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Chapter 17: Problem 16

In Problems 9-22, sketch the set of points in the complex plane satisfying the given inequality. Determine whether the set is a domain. $$ \operatorname{Im}(1 / z)<\frac{1}{2} $$

Short Answer

Expert verified
The set is the region outside the circle centered at (0, -1) with radius 1, and it is a domain.

Step by step solution

01

Understanding the inequality

The given inequality is \( \operatorname{Im}(1/z) < \frac{1}{2} \) which involves the imaginary part of the reciprocal of a complex number \( z \). Our task is to find all complex numbers \( z = x + yi \) that satisfy this condition.
02

Finding 1/z

Set \( z = x + yi \). The reciprocal of \( z \) equals \( \frac{1}{z} = \frac{x - yi}{x^2 + y^2} \). This expression separates into real and imaginary components as \( \frac{x}{x^2 + y^2} - i \frac{y}{x^2 + y^2} \). The imaginary part is \( \operatorname{Im}(1/z) = -\frac{y}{x^2 + y^2} \).
03

Applying the inequality

Substitute the imaginary component into the inequality: \( -\frac{y}{x^2 + y^2} < \frac{1}{2} \). This requires finding conditions on \( x \) and \( y \) such that the inequality holds true.
04

Solving the inequality

Rearrange \( -\frac{y}{x^2 + y^2} < \frac{1}{2} \) to \( -2y < x^2 + y^2 \). After simplifying, this becomes \( x^2 + (y + 1)^2 > 1 \). This equation describes the exterior of a circle centered at (0, -1) with a radius of 1.
05

Sketching the region

In the complex plane, sketch the circle centered at (0, -1) with radius 1. The region satisfying the inequality is the exterior of this circle, not including the circumference.
06

Checking if the set is a domain

A domain in the complex plane is an open and connected set. The exterior of a circle (excluding the boundary) is open and each point is connected to another through paths within the set. Therefore, the described region is indeed a domain.

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Key Concepts

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

Complex Plane
The complex plane is an essential concept in complex analysis. It provides a way to visualize complex numbers as points in a two-dimensional plane. Each complex number can be expressed as \( z = x + yi \), where \( x \) is the real part and \( y \) is the imaginary part. This representation allows us to plot complex numbers similarly to how we plot points in the Cartesian coordinate system.

In the complex plane:
  • The horizontal axis represents the real part of the complex numbers.
  • The vertical axis represents the imaginary part.
Understanding this visual representation helps in solving complex inequalities and determining domains like in the given exercise.
Imaginary Part
The imaginary part of a complex number \( z = x + yi \) is the coefficient \( y \) multiplying the imaginary unit \( i \). In mathematical terms, it is denoted by \( \operatorname{Im}(z) = y \). This part is crucial when dealing with complex inequalities, especially ones like \( \operatorname{Im}(1/z) < \frac{1}{2} \).

When finding the imaginary component of \( \frac{1}{z} \), after rewriting the reciprocal:
  • The expression becomes \( -\frac{y}{x^2 + y^2} \).
  • This is essential for applying and solving inequality conditions.
Mastering the imaginary aspect is key to navigating complex relationships within the complex plane.
Inequality in Complex Numbers
Inequalities in complex numbers involve determining regions in the complex plane that satisfy a certain condition. For instance, the inequality \( -\frac{y}{x^2 + y^2} < \frac{1}{2} \) in the exercise specifies a condition for the imaginary part of the reciprocal of \( z \).

Breaking down this inequality:
  • Rearrange to \( -2y < x^2 + y^2 \).
  • Further simplify to \( x^2 + (y + 1)^2 > 1 \).
This describes the points outside of a circle centered at (0, -1) with radius 1. Understanding how to translate complex inequalities into geometric figures on the complex plane helps solve problems efficiently.
Domain in Complex Plane
Domain refers to a specific set of points in the complex plane that satisfy certain conditions and is open and connected. In the context of the inequality \( x^2 + (y + 1)^2 > 1 \), we are looking at the set excluding the circle's boundary, yet still considered to be a domain.

Features of a domain in this context include:
  • Open: The region doesn't include the boundary (the circle's circumference).
  • Connected: Any two points within this region can be linked without stepping out of it.
Recognizing and verifying domains in the complex plane is crucial as it ensures all conditions are met, as demonstrated in the exercise solution.

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Most popular questions from this chapter

Write the given number in the form \(a+i b\). $$ \frac{i}{1+i} $$

Write the given number in the form \(a+i b\). $$ (1+i)^{2}(1-i)^{3} $$

In Problems 15 and 16, find \(z_{1} z_{2}\) and \(z_{1} / z_{2} .\) Write the number in the form \(a+i b\). $$ z_{1}=2\left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right), z_{2}=4\left(\cos \frac{3 \pi}{8}+i \sin \frac{3 \pi}{8}\right) $$

Sketch the set of points in the complex plane satisfying the given inequality. Determine whether the set is a domain. $$ |\arg (z)|<\pi / 4 $$

Write the given complex number in polar form. \(5-5 i\)
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