Problem 21
Question
\(17-24=\) Sketch the set in the complex plane. $$ \\{z| | z |<2\\} $$
Step-by-Step Solution
Verified Answer
The set is the interior of a circle with radius 2 centered at the origin on the complex plane.
1Step 1: Review the Exercise
The problem requires us to sketch a set on the complex plane. Specifically, it asks us to draw all complex numbers \( z \) such that the modulus of \( z \) is less than 2: \( |z| < 2 \).
2Step 2: Understand the Equation
The inequality \( |z| < 2 \) refers to all complex numbers whose distance from the origin in the complex plane is less than 2. In the complex plane, \( z = x + yi \) has a modulus \(|z| = \sqrt{x^2 + y^2}\).
3Step 3: Identify the Geometric Representation
The inequality \( |z| < 2 \) represents a circle centered at the origin (0,0) with radius less than 2. Specifically, it defines the interior of the circle, without including the boundary.
4Step 4: Sketch the Set
To sketch the set, draw a circle with radius 2 centered at (0,0) on the complex plane. Shade the interior of this circle, excluding the edge, to represent all points \( z = x + yi \) where \( \sqrt{x^2 + y^2} < 2 \). The boundary is represented as a dashed or light line to indicate that it is not included.
5Step 5: Verify the Sketch
Ensure that only the interior of the circle is shaded and that no points on the boundary are considered part of the set. Check that the radius from the origin to any point in the set is always less than 2.
Key Concepts
Modulus of a complex numberGeometric representationInequality in complex numbers
Modulus of a complex number
The modulus of a complex number is a fundamental concept when dealing with complex numbers. The modulus, sometimes referred to as the absolute value, measures the distance of the complex number from the origin on the complex plane. If you have a complex number \[ z = x + yi \]where \( x \) is the real part and \( y \) is the imaginary part, the modulus is calculated using:\[ |z| = \sqrt{x^2 + y^2} \] This formula can be visualized like the Pythagorean theorem, where \( x \) represents one leg, \( y \) the other leg, and the modulus \( |z| \) is the hypotenuse.Understanding modulus is crucial as it tells you how "far away" a complex number is from the origin \((0, 0)\). This comes especially handy when solving inequalities or analyzing distances between complex numbers.
Geometric representation
In the context of complex numbers, geometric representation plays a vital role in turning abstract numbers into visual insights. On the complex plane, any complex number \( z = x + yi \) is represented as a point \( (x, y) \).The inequality \( |z| < 2 \) translates into a particular shape on the complex plane. Specifically, it describes all points within a circle centered at the origin \((0, 0)\) with a radius less than 2.
- The center: where the circle meets the origin.
- The radius: the maximum modulus value, but exclusive of the boundary.
Inequality in complex numbers
Inequalities involving complex numbers, such as \( |z| < 2 \), help us identify specific regions on the complex plane. Much like inequalities with real numbers, they describe a set or range of values.In this case,
- The inequality \( |z| < 2 \) means we are considering only those complex numbers whose modulus is less than 2.
- It excludes any complex number whose modulus equals 2, resulting in a non-inclusive boundary.
- This inequality defines the interior of a circle rather than just a line interval.
- The shaded region inside the circle clearly indicates which points are included in the set.
- The boundary remains unshaded, showing it is not part of the solution set.
Other exercises in this chapter
Problem 21
Find \(2 \mathbf{u},-3 \mathbf{v}, \mathbf{u}+\mathbf{v},\) and \(3 \mathbf{u}-4 \mathbf{v}\) for the given vectors \(\mathbf{u}\) and \(\mathbf{v} .\) $$ \math
View solution Problem 21
19-22 Find the component of \(\mathbf{u}\) along \(\mathbf{v}\). $$\mathbf{u}=7 \mathbf{i}-24 \mathbf{j}, \quad \mathbf{v}=\mathbf{j}$$
View solution Problem 21
15–36 Sketch the graph of the polar equation. $$r=-2 \cos \theta$$
View solution Problem 22
Find \(2 \mathbf{u},-3 \mathbf{v}, \mathbf{u}+\mathbf{v},\) and \(3 \mathbf{u}-4 \mathbf{v}\) for the given vectors \(\mathbf{u}\) and \(\mathbf{v} .\) $$ \math
View solution