Problem 26
Question
Sketch the set in the complex plane. $$\\{z|2 \leq| z | \leq 5\\}$$
Step-by-Step Solution
Verified Answer
The set is an annular region in the complex plane between circles of radius 2 and 5 centered at the origin.
1Step 1: Understanding the Set
The given set is \(\{z | 2 \leq |z| \leq 5\}\). This represents the complex numbers \(z\) whose modulus \(|z|\) is between 2 and 5, inclusive. In geometric terms, these are the points in the complex plane with a distance from the origin between 2 and 5.
2Step 2: Identify the Boundary Circles
The modulus \(|z|\) represents the distance from the origin to the point \(z\) in the complex plane. The conditions \(2 \leq |z|\) and \(|z| \leq 5\) describe two circles: one with a radius of 2, and another with a radius of 5, both centered at the origin.
3Step 3: Determine Composition of Set
The set is the region between the two circles determined in Step 2. It includes the points on both circles because the inequalities are non-strict (\(\leq\) and \(\geq\)). This annular region is a closed disk between the two circles.
4Step 4: Sketch the Region
Draw a circle centered at the origin with radius 2 to represent \(|z| = 2\). Next, draw another circle centered at the origin with radius 5 to represent \(|z| = 5\). The set \(\{z|2 \leq| z |\leq 5\}\) is the shaded region that lies between these two circles, including the boundaries.
Key Concepts
ModulusBoundary CirclesAnnular Region
Modulus
In the complex plane, the term "modulus" refers to the magnitude or absolute value of a complex number. If we have a complex number of the form \( z = a + bi \), its modulus is given by \( |z| = \sqrt{a^2 + b^2} \). This formula helps us determine the distance from the origin (point \((0,0)\)) to the point \((a,b)\) where the complex number \(z\) is represented on the plane.
When considering a complex set like \( \{z | 2 \leq |z| \leq 5\} \), the modulus acts as a filter. It only includes those complex numbers whose distance from the origin is between 2 and 5 units. Therefore, understanding modulus is crucial for identifying and sketching regions in the complex plane.
When considering a complex set like \( \{z | 2 \leq |z| \leq 5\} \), the modulus acts as a filter. It only includes those complex numbers whose distance from the origin is between 2 and 5 units. Therefore, understanding modulus is crucial for identifying and sketching regions in the complex plane.
Boundary Circles
Boundary circles in the complex plane are circles that define the limits of a region based on the modulus of complex numbers. For the set \( \{z | 2 \leq |z| \leq 5\} \), there are two key boundary circles:
It's significant to remember that these boundary circles are inclusive, as indicated by the \(\leq\) inequality, meaning points on both circles are part of the set.
- The inner boundary circle has a radius of 2 centered at the origin. This circle represents points where the modulus \(|z| = 2\).
- The outer boundary circle has a radius of 5, also centered at the origin, representing points where \(|z| = 5\).
It's significant to remember that these boundary circles are inclusive, as indicated by the \(\leq\) inequality, meaning points on both circles are part of the set.
Annular Region
An annular region on the complex plane is a ring-shaped area defined by two concentric boundary circles. In the given exercise, the annular region is defined by the set \( \{z | 2 \leq |z| \leq 5\} \). This region includes:
- All points on the plane that lie between the two boundary circles described previously.
- Points that are exactly on the boundary circles, due to the use of non-strict inequalities \(\leq\) and \(\geq\).
Other exercises in this chapter
Problem 25
The position of an object in circular motion is modeled by the given parametric equations. Describe the path of the object by stating the radius of the circle,
View solution Problem 25
Sketch a graph of the polar equation. $$r=2-2 \cos \theta$$
View solution Problem 26
The position of an object in circular motion is modeled by the given parametric equations. Describe the path of the object by stating the radius of the circle,
View solution Problem 26
Sketch a graph of the polar equation. $$r=1+\sin \theta$$
View solution