Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Express the given complex number in the form \(R(\cos \theta+\) \(i \sin \theta)=R e^{i \theta}\) $$ -1-i $$

Short Answer

Expert verified
Question: Express the complex number \(-1-i\) in the form \(R(\cos \theta + i \sin \theta)=R e^{i \theta}\). Answer: \(\sqrt{2} e^{i\frac{5\pi}{4}}\)

Step by step solution

01

Find the magnitude (R)

To find the magnitude of \(-1-i\), calculate the square root of the sum of the squares of the real and imaginary parts. $$ R = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} $$
02

Find the argument θ

To find the angle (argument) θ, use the atan2 function, which gives values in the correct quadrant. The atan2 function takes the ratio of the imaginary part to the real part as input. $$ \theta = \operatorname{atan2}(-1,-1) = \frac{5\pi}{4} $$
03

Express the complex number in polar form

Now that we have the magnitude (R) and the argument (angle θ), we can express the complex number in polar form as: $$ R(\cos \theta + i \sin \theta)=\sqrt{2}(\cos\frac{5 \pi}{4} + i\sin\frac{5 \pi}{4}) $$
04

Express the complex number in exponential form

Finally, we can express the complex number in exponential form using Euler's formula, \(Re^{i\theta}\) : $$ \sqrt{2} e^{i\frac{5\pi}{4}} $$ Thus, the given complex number \(-1-i\) can be expressed as \(\sqrt{2} e^{i\frac{5\pi}{4}}\).

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 Vaia!

Key Concepts

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

Polar Form
Complex numbers can be represented in multiple ways, one of which is the polar form. The polar form expresses a complex number using its magnitude and direction, much like how you might describe a vector.
For a complex number in the form of polar coordinates, you will see it written as \( R(\cos \theta + i\sin \theta) \), where:
  • \( R \): The magnitude of the complex number.
  • \( \theta \): The angle (in radians) that the line makes with the positive real axis, also known as the argument.
In the given exercise, the complex number \(-1-i\) is expressed in polar form as \( \sqrt{2}(\cos \frac{5\pi}{4} + i\sin\frac{5\pi}{4}) \). This indicates a magnitude of \( \sqrt{2} \) and an angle of \( \frac{5\pi}{4} \). Recognizing this form can be helpful especially in multiplying complex numbers, as it's often easier to handle angles than handling separate real and imaginary components.
Magnitude and Argument
The magnitude and argument of a complex number provide both a sense of scale and direction on the complex plane.
The magnitude, \( R \), gives you the distance from the origin to the point in the complex plane, calculated using the Pythagorean theorem. For \(-1-i\), it was computed as \( \sqrt{2} \). This involves adding the squares of the real part and the imaginary part and then taking the square root of the result.
The argument, \( \theta \), represents the angle made with the positive x-axis, which shows the direction of the complex number. One effective method for finding the argument is by utilizing the \( \operatorname{atan2} \) function, which correctly considers the sign of both parts to place the angle in the correct quadrant. For \(-1-i\), it gives an angle of \( \frac{5\pi}{4} \).
  • **Magnitude (\( R \)):** The "length" of the vector representing the complex number.
  • **Argument (\( \theta \)):** Also known as the angle or phase, guides the "direction" of the vector.
Understanding both parameters is key to converting and working with complex numbers in different forms.
Exponential Form
The exponential form of complex numbers is a powerful way to represent them, leveraging Euler's formula, which links exponentials to trigonometric functions. The exponential form of a complex number \( z = R e^{i\theta} \) makes it straightforward to perform many mathematical operations, like multiplication or finding powers and roots of complex numbers.
Using Euler's formula \( e^{i\theta} = \cos \theta + i\sin \theta \), you can realize that this form efficiently marries the magnitude (\( R \)) of the complex number with its angle (\( \theta \)). For the complex number \(-1-i\), the exponential form is \( \sqrt{2}e^{i\frac{5\pi}{4}} \).
  • Exponential form encapsulates both components (magnitude and argument) in a compact expression.
  • It's highly useful in advanced operations and is often preferred in fields such as electrical engineering and physics.
Choosing exponential form can simplify calculations, and it offers a neat symmetry that blends algebra and geometry seamlessly.

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

determine a suitable form for Y(\(t\)) if the method of undetermined coefficients is to be used. Do not evaluate the constants. \(y^{\prime \prime \prime}-2 y^{\prime \prime}+y^{\prime}=t^{3}+2 e^{t}\)

Find the general solution of the given differential equation. $$ y^{\mathrm{iv}}+2 y^{\prime \prime}+y=0 $$

Find the general solution of the given differential equation. $$ y^{\prime \prime \prime}+5 y^{\prime \prime}+6 y^{\prime}+2 y=0 $$

Determine the general solution of the given differential equation. \(y^{\mathrm{iv}}-4 y^{\prime \prime}=t^{2}+e^{t}\)

We consider another way of arriving at the proper form of \(Y(t)\) for use in the method of undetermined coefficients. The procedure is based on the observation that exponential, polynomial, or sinusoidal terms (or sums and products of such terms) can be viewed as solutions of certain linear homogeneous differential equations with constant coefficients. It is convenient to use the symbol \(D\) for \(d / d t\). Then, for example, \(e^{-t}\) is a solution of \((D+1) y=0 ;\) the differential operator \(D+1\) is said to annihilate, or to be an annihilator of, \(e^{-t}\). Similarly, \(D^{2}+4\) is an annihilator of \(\sin 2 t\) or \(\cos 2 t,\) \((D-3)^{2}=D^{2}-6 D+9\) is an annihilator of \(e^{3 t}\) or \(t e^{3 t},\) and so forth. Show that linear differential operators with constant coefficients obey the commutative law, that is, $$ (D-a)(D-b) f=(D-b)(D-a) f $$ for any twice differentiable function \(f\) and any constants \(a\) and \(b .\) The result extends at once to any finite number of factors.

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free