/*! 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 3 If \(z_{1}=5+3 \mathrm{j}\) and ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 3

If \(z_{1}=5+3 \mathrm{j}\) and \(z_{2}=3+2 \mathrm{j}\) find \(z_{1} z_{2}\) and \(\frac{z_{1}}{z_{2}}\).

Short Answer

Expert verified
\(z_1 z_2 = 9 + 19j\); \(z_1/z_2 \approx 1.6154 - 0.0769j\).

Step by step solution

01

Understand the Problem

We are given two complex numbers, \(z_1 = 5 + 3j\) and \(z_2 = 3 + 2j\). Our task is to multiply these two complex numbers and then divide the first one by the second.
02

Multiply Complex Numbers

To multiply two complex numbers, \(z_1 = a + bj\) and \(z_2 = c + dj\), use the formula: \((a + bj)(c + dj) = ac + (ad)j + (bc)j + (bd)j^2\). Given \(z_1 = 5 + 3j\) and \(z_2 = 3 + 2j\), compute:\[ z_1 z_2 = (5 \times 3) + (5 \times 2)j + (3 \times 3)j + (3 \times 2)j^2 \]Recall \(j^2 = -1\), so:\[ z_1 z_2 = 15 + 10j + 9j + 6(-1) \]\[ z_1 z_2 = 15 + 19j - 6 \]\[ z_1 z_2 = 9 + 19j \]
03

Divide Complex Numbers

To divide the complex number \(z_1\) by \(z_2\), we use the formula: \(\frac{a+bj}{c+dj} = \frac{(a+bj)(c-dj)}{c^2 + d^2}\). First, find \(c^2 + d^2\) where \(c = 3\) and \(d = 2\):\[ c^2 + d^2 = 3^2 + 2^2 = 9 + 4 = 13 \]Now, multiply \(z_1\) by the conjugate of \(z_2\):\(z_2^* = 3 - 2j\):\[ (5 + 3j)(3 - 2j) = 15 - 10j + 9j - 6j^2 \]\[ = 15 - 10j + 9j + 6 \]\[ = 21 - j \]Thus, the division is:\[ \frac{z_1}{z_2} = \frac{21-j}{13} = \frac{21}{13} - \frac{1}{13}j \]\[ z_1/z_2 = 1.6154 - 0.0769j \]
04

Conclusion

We found that the product of \(z_1\) and \(z_2\) is \(9 + 19j\), and the quotient \(\frac{z_1}{z_2}\) is approximately \(1.6154 - 0.0769j\).

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!

Key Concepts

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

Multiplying Complex Numbers
Complex numbers of the form \(a + bj\) can be multiplied using a straightforward method similar to distributing in algebra. When multiplying two complex numbers, \((a + bj)(c + dj)\), you'll apply the distributive property: multiply each part of one complex number by each part of the other, then combine like terms. Here's a step-by-step guide:
  • Multiply the real parts \((a \times c)\).
  • Multiply the outer terms \((a \times dj)\) and the inner terms \((bj \times c)\) to get the imaginary parts. Combine them to form a single imaginary number.
  • Multiply the imaginary parts \((bj \times dj)\), remembering that \(j^2 = -1\). Multiply the product by \(-1\) to convert it into a real number.
For example, considering \(z_1 = 5 + 3j\) and \(z_2 = 3 + 2j\), the result is \(9 + 19j\), as processed through the distributed steps above. This method ensures the calculations are organized and correct.
Dividing Complex Numbers
Dividing complex numbers involves working with their conjugates to eliminate the imaginary part in the denominator. The complex conjugate \(c - dj\) of \(c + dj\) plays a crucial role here.
  • To divide \(z_1 = a + bj\) by \(z_2 = c + dj\), multiply both numerator and denominator by the conjugate of \(z_2\), resulting in a rational expression.
  • The denominator becomes a real number because \((c+dj)(c-dj) = c^2 + d^2\).
  • Compute the new numerator \((a + bj)(c - dj)\) using the multiplication method explained earlier. Then, divide each part of the resultant complex number by the real number denominator.
Following these steps for \(z_1/z_2\) results in \(1.6154 - 0.0769j\). Mastering this technique gives you clarity in dealing with complex number division.
Complex Conjugates
The complex conjugate of a complex number \(z = a + bj\) is given by \(z^* = a - bj\). Conjugates are useful not just for division, but also in simplifying the process of working out complex operations such as finding the magnitude or simplifying imaginary scenarios.
  • Computing \(zz^*\) results in a real number, which is the sum of the squares of the real and imaginary parts \(a^2 + b^2\).
  • The use of the conjugate helps simplify the division of complex numbers by eliminating the imaginary component from the denominator.
  • When finding the length or magnitude of a complex number represented as a vector, conjugates come in handy.
In this exercise, the conjugate \(3 - 2j\) was used to rationalize the denominator in the division step, ultimately aiding in achieving the final result.

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

Solve the quadratic equation $$ 5 x^{2}-11 x+13=0 $$

Express \(\cos \omega t\) in terms of exponential functions.

Sketch an Argand diagram and upon it mark the poles of the rational function $$ G(s)=\frac{3(s-3)}{(s+1)(s+1+3 \mathrm{j})(s+1-3 \mathrm{j})} $$

Use De Moivre's theorem to simplify (a) \((\cos 3 \theta+\mathrm{j} \sin 3 \theta)(\cos 4 \theta+\mathrm{j} \sin 4 \theta)\) (b) \(\frac{\cos 8 \theta+\mathrm{j} \sin 8 \theta}{\cos 2 \theta-\mathrm{j} \sin 2 \theta}\)

A resistor of resistance, \(R\), and an inductor of inductance, \(L\), are connected in series. If \(R=9\) and \(L=2 \mathrm{j}\) find (a) the total impedance, \(Z\) (b) the admittance.
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Dynamics

Read Explanation

Space Physics

Read Explanation

Medical Physics

Read Explanation

Physics of Motion

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.