/*! 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 1 Suppose \(a\) and \(b\) are real... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 1

Suppose \(a\) and \(b\) are real numbers, not both 0. Find real numbers \(c\) and \(d\) such that \\[ 1 /(a+b i)=c+d i \\]

Short Answer

Expert verified
The short answer to the given problem can be expressed as: \(c = \frac{a}{a^2 + b^2}\) and \(d = -\frac{b}{a^2 + b^2}\)

Step by step solution

01

Write the given expression in the form of a complex number

The given expression is: \[ \frac{1}{a+bi} = c + di \]
02

Multiply the numerator and the denominator by the conjugate of the denominator

We will multiply both the numerator and the denominator by the complex conjugate of the denominator, which is \((a - bi)\). This is done to eliminate the imaginary part in the denominator: \[ \frac{1}{a+bi} \times \frac{a-bi}{a-bi} = \frac{a-bi}{(a+bi)(a-bi)} \]
03

Simplify the expression on the right-hand side

Now we need to multiply the denominators and simplify the expression: \[ \frac{a-bi}{a^2 - (bi)^2} = \frac{a-bi}{a^2 + b^2} \]
04

Separate the real and imaginary parts

Separate the real and imaginary parts of the complex number: \[ \frac{a-bi}{a^2 + b^2} = \frac{a}{a^2 + b^2} + \frac{-b}{a^2 + b^2}i \] Finally, we can write the values of \(c\) and \(d\):
05

Write the values of c and d

Comparing with the original expression \((c + di)\), we can deduce the values of \(c\) and \(d\): \[ c = \frac{a}{a^2 + b^2},\quad d = - \frac{b}{a^2 + b^2} \] Now we have found the real numbers \(c\) and \(d\) for the given complex numbers \(a\) and \(b\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Prove that \(-(-v)=v\) for every \(v \in V\)

Does the operation of addition on the subspaces of \(V\) have an additive identity? Which subspaces have additive inverses?

Prove that the intersection of any collection of subspaces of \(V\) is a subspace of \(V\)

Suppose that \(U\) is a subspace of \(V\). What is \(U+U ?\)

Prove or give a counterexample: if \(U_{1}, U_{2}, W\) are subspaces of \(V\) such that \\[ U_{1}+W=U_{2}+W \\] \(\operatorname{then} U_{1}=U_{2}\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.