Problem 10
Question
If \(w=z+\frac{1}{z}\) and \(z=e^{i t}, w=e^{i t}+e^{-i t}=2 \cos t .\) Therefore \(u=2 \cos t\) and \(v=0\) and so the image is the closed interval [-2,2] on the \(u\) -axis.
Step-by-Step Solution
Verified Answer
The image is the closed interval [-2, 2] on the \( u \)-axis.
1Step 1: Understand the Given Expressions
The problem involves working with complex numbers in the form of exponential functions. We have two expressions: \( z = e^{it} \) and \( w = z + \frac{1}{z} \).
2Step 2: Simplify the Expression for w
Substitute \( z = e^{it} \) into \( w \):\[ w = e^{it} + \frac{1}{e^{it}} = e^{it} + e^{-it} \]Simplify using trigonometric identities, which gives: \[ w = 2 \cos t \]
3Step 3: Identify the Real and Imaginary Parts
Since \( w = 2 \cos t \) is expressed purely in terms of cosine, it's clear that this expression is purely real, with no imaginary component. Thus, \( u = 2 \cos t \) and \( v = 0 \).
4Step 4: Determine the Range of the Function
The range of \( 2 \cos t \) as \( t \) varies across all real numbers (i.e., as it completes one full circle in radians from \(0\) to \(2\pi\)) is the interval \([-2, 2]\) on the real axis, since the cosine function oscillates between -1 and 1. Therefore, the image of \( w \) is the real interval \([-2, 2]\).
Key Concepts
Exponential FunctionsTrigonometric IdentitiesReal and Imaginary PartsRange of Function
Exponential Functions
Exponential functions are an integral part of understanding complex numbers and their operations. In our exercise, we encounter the exponential function in the form of a complex number: \( z = e^{it} \). Here, \( e^{it} \) is a representation of complex numbers on the unit circle in the complex plane.
This function essentially maps the angle \( t \) as it varies, onto a circle of radius 1 centered at the origin.
Understanding how exponential functions behave with complex numbers is critical. It helps us transform and interpret values easily in the polar form, which is extremely helpful in trigonometric contexts.
This function essentially maps the angle \( t \) as it varies, onto a circle of radius 1 centered at the origin.
Understanding how exponential functions behave with complex numbers is critical. It helps us transform and interpret values easily in the polar form, which is extremely helpful in trigonometric contexts.
- The exponential form \( e^{it} \) shows a convenient and powerful way to depict rotations and periodicity of complex numbers.
- It ties closely with Euler's formula which connects complex exponentials to trigonometry.
- This approach simplifies many operations, particularly multiplication and division of complex numbers.
Trigonometric Identities
Trigonometric identities are vital in simplifying expressions involving complex numbers, especially those in exponential form. In our case, the expression we encounter, \( w = z + \frac{1}{z} = e^{it} + e^{-it} \), can be simplified using the well-known identity of trigonometry.
Using the trigonometric identity \( e^{it} + e^{-it} = 2\cos t \), we convert the expression into a purely trigonometric form.
Using the trigonometric identity \( e^{it} + e^{-it} = 2\cos t \), we convert the expression into a purely trigonometric form.
- This simplification is valuable as it translates complex expressions into more manageable real-valued functions.
- Understanding how this transformation works depends on grasping basic trigonometric properties, such as that cosine values oscillate between -1 and 1.
- This conversion underscores the connectivity between exponential and trigonometric functions, demonstrated profoundly in Euler's formula.
Real and Imaginary Parts
After simplifying our expression, \( w = 2 \cos t \), it is essential to identify the real and imaginary components. Here, the outcome is straightforward: the expression fully relies on cosine, which is a real function, leaving no room for imaginary parts.
In terms of complex numbers, we often represent them in the form \( w = u + iv \), where \( u \) represents the real part, and \( v \) the imaginary part.
In terms of complex numbers, we often represent them in the form \( w = u + iv \), where \( u \) represents the real part, and \( v \) the imaginary part.
- For \( w = 2 \cos t \), it implies \( u = 2 \cos t \) and \( v = 0 \).
- Recognizing when an expression is purely real (or purely imaginary) is crucial in complex number operations and simplifies various analytical approaches.
- This concept helps in plotting complex numbers accurately on the complex plane.
Range of Function
Understanding the range of a function, particularly in the context of trigonometric functions linked with complex numbers, is crucial. In the phase of our exercise, determining the range of \( w = 2 \cos t \) becomes inevitable.
As \( t \) spans from \(0\) to \(2\pi\), the cosine function itself varies between -1 and 1. Therefore, the output, \( 2 \cos t \), must lie within the interval \([-2, 2]\).
As \( t \) spans from \(0\) to \(2\pi\), the cosine function itself varies between -1 and 1. Therefore, the output, \( 2 \cos t \), must lie within the interval \([-2, 2]\).
- This range clearly maps the oscillation and behaviors of trigonometric components joined with complex numbers.
- It also helps us graph and interpret the function's image on the real axis.
- Understanding the interval where values reside enhances one's ability to predict and generalize results in further calculations and extensions into more complex functions.
Other exercises in this chapter
Problem 10
(a) \(\operatorname{since} G\left(r e^{i \theta}\right)=r^{2 / 3} e^{i 2 \theta / 3}, \psi(r, \theta)=\operatorname{Im}\left(G\left(r e^{i \theta}\right)\right)
View solution Problem 10
Using \(\mathrm{H}-5, f(z)=\left(\frac{1+z}{1-z}\right)^{2}\) maps \(R\) onto the upper half-plane \(R^{\prime} .\) The corre sponding Dirichlet problem in \(R^
View solution Problem 11
(a) \(\psi(x, y)=\operatorname{Im}(\sin z)=\cos x \sinh y\) and \(\psi(x, y)=0\) when \(x=\pm \pi / 2\) or when \(y=0\) (b) \(\mathbf{V}=\overline{G^{\prime}(z)
View solution Problem 11
\(S(w)=\frac{\left(w-w_{1}\right)\left(w_{2}-w_{3}\right)}{\left(w-w_{3}\right)\left(w_{2}-w_{1}\right)}\) maps \(w_{1}, w_{2}, w_{3}\) to \(0,1, \infty\) and s
View solution