Problem 76 - Precalculus Student Solutions Manual 5th

Question

Find each quotient. Write the answer in standard form \(a+b i .\) $$\frac{-6}{i}$$

Step-by-Step Solution

Answer

The quotient is \(0 + 6i\).

Step 1: Express the denominator in terms of its conjugate

To simplify the given complex division, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of the denominator, which is \(i\), is \(-i\). So, we have: \frac{-6}{i} \times \frac{-i}{-i}\.

Step 2: Perform the multiplication

Multiply the numerator and the denominator by \(-i\). This gives: \frac{-6 \times (-i)}{i \times (-i)} = \frac{6i}{-i^2}\.

Step 3: Simplify the denominator

Recall that \(i^2 = -1\). Therefore, \(-i^2 = -(-1) = 1\). So, the expression simplifies to: \frac{6i}{1} = 6i\.

Step 4: Write the result in standard form

The standard form of a complex number is \(a + bi\). Here, the expression \6i\ can be written as \0 + 6i\. Therefore, the quotient in standard form is: \(0 + 6i\).

Key Concepts

multiplying by conjugate

When working with complex numbers, simplifying division often involves multiplying by the conjugate of the denominator. The conjugate of a complex number is created by changing the sign of the imaginary part. For instance, if we have a denominator of \ \( z = a + bi \ \), its conjugate is \ \( \bar{z} = a - bi \ \).

The main reason for using the conjugate is to eliminate the imaginary part of the denominator. This process converts a complex number into a real number, which simplifies division:

• Multiply both the numerator and the denominator by the conjugate of the denominator.
• The imaginary units should cancel out, leaving a real number in the denominator.

In this exercise, the conjugate of \ \( i \ \) is \ \( -i \ \). So, we multiplied by \ \( \frac{-i}{-i} \). Performing multiplication involves distributing each part:
\ \(\frac{-6}{i} \times \frac{-i}{-i} = \frac{-6 \times -i}{i \times -i} = \frac{6i}{-i^2} \).

standard form of complex number

A complex number is usually written in the form \ \(a + bi \ \), known as the standard form. In this form, \ \( a \ \) and \ \( b \ \) are real numbers, and \ \( i \ \) is the imaginary unit. The standard form clearly represents both the real and imaginary parts:

• \ \( a \ \) is the real part.
• \ \( b i \ \) is the imaginary part.

For instance, if we encounter a quotient like \ \( \frac{-6}{i} \ \) and after working through step-by-step operations, we find \ \( 6i \ \), we should express it in standard form.
In the final step of the given solution, we rewrite \ \( 6i \ \) as \ \( 0 + 6i \ \):

• The real part \ \( a \ \) is \ \( 0 \ \).
• The imaginary part \ \( bi \ \) is \ \( 6i \ \).

This allows us to write the complex number in a more recognizable and standardized form.

imaginary unit

The imaginary unit, denoted as \ \(i \ \)​, is defined as the square root of \ \(-1 \ \): \ \( i = \sqrt{-1} \ \). It's a fundamental concept in complex numbers because it makes it possible to work with the square roots of negative numbers.

Some properties of the imaginary unit include:

• \ \(i^2 = -1 \ \).
• \ \(i^3 = -i \ \).
• \ \(i^4 = 1 \ \).

Remembering these properties is crucial when handling complex number operations, such as multiplication and division.

In the provided exercise, knowing the property \ \( i^2 = -1 \ \) helps significantly:
When we multiply the numerator and the denominator by \ \( -i \ \), we end up with \ \( \frac{6i}{-i^2} \ \). Since \ \( i^2 = -1 \ \), we substitute and get \ \( -i^2 = -(-1) = 1 \ \). This turns the fraction into \ \(6i/1 \ \), simplifying our calculations and allowing us to write the final answer in standard form.