/*! 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 19 Plot the parametric curve \(C\) ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Plot the parametric curve \(C\) given by \(z(t)\) and describe the curve in words, (b) find a parametrization of the image, \(C^{\prime}\), of \(C\) under the given complex mapping \(w=f(z)\), and (c) plot \(C^{\prime}\) and describe this curve in words.\(z(t)=t, 0 \leq t \leq 2 ; f(z)=e^{i \pi z}\)

Short Answer

Expert verified
C is a line segment from 0 to 2 on the real axis. C' is the upper half of the unit circle.

Step by step solution

01

Understand the Parametric Curve

The parametric curve is given by the equation \( z(t) = t \) where \( 0 \leq t \leq 2 \). This means the curve is a line segment on the real axis starting at \( t=0 \) and ending at \( t=2 \). This can be visualized as a straight horizontal line from \( 0 \) to \( 2 \) on the real line.
02

Describe the Curve C

Since \( z(t) = t \), where \( t \) ranges from 0 to 2, the curve \( C \) is simply a line segment along the real part of the complex plane from \( 0 \) to \( 2 \). Thus, it covers the points \( 0, 1, 2 \) on the real axis.
03

Apply the Complex Mapping

The complex mapping \( w = f(z) = e^{i \pi z} \) is applied to the parametric curve. This mapping transforms \( z(t) \) to \( w(t) = e^{i \pi t} \).
04

Parameterize the Image Curve C'

The image of the curve \( C \) under the mapping \( f \) is parameterized by \( w(t) = e^{i \pi t} \). As \( t \) changes from 0 to 2, \( w(t) \) traces out a curve on the complex unit circle. Specifically, it starts at \( e^{0} = 1 \) and moves counterclockwise to \( e^{i \pi} = -1 \), completing a semicircle.
05

Plot and Describe the Image Curve C'

The curve \( C' \), parameterized by \( w(t) = e^{i \pi t} \) for \( 0 \leq t \leq 2 \), lies on the unit circle in the complex plane. When plotted, \( C' \) appears as the upper half of the unit circle, beginning from \( 1 \) and ending at \( -1 \) on the real axis.

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 91Ó°ÊÓ!

Key Concepts

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

Parametric Curves
Parametric curves are a powerful way to represent curves by using parameters. In complex analysis, we often use parameters like \( t \) to describe a path or curve within the complex plane. A parametric curve is described by an equation that maps a range or set of parameters to the points on the curve.In this exercise, the parametric curve is defined by the equation \( z(t) = t \), where \( 0 \leq t \leq 2 \). This shows that our curve simply moves along the real axis, starting at \( t = 0 \) and ending at \( t = 2 \). This method is advantageous as it allows us to continuously trace the curve's path by just adjusting the parameter \( t \).
  • This particular parametric equation results in a line segment.
  • It maps each value of \( t \) directly onto the real axis, between 0 and 2.
Parametric curves like these provide a simple, flexible approach to curve representation, essential for transforming curves through complex mappings.
Complex Mapping
Complex mapping is a fascinating concept in complex analysis. It involves transforming points from one curve to another within the complex plane using a function. Here, the function used is \( f(z) = e^{i \pi z} \).The transformation \( f(z) = e^{i \pi z} \) affects the original parametric curve \( z(t) \) by translating each point into a new location, \( w(t) \), on the complex plane. This process is key in understanding how expressions and operations can morph the shape of curves or paths.
  • Each input \( z \) is moved to a new point defined by \( w \).
  • The mapping \( e^{i \pi z} \) specifically converts a linear path into a circular path.
This type of mapping produces dynamic effects, such as transforming linear paths into loops or spirals, and is widely used in mathematics and physics.
Unit Circle
The unit circle is a fundamental concept both in trigonometry and complex analysis. It represents a circle with radius 1, centered at the origin of the complex plane. The unit circle is particularly useful for visualizing complex functions and transformations.In this context, when the parametric line \( z(t) = t \) is transformed by the function \( f(z) = e^{i \pi z} \), it traces the upper half of the unit circle. Starting at \( 1 \) when \( t = 0 \) and ending at \( -1 \) when \( t = 2 \), this semi-circle is a direct result of the complex exponential function.
  • Any exponent of the form \( e^{i \theta} \) lies on the unit circle.
  • For \( \theta = \pi t \), as \( t \) increases, you trace out a path along the unit circle.
The unit circle thus acts as a boundary for any complex exponential expression, helping us visualize phase and angle changes in complex functions.
Real Axis
The real axis is the horizontal axis of the complex plane. It effectively represents all the real numbers from negative to positive infinity, and it helps define the complex plane alongside the imaginary axis.In this problem, the parametric curve \( z(t) = t \) is entirely confined to the real axis for \( 0 \leq t \leq 2 \). This means the curve is simply a real number progression. Post-mapping, the importance of the real axis shifts to signify endpoints on the unit circle.
  • On the initial curve \( z(t) \), it starts at 0 and ends at 2, following the real axis.
  • In the transformed curve \( C' \), the real axis marks critical points of the unit circle at 1 and -1.
The real axis serves as a foundation, offering a straightforward scaling system for initial and transformed complex values.

