Problem 9
Question
Consider the following sets.
(i) \(\\{z: \operatorname{Re}(z)>1\\}\).
(ii) \(\\{z:-1<\operatorname{lm}(z) \leq 2\\}\).
(iii) \(\\{z:|z-2-i| \leq 2\\}\).
(iv) \(\\{z:|z+3 i|>1\\}\).
(v) \(\left\\{r e^{i \theta}: 0
Step-by-Step Solution
Verified Answer
See the descriptions in Step 8 for classification details of each set.
1Step 1: Sketch Set (i)
The set \( \{z : \operatorname{Re}(z) > 1 \} \) represents all complex numbers whose real part is greater than 1. This is a vertical half-plane to the right of the line \( \operatorname{Re}(z) = 1 \).
2Step 2: Sketch Set (ii)
The set \( \{z : -1 < \operatorname{lm}(z) \leq 2 \} \) includes all complex numbers with imaginary parts satisfying \(-1 < \operatorname{lm}(z) \leq 2\). This is a horizontal strip between the lines \( \operatorname{lm}(z) = -1 \) and \( \operatorname{lm}(z) = 2 \) and includes the top line but not the bottom.
3Step 3: Sketch Set (iii)
The set \( \{z : |z - 2 - i| \leq 2 \} \) is a closed disk centered at \((2,1)\) with radius 2. This includes all points inside and on the boundary of the circle.
4Step 4: Sketch Set (iv)
The set \( \{z : |z + 3i| > 1 \} \) is the exterior of an open circle centered at \((0, -3)\) with radius 1. It does not include the boundary.
5Step 5: Sketch Set (v)
The set \( \{r e^{i \theta} : 0 < r < 1, -\frac{\pi}{2} < \theta < \frac{\pi}{2} \} \) is a sector of a disk in the upper half-plane, extending from the positive real axis to the negative real axis but excluding them.
6Step 6: Sketch Set (vi)
The set \( \{r e^{i \theta} : r > 1, \frac{\pi}{4} < \theta < \frac{\pi}{3} \} \) represents a sector of an unbounded region in the first quadrant between \(\theta = \pi/4\) and \(\theta = \pi/3\), extending to infinity.
7Step 7: Sketch Set (vii)
The set \( \{z : |z| < 1 \) or \( |z - 4| < 1 \} \) consists of two open disks, one centered at the origin with radius 1 and another centered at \((4,0)\) with radius 1.
8Step 8: Analyze and Classify Sets
(i) Open, connected, not bounded, domain.
(ii) Not open, connected, not bounded, not a domain.
(iii) Closed, connected, bounded, closed region.
(iv) Open, connected, not bounded, domain.
(v) Open, connected, bounded, region.
(vi) Open, connected, not bounded, region.
(vii) Open, not connected, bounded, not a domain.
Key Concepts
Complex NumbersSets in Complex PlaneDomains in Complex AnalysisRegion and Boundedness in Complex Analysis
Complex Numbers
Complex numbers extend the concept of the traditional number system by incorporating a real part and an imaginary part. They are generally represented as \( z = x + yi \), where \( x \) is the real part and \( y \) is the imaginary part. Here, \( i \) is the imaginary unit with the property \( i^2 = -1 \). This extension allows us to perform operations involving square roots of negative numbers, something not possible with real numbers alone.
As we explore the complex plane, we see that it gives us a two-dimensional perspective on numbers that allows deeper insights into mathematical concepts such as roots and trigonometric functions.
- The real part of a complex number is denoted as \( \operatorname{Re}(z) = x \).
- The imaginary part is denoted as \( \operatorname{Im}(z) = y \).
As we explore the complex plane, we see that it gives us a two-dimensional perspective on numbers that allows deeper insights into mathematical concepts such as roots and trigonometric functions.
Sets in Complex Plane
The complex plane, also known as the Argand plane, is a two-dimensional plane for representing complex numbers. It treats the real part of the complex number as the x-coordinate and the imaginary part as the y-coordinate, making it possible to graphically display and explore complex number operations.
Visualization of such sets helps us understand their properties, such as openness, connectedness, and boundedness, which are essential in more advanced topics like topology and analysis.
- The complex plane allows for the geometric interpretation of complex arithmetic.
- It provides a visual way to understand complex functions and transformations.
Visualization of such sets helps us understand their properties, such as openness, connectedness, and boundedness, which are essential in more advanced topics like topology and analysis.
Domains in Complex Analysis
In complex analysis, a domain is a specific type of open set in the complex plane. A set is considered open if for every point within the set, there exists a neighborhood of points also contained within the set. Moreover, a domain must also be connected, meaning any two points in the set can be joined by a continuous path that lies entirely within the set.
- An example of a domain is the set \( \{z : \operatorname{Re}(z) > 1 \} \), as it is both open and connected.
- Domains are important because many properties and theorems in complex analysis (such as Cauchy's theorem) rely on functions defined over domains.
Region and Boundedness in Complex Analysis
A region in the context of complex analysis refers to an open connected set, and it might include some or all of its boundary, forming a closed region if it does so.
Regions can be:
These classifications help in understanding the constraints and limitations of complex functions, especially when considering convergence and continuity, which are integral to the field of complex analysis.
Regions can be:
- Bounded: The set can be contained within a disk of finite radius.
- Unbounded: The set extends infinitely in one or more directions.
These classifications help in understanding the constraints and limitations of complex functions, especially when considering convergence and continuity, which are integral to the field of complex analysis.
Other exercises in this chapter
Problem 9
Describe the set of complex numbers for which \(\operatorname{Arg}\left(\frac{1}{x}\right) \neq-\operatorname{Arg}(z) .\) Prove your assertion.
View solution Problem 9
Find the three solutions to \(z^{\frac{3}{2}}=4 \sqrt{2}+i 4 \sqrt{2}\).
View solution Problem 10
Explain why the complex number \((0,0)\) (which, you recall, we identify with the real number 0 ) has no multiplicative inverse.
View solution Problem 10
Prove that \(|z|=0\) iff \(z=0\).
View solution