/*! 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 5 Find the image of the closed dis... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 5

Find the image of the closed disk \(|z| \leq 1\) under the given linear mapping \(w=f(z)\) and (b) represent the linear mapping with a sequence of plots as in Figure \(2.14\).\(f(z)=2 z-i\)

Short Answer

Expert verified
The image is a disk centered at \(-i\) with radius 2.

Step by step solution

01

Understand the Mapping

The function given is a linear mapping of the form \( f(z) = 2z - i \). We need to determine how this function transforms the closed disk \( |z| \leq 1 \).
02

Apply Mapping to the Center of the Disk

The center of the disk \( |z| \leq 1 \) is the origin \( z = 0 \). Applying the function, we have \( f(0) = 2(0) - i = -i \). This means the center of the disk is mapped to \( -i \) on the complex plane.
03

Analyze the Effect on Boundary

For the boundary \( |z| = 1 \), write \( z = e^{i\theta} \). Applying the map, we get \( f(e^{i\theta}) = 2e^{i\theta} - i \). This traces out a new boundary path in the complex plane.
04

Determine the New Radius

Since \( f(z) = 2z - i \) doubles the magnitude of \( z \), the radius of the image is doubled. The original disk \( |z| \leq 1 \) has radius 1, so the image will have radius 2.
05

Determine the Shape and Position of the Image

Initially, the disk \( |z| \leq 1 \) is transformed into a disk \( |w + i| \leq 2 \), which implies \( |w| \leq 2 - 1 = 1 \). Thus, the new disk is centered at \( -i \) with a radius of 2.
06

Conclude the Image of the Disk

The image of the closed disk \( |z| \leq 1 \) under the map is a larger disk centered at \( -i \) with radius 2.

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.

Linear Mapping
In complex analysis, linear mapping refers to a transformation of a complex number, usually represented as a function like \( f(z) = az + b \). This type of function can be visualized as a combination of two operations: stretching or compressing and translating. Here, \( a \) affects the magnitude and direction, while \( b \) shifts the entire figure within the complex plane.
  • Stretching/Compressing: The value of \( a \) dictates how much the complex number is stretched or compressed. If \( |a| > 1 \), the image expands; if \( |a| < 1 \), it contracts.
  • Translation: The term \( b \) shifts the transformation. Adding \( b \) moves the entire image to a new location, determined by \( b \) itself.

For example, in the function \( f(z) = 2z - i \), \( 2 \) is the linear coefficient that doubles the size of the figure, and \(-i\) translates the image downward by one unit. Together, these actions reshape the initial closed disk into a new configuration within the complex plane.
Complex Plane
The complex plane is a two-dimensional plot where the horizontal axis represents the real part of a complex number and the vertical axis represents the imaginary part. Each point on this plane corresponds to a complex number of the form \( z = x + yi \), where \( x \) is the real part and \( yi \) is the imaginary part.
  • Imaginary Axis: The vertical line represents multiples of \( i \), the imaginary unit.
  • Real Axis: The horizontal line where imaginary parts are zero, showing real numbers.
  • Modulus and Argument: The distance of a point from the origin is its magnitude, \( |z| \), whereas its angle from the positive real axis is called the argument.

Transformations like linear mappings move figures around this plane. Our closed disk \(|z| \leq 1\) in initial form sits centered at the origin. Under the mapping \(f(z) = 2z - i\), it shifts to a new location with different size, aligning with linear transformation properties.
Closed Disk
A closed disk in the complex plane is a set of all points that are within or on the boundary of a circle. Defined mathematically as \( |z - z_0| \leq R \), where \( z_0 \) is the center and \( R \) is the radius, it encompasses all points no farther from \( z_0 \) than \( R \).
  • Centre: The point \( z_0 \) defines the location of the disk. For \( |z| \leq 1 \), this is at the origin.
  • Radius: The value \( R \) determines how large the disk is. Initially, \( R = 1 \).
  • Boundary: The perimeter represented by \( |z| = 1 \) is included in the disk.

When undergoing a transformation via liner mapping, such as \( f(z) = 2z - i \), these disks can change in size and position. The given mapping modifies the original disk \( |z| \leq 1 \) to a disk centered at \(-i\) with a doubled radius of 2. This change represents how the disk's radius increases and its position shifts within the complex plane.

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

Consider the complex function \(f(z)=2 i z^{2}-i\) defined on the quarter disk \(|z| \leq 2,0 \leq \arg (z) \leq \pi / 2\) (a) Use mappings to determine upper and lower bounds on the modulus of \(f(z)=2 i z^{2}-i . \quad\) That is, find real values \(L\) and \(M\) such that \(L \leq\left|2 i z^{2}-i\right| \leq M\) (b) Find values of \(z\) that achieve your bounds in (a). In other words, find \(z_{0}\) and \(z_{1}\) such that \(\left|f\left(z_{0}\right)\right|=L\) and \(\left|f\left(z_{1}\right)\right|=M\).

Use a CAS to show that the given function is not continuous inside the unit circle by plotting the image of the given continuous parametric curve. (Be careful, Mathematica and Maple plots can sometimes be misleading.)\(f(z)=|z-1| \operatorname{Arg}(-z)+i \operatorname{Arg}(i z), z(t)=\frac{1}{2} e^{i t}, 0 \leq t \leq 2 \pi\)

In this problem we show that a linear mapping is uniquely determined by the images of two points. (a) Let \(f(z)=a z+b\) be a complex linear function with \(a \neq 0\) and assume that \(f\left(z_{1}\right)=w_{1}\) and \(f\left(z_{2}\right)=w_{2} .\) Find two formulas that express \(a\) and \(b\) in terms of \(z_{1}, z_{2}, w_{1}\), and \(w_{2} .\) Explain why these formulas imply that the linear mapping \(f\) is uniquely determined by the images of two points. (b) Show that a linear function is not uniquely determined by the image of one point. That is, find two different linear functions \(f_{1}\) and \(f_{2}\) that agree at one point.

Consider the multiple-valued function \(F(z)=(z-1+i)^{1 / 2}\). (a) What is the branch point of \(F ?\) Explain. (b) Explicitly define two distinct branches of \(f_{1}\) and \(f_{2}\) of \(F .\) In each case, state the branch cut.

Find the image of the given set under the reciprocal mapping \(w=1 / z\) on the extended complex plane.the region \(1 \leq|z| \leq 4,0 \leq \arg (z) \leq 2 \pi / 3\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

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.