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 \(f(z)=\frac{1+\imath}{z}+2\) defined on the annulus \(1 \leq|z| \leq 2\) (a) Use mappings to determine upper and lower bounds on the modulus of \(f(z)=\frac{1+i}{z}+2 .\) That is, find real values \(L\) and \(M\) such that \(L \leq\left|\frac{1+i}{z}+2\right| \leq M\) (b) Find values of \(z\) that attain your bounds in (a). In other words, find \(z_{0}\) and \(z_{1}\) such that \(z_{0}\) and \(z_{1}\) are in the annulus \(1 \leq|z| \leq 2\) and \(\left|f\left(z_{0}\right)\right|=L\) and \(\left|f\left(z_{1}\right)\right|=M\).

Consider the mapping \(h(z)=\frac{3 i}{z^{2}}+1+i\) defined on the extended complex plane. (a) Write \(h\) as a composition of a linear, the reciprocal, and the squaring function. (b) Determine the image of the circle \(\left|z+\frac{1}{2} i\right|=\frac{1}{2}\) under the mapping \(w=h(z)\) (c) Determine the image of the circle \(|z-1|=1\) under the mapping \(w=h(z)\).

Find the image of the ray \(\arg (z)=\pi / 6\) under each of the following mappings. (a) \(f(z)=z^{3}\) (b) \(f(z)=z^{4}\) (c) \(f(z)=z^{5}\)

Find the streamlines of the planar flow associated with the given complex function \(f\) and (b) sketch the streamlines.\(f(z)=\frac{1}{\bar{z}}\)

Groups of Isometries In this project we investigate the relationship between complex analysis and the Euclidean geometry of the Cartesian plane. The Euclidean distance between two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) in the Cartiesian plane is $$ d\left(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right)=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} $$ Of course, if we consider the complex representations \(z_{1}=x_{1}+i y_{1}\) and \(z_{2}=\) \(x_{2}+i y_{2}\) of these points, then the Euclidean distance is given by the modulus $$ d\left(z_{1}, z_{2}\right)=\left|z_{2}-z_{1}\right| $$ A function from the plane to the plane that preserves the Euclidean distance between every pair of points is called a Euclidean isometry of the plane. In particular, a complex mapping \(w=f(z)\) is a Euclidean isometry of the plane if $$ \left|z_{2}-z_{1}\right|=\left|f\left(z_{1}\right)-f\left(z_{2}\right)\right| $$ for every pair of complex numbers \(z_{1}\) and \(z_{2}\).(a) Prove that every linear mapping of the form \(f(z)=a z+b\) where \(|a|=1\) is a Euclidean isometry. A group is an algebraic structure that occurs in many areas of mathematics. A group is a set \(G\) together with a special type of function \(*\) from \(G \times G\) to \(G\). The function \(*\) is called a binary operation on \(G\), and it is customary to use the notation \(a * b\) instead of \(*(a, b)\) to represent a value of \(* .\) We now give the formal definition of a group. A group is a set \(G\) together with a binary operation \(*\) on \(G\), which satisfies the following three properties: (i) for all elements \(a, b\), and \(c\) in \(G, a *(b * c)=(a * b) * c\),(b) Prove that composition of functions is a binary operation on Isom \(_{+}(\mathbf{E})\). That is, prove that if \(f\) and \(g\) are functions in Isom \(_{+}(\mathbf{E})\), then the function \(f \circ g\) defined by \(f \circ g(z)=f(g(z))\) is an element in Isom \(_{+}(\mathbf{E})\). (c) Prove that the set Isom \(_{+}(\mathbf{E})\) with composition satisfies property \((i)\) of a group. (d) Prove that the set \(\operatorname{Isom}_{+}(\mathbf{E})\) with composition satisfies property \((i i)\) of a group. That is, show that there exists a function \(e\) in Isom \(_{+}(\mathbf{E})\) such that \(e \circ f=f \circ e=f\) for all functions \(f\) in Isom \(_{+}(\mathbf{E})\). (e) Prove that the set Isom \(_{+}(\mathbf{E})\) with composition satisfies property \((i i i)\) of a group.

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.