Problem 20
Question
In Problems 9-22, sketch the set of points in the complex plane satisfying the given inequality. Determine whether the set is a domain. $$ |z-i|>0 $$
Step-by-Step Solution
Verified Answer
The set is the entire complex plane except the point \((0,1)\), and it is a domain.
1Step 1: Understand the Inequality
The inequality states \(|z-i| > 0\). Here, \(z\) is a complex number represented as \(z = x + yi\), where \(x\) and \(y\) are real numbers. The term \(i\) is the imaginary unit which represents the point \(0 + 1i\) on the complex plane. The given inequality asks for all points such that the distance from \(z\) to \(i\) is greater than zero.
2Step 2: Interpret the Distance Condition
The distance \(|z-i|\) between two complex numbers \(z = x + yi\) and \(i = 0 + 1i\) is given by the modulus \(|z-i| = \sqrt{x^2 + (y-1)^2}\). The inequality \(|z-i| > 0\) is satisfied for all points \(z\) except the point where \(z = i\), meaning \(x = 0\) and \(y = 1\).
3Step 3: Sketch the Inequality on the Complex Plane
To sketch the points satisfying \(|z-i|>0\), plot the point \((0, 1)\) on the complex plane. All points in the complex plane except this one represent a distance from \(i\) that is greater than zero. Therefore, the solution includes the entire complex plane, excluding the point \((0, 1)\).
4Step 4: Determine if the Set is a Domain
A domain in the complex plane is an open, connected set. Since the set includes all points except one, and the complex plane minus a single point is still open and connected, the set is indeed a domain.
Key Concepts
Complex PlaneInequalitiesModulus of Complex NumberDomain in Complex Analysis
Complex Plane
In complex analysis, the complex plane is a vital concept that represents the entire set of complex numbers visually. It is akin to a 2D coordinate system where the horizontal axis represents the real part of a complex number, and the vertical axis represents the imaginary part.
To understand this idea:
To understand this idea:
- Real numbers are plotted along the horizontal axis.
- Imaginary numbers are plotted along the vertical axis.
- Every point in the plane corresponds to a complex number in the form of \(z = x + yi\), where \(x\) is the real part and \(y\) is the imaginary part.
Inequalities
Inequalities in the context of complex numbers are expressions that describe the relative position or size of complex numbers or sets in the complex plane. They tell us something about the "space" between numbers or how one number compares with another.
In our example, the inequality \(|z-i| > 0\) describes a condition:
In our example, the inequality \(|z-i| > 0\) describes a condition:
- This condition signifies all points \(z\) in the complex plane whose distance from the point \(i\) is greater than zero.
- Graphically, this inequality represents the entire complex plane except for the single point \(i\).
Modulus of Complex Number
The modulus of a complex number, symbolized as \(|z|\), is a measure of its magnitude or size. It represents the distance from the origin \(0, 0\) in the complex plane to the point \(x, y\) representing the complex number \(z = x + yi\).
The formula to calculate the modulus is: \[|z| = \sqrt{x^2 + y^2}\]In our specific exercise, the modulus comes into play when determining the distance of \(z\) from \(i\). For \(|z-i|\), it is calculated as:\[|z-i| = \sqrt{x^2 + (y-1)^2}\]
The formula to calculate the modulus is: \[|z| = \sqrt{x^2 + y^2}\]In our specific exercise, the modulus comes into play when determining the distance of \(z\) from \(i\). For \(|z-i|\), it is calculated as:\[|z-i| = \sqrt{x^2 + (y-1)^2}\]
- The modulus is always a non-negative value.
- If the modulus of a point from \(i\) is greater than zero, the point is not equal to \(i\).
Domain in Complex Analysis
A domain in complex analysis is a specific type of set in the complex plane, described by conditions that ensure certain properties are met. To be considered a domain, a set must be open and connected.
In our exercise, we needed to determine if the set satisfying \(|z-i|>0\) represents a domain. Let's break it down:
In our exercise, we needed to determine if the set satisfying \(|z-i|>0\) represents a domain. Let's break it down:
- **Open Set:** This means that every point within the set is an interior point. Removing a single point (in this case, \(i\)) from the complex plane does not affect its openness.
- **Connected Set:** This implies there is a path within the set between any two points in it without leaving the set. The entire complex plane (minus one point) remains connected because any point can reach another without stepping out of the set.
Other exercises in this chapter
Problem 20
In Problems 17-20, show that the given function is not analytic at any point, but is differentiable along the indicated curve(s). $$ f(z)=x^{2}-x+y+i\left(y^{2}
View solution Problem 20
In Problems 19-22, the given limit exists. Find its value. $$ \lim _{z \rightarrow 1-i} \frac{5 z-2 z+2}{z+1} $$
View solution Problem 20
In Problems 1-26, write the given number in the form \(a+i b\). $$ \frac{(4+5 i)+2 i^{3}}{(2+i)^{2}} $$
View solution Problem 21
Sketch the set of points in the complex plane satisfying the given inequality. Determine whether the set is a domain. $$ 2
View solution