roots:

\(\sqrt{-4} = \sqrt{-1 \times 4} = \sqrt{-1} \times \sqrt{4}\)

Step 3: Evaluate Each Square Root

Evaluate each square root separately. The square root of -1 is \(i\), and the square root of 4 is 2:

\(\sqrt{-1} \times \sqrt{4} = i \times 2 = 2i\)

Conclusion

So, the square root of -4 is \(2i\). This is a basic example of how to calculate the square root of a negative number.

Efficient Calculation

The calculation can be written in one efficient line as follows:

\(\sqrt{-4} = \sqrt{-1 \times 4} = \sqrt{-1} \times \sqrt{4} = i \times 2 = 2i\)

Calculating the Product of Square Roots of Negative Numbers

In this example, we will calculate the product of the square roots of -4 and -9. Understanding the square roots of negative numbers is a fundamental concept in complex numbers and imaginary numbers.

Step 1: Break Down the Numbers

Recognize that -4 can be written as -1 times 4 and -9 can be written as -1 times 9:

\(-4 = -1 \times 4, -9 = -1 \times 9\)

Step 2: Apply the Property of Square Roots

Apply the property of square roots that says the square root of a product is the product of the square roots:

\(\sqrt{-4} \times \sqrt{-9} = \sqrt{-1 \times 4} \times \sqrt{-1 \times 9} = \sqrt{-1} \times \sqrt{4} \times \sqrt{-1} \times \sqrt{9}\)

Step 3: Evaluate Each Square Root

Evaluate each square root separately. The square root of -1 is \(i\), the square root of 4 is 2, and the square root of 9 is 3:

\(\sqrt{-1} \times \sqrt{4} \times \sqrt{-1} \times \sqrt{9} = i \times 2 \times i \times 3\)

Step 4: Evaluate \(i \times i\)

Remember that \(i \times i = i^2\), which is equal to -1:

\(i \times 2 \times i \times 3 = i^2 \times 2 \times 3 = -1 \times 2 \times 3 = -6\)

Conclusion

So, the product of the square roots of -4 and -9 is \(-6\). This is a basic example of how to calculate the product of the square roots of negative numbers.

Efficient Calculation

The calculation can be written in one efficient line as follows:

\(\sqrt{-4} \times \sqrt{-9} = \sqrt{-1 \times 4} \times \sqrt{-1 \times 9} = \sqrt{-1} \times \sqrt{4} \times \sqrt{-1} \times \sqrt{9} = i \times 2 \times i \times 3 = i^2 \times 2 \times 3 = -1 \times 2 \times 3 = -6\)

Calculating the Sum of Square Roots of Negative Numbers

In this example, we will calculate the sum of the square roots of -4 and -16. Understanding the square roots of negative numbers is a fundamental concept in complex numbers and imaginary numbers.

Step 1: Break Down the Numbers

Recognize that -4 can be written as -1 times 4 and -16 can be written as -1 times 16:

\(-4 = -1 \times 4, -16 = -1 \times 16\)

Step 2: Apply the Property of Square Roots

Apply the property of square roots that says the square root of a product is the product of the square roots:

\(\sqrt{-4} + \sqrt{-16} = \sqrt{-1 \times 4} + \sqrt{-1 \times 16} = \sqrt{-1} \times \sqrt{4} + \sqrt{-1} \times \sqrt{16}\)

Step 3: Evaluate Each Square Root

Evaluate each square root separately. The square root of -1 is \(i\), the square root of 4 is 2, and the square root of 16 is 4:

\(\sqrt{-1} \times \sqrt{4} + \sqrt{-1} \times \sqrt{16} = i \times 2 + i \times 4\)

Step 4: Factor Out the Common Factor

Factor out the common factor \(i\):

\(i \times 2 + i \times 4 = i \times (2 + 4) = i \times 6 = 6i\)

Conclusion

So, the sum of the square roots of -4 and -16 is \(6i\). This is a basic example of how to calculate the sum of the square roots of negative numbers.

Efficient Calculation

The calculation can be written in one efficient line as follows:

\(\sqrt{-4} + \sqrt{-16} = \sqrt{-1 \times 4} + \sqrt{-1 \times 16} = \sqrt{-1} \times \sqrt{4} + \sqrt{-1} \times \sqrt{16} = i \times 2 + i \times 4 = i \times (2 + 4) = i \times 6 = 6i\)

Complex Number Division Examples

Example 1: Dividing by \( i \): \( \frac{1}{i} \)

When dividing by \( i \), we want to eliminate the imaginary unit from the denominator. This can be achieved by multiplying the numerator and denominator by \( i \):

Here, we multiplied the numerator and denominator by \( i \) to get rid of \( i \) in the denominator. This is a common technique used when dividing by complex numbers.

Next, we simplify the denominator. Remember that \( i^2 = -1 \), so we can replace \( i^2 \) with \( -1 \) in the denominator:

Finally, dividing \( i \) by \( -1 \) gives us the result in standard form:

Dividing Complex Numbers:

Let’s break down the process of dividing a complex number \(a+bi\) by another complex number \(c+di\):

Step 1: Multiply by the conjugate of the denominator

The goal here is to eliminate the imaginary part from the denominator. We can do this by multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of \(c+di\) is \(c-di\).

So, we have:

\[ \frac{a+bi}{c+di} \times \frac{c-di}{c-di} \]

Step 2: Simplify the numerator

We distribute the multiplication across the terms in the numerator:

\[ (a+bi) \times (c-di) = ac – adi + bci – bdi^2 \]

Remember that \(i^2 = -1\), so we can simplify this to:

\[ ac – adi + bci + bd = (ac+bd) + (bc-ad)i \]

Step 3: Simplify the denominator

We distribute the multiplication across the terms in the denominator:

\[ (c+di) \times (c-di) = c^2 – cdi + cdi – d^2i^2 \]

The \(cdi\) and \(-cdi\) terms cancel out, and \(i^2 = -1\), so we can simplify this to:

\[ c^2 + d^2 \]

Step 4: Divide both the real and imaginary parts of the numerator by the denominator

We divide the real part \(ac+bd\) and the imaginary part \(bc-ad\) by the denominator \(c^2 + d^2\):

\[ \frac{(ac+bd) + (bc-ad)i}{c^2 + d^2} = \frac{ac+bd}{c^2 + d^2} + \frac{bc-ad}{c^2 + d^2}i \]

This is the final result in standard form. Each step of the process is explained in detail, and the reason for each step is provided. This should help make the process of dividing by a complex number clearer.