Complex Plane Points with Re(z)≤0 and |z|=3

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Homework Help Overview

The discussion revolves around identifying sets of points in the complex plane that satisfy specific conditions related to the real and imaginary parts of complex numbers, as well as their magnitudes. The problem specifically addresses conditions such as Re(z) ≤ 0 and |z| = 3.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of the conditions on the complex plane, questioning how to express the sets of points mathematically. There are discussions about the geometric representation of circles and regions defined by inequalities.

Discussion Status

Participants are actively engaging with the problem, offering interpretations and questioning the clarity of the requirements. Some guidance has been provided regarding the geometric nature of the solutions, but there remains uncertainty about how to present the results and whether the interpretations align with the original problem statement.

Contextual Notes

There are indications of confusion regarding the expectations for presenting solutions, particularly in relation to drawing figures and understanding the implications of the inequalities and equations involved.

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Homework Statement



Find the set of points z=x+iy from the complex plane which satisfy this conditions:

a)Re(z)\leq0

b)|z|=3

Homework Equations


The Attempt at a Solution



I've solved the first one...

Re(z)=x
so x\leq0

x\in(-infinity,0]

But for the second one? Should I find x, y or should I find z?
 
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B) is just a circle of radius 3, so it would be the same values that satisfy x^2 + y^2 = 9. Although I don't think you can simplify the set of points much further than |z| = 3 anyway lol.
 
(b) has the same form as a problem you've solved already:

|z-a|=r
 
So what will be the set of points? :smile:

(-infinity,+infinity) ?
 
The set of points is what Gib Z said it is.
 
But in the task some is asked for some set of z points...

Is that set x^2+y^2=9[/tex]?
 
Yes that is correct.
 
And what about Re(z)\geq0 , Im(z)\geq0, |z|\leq2 ?

for Re(z), x \in (0,+\propto)

for Im(z) y \in (0,+\propto)

and for |z|\leq2

x^2+y^2=4

What to do next?
 
Last edited:
Physicsissuef said:
And what about Re(z)\geq0 , Im(z)\geq0, |z|\leq2 ?

for Re(z), x \in (0,+\propto)
Yes, which means the set of points is the right half plane (including the line y-axis).

for Im(z) y \in (0,+\propto)
Yes, which means the set of points is the upper half plane (including the x-axis)

and for |z|\leq2

x^2+y^2=4
No, you have forgotten the "<". The set of points is the disk with center at (0,0) and radius 2, including the boundary.

What to do next?
To do what?
 
  • #10
What are the set of points z for that conditions?

Is it x,y \in (0,2)?

Since |z|\leq4
 
  • #11
Physicsissuef,

You're asking a lot of similar questions that people are helping you with, but you don't really seem to understand the answers that people are providing or helping you find.

If you're satisfied with the earlier solution
<br /> x^2+y^2=9<br />

then what is troubling you about
<br /> x^2+y^2 \leq 4<br />

Can you say what format your teacher or professor is expecting the solution to have?
 
  • #12
Are the solutions:

x,y <br /> \in (0,2)<br />

?
I understand the answers... Just I need confirmation, that's alll...
 
  • #13
Very well then! Every 'answer' you have given is wrong. And since you have been told what the correct answers are, and say that you understand them, that mysifies me!
 
  • #14
:smile: Ok. I don't understand something.

Can somebody please exactly give me what is the set of z points?

Is it x^2+y^2 \leq 4 ?

Is it x <br /> \in (0,+\propto)<br />

Is it
<br /> \in (0,+\propto)<br />
 
Last edited:
  • #15
None of those are correct. Are you aware of the interpretation of complex numbers as points in the complex plane?
 
  • #16
For part a, the set of points is the left half of the complex plane, including the imaginary axis. For part b, the set of points is a circle of radius 3 around the origin.

Do you understand why?
 
  • #17
dx said:
None of those are correct. Are you aware of the interpretation of complex numbers as points in the complex plane?

So I should present the results in circular, right?
 
  • #18
Physicsissuef said:
So I should present the results in circular, right?

I don't understand this question.
 
  • #19
I asked my professor and he told me that I should draw a circle and present the results...
 
  • #20
Yes, the solution to part b is a circle in the complex plane. But the question is, do you understand why it is a circle?
 
  • #21
Physicsissuef said:
I asked my professor and he told me that I should draw a circle and present the results...

Thanks, that helps clear things up.
 
  • #22
dx said:
Yes, the solution to part b is a circle in the complex plane. But the question is, do you understand why it is a circle?

Yes, because z=a+bi, we can say it is point (a,b). It is circle because of |z|=r. So we have circular... But the question is, how we will present it in circular...
 
  • #23
I'm not sure what you mean by "present it in circular". Do you mean circular coordinates? If so, then are you familiar with the formula e^{i\theta} = cos(\theta) + i sin(\theta)?
 
  • #24
The professor said to draw a circle. It would have to be drawn on an x-y coordinate plane, with the radius and center location clearly indicated.

The problem seems to be that the professor's instructions are not being understood or explained clearly.

Edit added:
If we could see an example, either worked out in class by the professor or in the textbook, that would help us understand just what is needed in terms of presenting the results.
 
Last edited:
  • #25
Another question just occurred to me. Is the following 3 separate distinct questions, or is it all part of the same question (i.e., find (x,y) that satisfies all 3 statements)?

Physicsissuef said:
And what about Re(z)\geq0 , Im(z)\geq0, |z|\leq2 ?

for Re(z), x \in (0,+\propto)

for Im(z) y \in (0,+\propto)

and for |z|\leq2

x^2+y^2=4

What to do next?
 
  • #26
Yes, those are three distinct question- and you have already pretty much been told the answers!''

See response #9.
 
  • #27
Halls, I was trying to ask the OP that question.
 
  • #28
Last edited by a moderator:
  • #29
The pictures help to see what is required here, thanks.

You just need to draw the appropriate circle, and only include the part that satisfies the other conditions:

x \geq 0 and y \geq 0

So it won't be a full circle.
 

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