/*! 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 15 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.\(z(t)=2(1-t)+i t, 0 \leq t \leq 1 ; f(z)=3 z\)

Short Answer

Expert verified
C is a line segment from 2 to i; C' is a line segment from 6 to 3i under mapping w=3z.

Step by step solution

01

Identify the Real and Imaginary Components

First, we write the parametric curve \( z(t) \) given by \( z(t) = x(t) + i y(t) \), where \( x(t) = 2(1-t) \) and \( y(t) = t \). This gives us the real part \( x(t) = 2 - 2t \) and the imaginary part \( y(t) = t \).
02

Plot the Curve C

Using the parameter \( t \) range from 0 to 1, we plot the curve in the complex plane where the x-axis represents the real part and the y-axis represents the imaginary part. This linear curve starts at point \( (2, 0) \) and ends at \( (0, 1) \). It is a line segment from \( 2 + 0i \) to \( 0 + i \).
03

Apply the Complex Mapping

Now, we find the image of the curve under the mapping \( w = f(z) \) where \( f(z) = 3z \). Therefore, \( w(t) = 3z(t) = 3[2(1-t) + it] = 6(1-t) + 3it \).
04

Find the Real and Imaginary Components of w(t)

Rewrite \( w(t) = u(t) + iv(t) \), where \( u(t) = 6 - 6t \) and \( v(t) = 3t \). This gives us the real and imaginary components for the mapping \( w(t) \).
05

Plot the Curve C'

We plot \( C' \) in the complex plane again with the x-axis for \( u(t) \) and the y-axis for \( v(t) \). Starting from point \( (6, 0) \) when \( t = 0 \) and ending at \( (0, 3) \) when \( t = 1 \), this curve is another straight line.
06

Describe Curve C'

The curve C' is a straight line segment from \( 6 + 0i \) to \( 0 + 3i \). This line is longer and steeper compared to the original curve C due to the multiplication factor of 3 applied to both the real and imaginary components.

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 Curve
In mathematics, a parametric curve is represented by equations that express the coordinates of points as functions of a variable called a parameter. Instead of defining a curve implicitly or explicitly through a function of two variables, parametric equations make it possible to express the curve using one or more parameters.

The original exercise gives us a parametric curve defined as:
  • Real component: \( x(t) = 2(1-t) \)
  • Imaginary component: \( y(t) = t \)
This curve is specifically in the complex number context, where the real part is seen as the x-coordinate and the imaginary part as the y-coordinate. By varying the parameter \( t \) from 0 to 1, the curve forms a line segment starting at the point \((2, 0)\) and ending at \((0, 1)\) in the complex plane.

Understanding parametric curves is crucial as they allow us to describe a wide range of paths and shapes by controlling the parameter, providing flexibility not available with other mathematical representations. It simplifies many geometric problems by treating curves as continuous processes that trace the path of a moving point.
Complex Mapping
Complex mapping or transformations involve applying a function to complex numbers to produce new complex numbers, essentially transforming figures or regions in the complex plane. A complex function like \( f(z) = 3z \) can change both the direction and scale of a curve.

In the exercise provided, the original curve \( C \) is transformed using the mapping \( f(z) = 3z \). Upon applying this mapping to the curve given by \( z(t) = 2(1-t) + it \), we calculate its image as:
  • For real part: \( u(t) = 6 - 6t \)
  • For imaginary part: \( v(t) = 3t \)
The effect of the mapping \( w = f(z) \) is a scaling transformation that stretches the original curve by a factor of 3. This results in a new line segment that is three times the original length, which starts at the point \((6, 0)\) and ends at \((0, 3)\).

Grasping complex mapping is crucial in fields like electrical engineering and quantum physics, where transformations in complex planes simplify the understanding and solving of complex differential equations and other analytical problems.
Complex Plane Plotting
Plotting in the complex plane involves translating complex numbers into a two-dimensional space, where the x-axis corresponds to the real part and the y-axis corresponds to the imaginary part of complex numbers. This allows us to visually interpret complex functions and transformations.

In this exercise, the task is to plot both the original curve \( C \) and its image \( C' \) under the mapping \( f(z) = 3z \). The original curve \( C \) represents a line in the complex plane moving from \((2, 0)\) to \((0, 1)\). The image curve \( C' \), resulting from the complex mapping, stretches to form a line from \((6, 0)\) to \((0, 3)\).

Here are some important aspects to consider:
  • Starting and ending points of the curves provide clear reference points for drawing.
  • The direction and length of these lines can indicate the effect of complex transformations applied.
This visualization supports students in understanding how manipulatable complex transformations are, particularly how a linear transformation scales and may change the direction of the curve. It provides a concrete way to see otherwise abstract mathematical principals in action.

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 multiple-valued function \(F(z)=\left(z^{2}+1\right)^{1 / 2}\). What are the branch points (there are two of them) of \(F ?\) Explain.

the polygon with vertices \(0,1,1+i\), and \(-1+i\)

Show that the image of the line \(x=k, x \neq 0\), under the reciprocal map defined on the extended complex plane is the circle \(\left|w-\frac{1}{2 k}\right|=\left|\frac{1}{2 k}\right|\).

Consider the complex function \(f(z)=\frac{1}{z}+i\) defined on the half-plane \(x \geq 2\). (a) Use mappings to determine an upper bound \(M\) on the modulus of \(f(z)\). (b) Find a value of \(z\) that attains your bound in (a). That is, find \(z_{0}\) such that \(z_{0}\) is in the half-plane \(x \geq 2\) and \(\left|f\left(z_{0}\right)\right|=M\)

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.