/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Compute each quotient. $$\frac... [FREE SOLUTION] | 91Ó°ÊÓ

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Compute each quotient. $$\frac{1}{i}$$

Short Answer

Expert verified
The quotient is \( -i \).

Step by step solution

01

Understand the Problem

The problem asks us to find the quotient of \( \frac{1}{i} \), where \( i \) is the imaginary unit with the property that \( i^2 = -1 \). To simplify complex fractions involving \( i \), we often multiply by the conjugate to eliminate \( i \) in the denominator.
02

Multiply by the Conjugate

The conjugate of \( i \) is \( -i \). Multiply the numerator and denominator of \( \frac{1}{i} \) by \( -i \):\[ \frac{1}{i} \times \frac{-i}{-i} = \frac{-i}{-i^2} \]
03

Simplify the Expression

Simplify the denominator \( -i^2 \). Since \( i^2 = -1 \), we have \( -i^2 = -(-1) = 1 \). Therefore, the expression becomes:\[ \frac{-i}{1} = -i \]
04

Final Result

The quotient of \( \frac{1}{i} \) simplifies to \( -i \). Thus, whenever you divide by \( i \), you multiply the number by \( -i \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Fractions
A complex fraction is a type of fraction where the numerator, the denominator, or both contain a fraction themselves. For example, when you see something like \( \frac{1}{i} \), where \( i \) is an imaginary number, it qualifies as a complex fraction because the denominator includes something more than just a simple integer or fractional number.

To simplify complex fractions, the usual goal is to rid the fraction of any imaginary unit in the denominator. We perform this by multiplying both the numerator and the denominator by a value that will cancel out the imaginary component. In many cases, this involves using the conjugate of the denominator, or in simpler terms, a value that turns the imaginary part into something real when both are multiplied together.
  • Think of complex fractions as a way to practice making the math more approachable by turning dense, complicated-looking expressions into more manageable ones.
Multiplying by the Conjugate
When working with complex numbers, you often encounter the concept of the conjugate. The conjugate of a complex number is formed by changing the sign in front of the imaginary part. So, if you have \( i \), its conjugate is \( -i \). Multiplying by the conjugate is particularly useful when you need to remove the imaginary number from the denominator of a fraction.

Let's take a specific example: you've got \( \frac{1}{i} \), and to simplify it, you need to multiply by its conjugate, \( -i \). This turns your expression into:\[\frac{1}{i} \times \frac{-i}{-i} = \frac{-i}{-i^2}\]The reason we multiply fractions in this way is to make the denominator a real number by using the property \( i^2 = -1 \). Thus, this multiplication process turns the imaginary denominator into a real, simplifying the entire fraction in one smooth, clear step.
Imaginary Unit
The imaginary unit, often represented by \( i \), is a fundamental concept in complex numbers. The property that makes \( i \) unique is that \( i^2 = -1 \). This means using \( i \) allows us to handle and express numbers that are not real.
  • Any expression like \( i \cdot i = i^2 \) naturally results in \(-1\).
  • This property is crucial for operations involving complex fractions where denominators often need to be turned into real numbers.
By multiplying or dividing by \( i \), you frequently rely on this property to simplify expressions. For example, simplifying \( \frac{1}{i} \) involves multiplying by \(-i\), leading a real result \(-i\) since \(-i \times i = -i^2 = 1\). Understanding this allows for clearer manipulations of complex expressions, making it a cornerstone of algebra involving imaginary numbers.

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Most popular questions from this chapter

Determine the partial fraction decomposition for each of the given rational expressions. Hint: In Exercises \(17,18,\) and \(26,\) use the rational roots theorem to factor the denominator. $$\frac{2 x}{32 x^{2}-12 x+1}$$

You are given an improper rational expression. First, use long division to rewrite the expression in the form (polynomial) \(+\) (proper rational expression) Next, obtain the partial fraction decomposition for the proper rational expression. Finally, rewrite the given improper rational expression in the form (polynomial) \(+\) (partial fractions) $$\frac{x^{6}+3 x^{5}+9 x^{3}+26 x^{2}+3 x+8}{x^{3}+8}$$

Express the polynomial \(f(x)\) in the form \(a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\). (a) Find a quadratic function that has a maximum value of 2 and that has -2 and 4 as zeros. (b) Use a graphing utility to check that your answer in part (a) appears to be correct.

First graph the two functions. Then use the method of successive approximations to locate, between successive thousandths, the \(x\)-coordinate of the point where the graphs intersect.Use a graphing utility to draw the graphs as well as to check your final answer. $$y=e^{-x} ; y=\ln x \quad$$ Remark: The method of successive approximations is not restricted to polynomial functions.

Each polynomial equation has exactly one negative root. (a) Use a graphing utility to determine successive integer bounds for the root. (b) Use the method of successive approximations to locate the root between successive thousandths. (Make use of the graphing utility to generate the required tables. ) $$\frac{x^{5}}{10,000}-\frac{x^{3}}{50}+\frac{x}{1250}+\frac{1}{2000}=0$$

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