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 given complex number \(1+i\) in polar form. Answer: The polar form of the complex number \(1+i\) is \(\sqrt{2}e^{i\frac{\pi}{4}}\).

Step by step solution

01

Find Modulus (R) of the complex number

The modulus of a complex number of the form \(a+bi\) is given by \(R = \sqrt{a^2+b^2}\). In our case, \(a = 1\) and \(b = 1\). So, $$ R = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}. $$
02

Find argument (angle \(\theta\)) of the complex number

To find the argument of a complex number, we use the formula \(\theta = \text{atan2}(b,a)\), where \(\text{atan2}(y,x)\) is the two-argument arctangent function that returns the angle in radians between the positive x-axis and the point \((x,y)\). For our complex number, \(a = 1\) and \(b = 1\), so $$ \theta = \text{atan2}(1,1) = \frac{\pi}{4}. $$
03

Write the complex number in polar form

Now that we have found the modulus (R) and the argument (angle \(\theta\)), we can write the complex number in the polar form: $$ 1+i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = \sqrt{2}e^{i \frac{\pi}{4}}. $$

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.

Modulus of a Complex Number
Understanding the modulus of a complex number is essential when it comes to grasping the basics of complex analysis. It can be thought of as the distance of the complex number from the origin in the complex plane. To calculate the modulus of a complex number given by the expression \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, we use the formula \[ R = \sqrt{a^2 + b^2} \]. This formula is actually derived from the Pythagorean theorem, as the complex number in question can be represented as a point \( (a, b) \) in the plane.

For instance, with the complex number \(1 + i\), the modulus is computed by squaring both the real part (1) and imaginary part (1), adding these together, and then taking the square root. Thus, \[ R = \sqrt{1^2 + 1^2} = \sqrt{2}\]. The result, \(\sqrt{2}\), represents the length of the vector pointing from the origin to the point \(1, 1\) on the complex plane.
Argument of a Complex Number
The argument of a complex number is the measure of the angle it makes with the positive real-axis, counter-clockwise, often denoted as \(\theta\) or arg. For a complex number \(a + bi\), the argument can be found by using trigonometry, and it is typically determined using the arctangent function. The preferred function in computing this is \(\text{atan2}(b, a)\), which is the two-argument version of arctangent that takes into account the signs of both \(a\) and \(b\) to provide the correct quadrant for the angle.

Applying this to our example \(1 + i\), we have \(a = b = 1\). Thus, \[ \theta = \text{atan2}(1, 1) = \frac{\pi}{4} \]. This tells us that the line from the origin to the point \(1, 1\) in the complex plane forms a \(45^\circ\) angle with the positive real-axis. Whenever you are calculating the argument, it's important to figure out which quadrant the complex number is in to determine the correct direction of the angle.
Polar and Exponential Form of Complex Numbers
Converting a complex number to its polar form not only simplifies multiplication and division but also provides deeper insight into the properties of the number. The polar form is expressed as \(R(\cos \theta + i \sin \theta)\), where \(R\) is the modulus and \(\theta\) is the argument of the complex number. This form can also be expressed using Euler's formula as \(Re^{i\theta}\), where \(e\) is the base of the natural logarithm. Euler's formula connects the seemingly unrelated exponential functions to trigonometric functions.

For the given example \(1 + i\), after calculating the modulus and the argument, we can represent it in its polar form as \[ \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) \] or, equivalently, in the exponential form as \[ \sqrt{2}e^{i \frac{\pi}{4}} \]. The exponential form is particularly useful when dealing with complex number powers and roots, offering a straightforward approach to these calculations by manipulating the modulus and argument.

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

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. Consider the problem of finding the form of the particular solution \(Y(t)\) of $$ (D-2)^{3}(D+1) Y=3 e^{2 t}-t e^{-t} $$ where the left side of the equation is written in a form corresponding to the factorization of the characteristic polynomial. (a) Show that \(D-2\) and \((D+1)^{2},\) respectively, are annihilators of the terms on the right side of Eq. and that the combined operator \((D-2)(D+1)^{2}\) annihilates both terms on the right side of Eq. (i) simultaneously. (b) Apply the operator \((D-2)(D+1)^{2}\) to Eq. (i) and use the result of Problem 20 to obtain $$ (D-2)^{4}(D+1)^{3} Y=0 $$ Thus \(Y\) is a solution of the homogeneous equation (ii). By solving Eq. (ii), show that $$ \begin{aligned} Y(t)=c_{1} e^{2 t} &+c_{2} t e^{2 t}+c_{3} t^{2} e^{2 t}+c_{4} t^{3} e^{2 t} \\ &+c_{5} e^{-t}+c_{6} t e^{-t}+c_{7} t^{2} e^{-t} \end{aligned} $$ where \(c_{1}, \ldots, c_{7}\) are constants, as yet undetermined. (c) Observe that \(e^{2 t}, t e^{2 t}, t^{2} e^{2 t},\) and \(e^{-t}\) are solutions of the homogeneous equation corresponding to Eq. (i); hence these terms are not useful in solving the nonhomogeneous equation. Therefore, choose \(c_{1}, c_{2}, c_{3},\) and \(c_{5}\) to be zero in Eq. (iii), so that $$ Y(t)=c_{4} t^{3} e^{2 t}+c_{6} t e^{-t}+c_{7} t^{2} e^{-t} $$ This is the form of the particular solution \(Y\) of Eq. (i). The values of the coefficients \(c_{4}, c_{6},\) and \(c_{7}\) can be found by substituting from Eq. (iv) in the differential equation (i).

Determine the general solution of the given differential equation. \(y^{\mathrm{iv}}+2 y^{\prime \prime}+y=3+\cos 2 t\)

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.

Use the method of variation of parameters to determine the general solution of the given differential equation. $$ y^{\prime \prime \prime}-y^{\prime \prime}+y^{\prime}-y=e^{-t} \sin t $$

Use the method of reduction of order (Problem 26) to solve the given differential equation. $$ t^{2}(t+3) y^{\prime \prime \prime}-3 t(t+2) y^{\prime \prime}+6(1+t) y^{\prime}-6 y=0, \quad t>0 ; \quad y_{1}(t)=t^{2}, \quad y_{2}(t)=t^{3} $$

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