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Chapter 2: Problem 1

Plot the images of the complex numbers \(z=1,1+i, 1-i\), and \(i\) under the given function \(f\) as position vectors, and (b) plot the images as vectors in the vector field associated with \(f\).\(f(z)=2 z-i\)

Short Answer

Expert verified
The images of the numbers are: 2-i, 2+i, 2-3i, and i as position vectors and vectors in the vector field.

Step by step solution

01

Understanding the Function

The function given is \(f(z) = 2z - i\). This means for any complex number \(z\), we first multiply it by 2 and then subtract \(i\) from the result. Our goal is to find the transformed images of the complex numbers \(z = 1, 1+i, 1-i,\) and \(i\).
02

Calculating f(z) for z=1

Substitute \(z = 1\) into the function: \(f(1) = 2(1) - i = 2 - i\). The image of \(z = 1\) under the function is the complex number \(2 - i\).
03

Calculating f(z) for z=1+i

Substitute \(z = 1+i\) into the function: \(f(1+i) = 2(1+i) - i = 2 + 2i - i = 2 + i\). The image of \(z = 1+i\) is the complex number \(2 + i\).
04

Calculating f(z) for z=1-i

Substitute \(z = 1-i\) into the function: \(f(1-i) = 2(1-i) - i = 2 - 2i - i = 2 - 3i\). The image of \(z = 1-i\) is the complex number \(2 - 3i\).
05

Calculating f(z) for z=i

Substitute \(z = i\) into the function: \(f(i) = 2i - i = i\). The image of \(z = i\) is the complex number \(i\).
06

Plotting Position Vectors

On the complex plane, plot the points corresponding to the images: \(2-i\), \(2+i\), \(2-3i\), and \(i\). These points should be plotted as position vectors from the origin (0,0).
07

Plotting Vectors in Vector Field

For each complex number transformation result, draw a vector on the complex plane starting at the origin pointing towards each image: \(2-i\), \(2+i\), \(2-3i\), and \(i\). This demonstrates the vector field behavior of the function \(f(z)\).

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Key Concepts

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

Complex Numbers
Complex numbers are an extension of the real numbers and include a real part and an imaginary part. They are usually expressed in the form of \( z = a + bi \), where \( a \) is the real part, and \( b \) is the imaginary part accompanied by the imaginary unit \( i \). The imaginary unit \( i \) is defined by the property \( i^2 = -1 \). This allows us to handle square roots of negative numbers effectively.
  • Real Part: The component \( a \) in the expression \( z = a + bi \).
  • Imaginary Part: The component \( bi \) in the expression.
Complex numbers can be visualized on a complex plane, where the horizontal axis represents the real part and the vertical axis represents the imaginary part.
Vector Field
A vector field is a representation that assigns a vector to every point in a given space. In the case of complex functions, this space is often the complex plane. Each point on the plane is associated with a complex number, and the complex function assigns a new complex number, effectively a vector, to each point.
  • The arrow of the vector indicates the direction of the transformation.
  • The length of the arrow can represent the magnitude of the transformation.
By plotting these vectors, we can visualize how a function transforms the entire space and observe patterns of motion and transformation. In our exercise, we see how the function \( f(z) = 2z - i \) modifies the vector field by transforming specific complex numbers.
Complex Plane
The complex plane is a two-dimensional plane used to represent complex numbers geometrically. It consists of a horizontal axis, called the real axis, and a vertical axis, called the imaginary axis.
  • Real Axis: The axis along which the real part of a complex number is measured.
  • Imaginary Axis: The axis along which the imaginary part of a complex number is measured.
Each complex number \( z = a + bi \) maps to a point \((a, b)\) in this plane. For example, the number \( 1 + i \) corresponds to the point \( (1, 1) \). When we apply the transformation \( f(z) = 2z - i \), each original point transforms to a new point, which is also plotted on the complex plane. Position vectors or arrows can be used from the origin to these new points to visualize the transformation.
Transformation
Transformation refers to the process of applying a function to a set of complex numbers to obtain a new set of numbers in the complex plane. This is what happens in our exercise when we use the function \( f(z) = 2z - i \) on the given points.
  • Mapping: Each complex number \( z \) is mapped to a new number \( f(z) \).
  • Steps: The transformation involves multiplying the original number by 2, then subtracting \( i \).
In visual terms, a transformation changes the position or orientation of points in the complex plane, and these changes can be represented with vectors showing the old position to the new position. Understanding transformations can provide insight into the behavior and geometry of complex functions and how they adjust the structure of the plane.

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Most popular questions from this chapter

We say that two mappings \(f\) and \(g\) commute if \(f \circ g(z)=g \circ f(z)\) for all \(z\). That is, two mappings commute if the order in which you compose them does not change the mapping. (a) Can a translation and a nonidentity rotation commute? (b) Can a translation and a nonidentity magnification commute? (c) Can a nonidentity rotation and a nonidentity magnification commute?

Find two different linear mappings that map the half-plane \(\operatorname{Re}(z) \geq 2\) onto the half-plane \(\operatorname{Re}(z) \geq 5\)

Let \(f\) be a complex function. Explain the relationship between the vector field associated with \(f(z)\) and the vector field associated with \(g(z)=i f(z) .\) Illustrate with sketches using a simple function for \(f\).

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.)\(f(z)=z+\operatorname{Arg}(z), z(t)=-\frac{1}{2}+\frac{1}{2} \sqrt{3} i t,-1 \leq t \leq 1\)

If \(A, B, C\), and \(D\) are real numbers, then the set of points in the plane satisfying the equation: $$ A\left(x^{2}+y^{2}\right)+B x+C y+D=0 $$ is called a generalized circle. (a) Show that if \(A=0\), then the generalized circle is a line. (b) Suppose that \(A \neq 0\) and let \(\Delta=B^{2}+C^{2}-4 A D\). Complete the square in \(x\) and \(y\) to show that a generalized circle is a circle centered at \(\left(\frac{-B}{2 A}, \frac{-C}{2 A}\right)\) with radius \(\frac{\sqrt{\Delta}}{2 A}\) provided \(\Delta>0 .\) (If \(\Delta<0\), the generalized circle is often called an imaginary circle.)
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