Set of points z=x+iy

  • #1

Homework Statement



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

a)Re(z)[itex]\leq[/itex]0

b)|z|=3

Homework Equations





The Attempt at a Solution



I've solved the first one...

Re(z)=x
so x[itex]\leq[/itex]0

x[itex]\in[/itex](-infinity,0]

But for the second one? Should I find x, y or should I find z?
 

Answers and Replies

  • #2
Gib 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.
 
  • #3
Redbelly98
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(b) has the same form as a problem you've solved already:

|z-a|=r
 
  • #4
So what will be the set of points? :smile:

(-infinity,+infinity) ?
 
  • #5
Redbelly98
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The set of points is what Gib Z said it is.
 
  • #6
But in the task some is asked for some set of z points...

Is that set [itex]x^2+y^2=9[/tex]?
 
  • #7
Defennder
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Yes that is correct.
 
  • #8
And what about Re(z)[itex]\geq[/itex]0 , Im(z)[itex]\geq[/itex]0, |z|[itex]\leq[/itex]2 ?

for Re(z), x [itex]\in (0,+\propto)[/itex]

for Im(z) y [itex]\in (0,+\propto)[/itex]

and for |z|[itex]\leq[/itex]2

[itex]x^2+y^2[/itex]=4

What to do next?
 
Last edited:
  • #9
HallsofIvy
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And what about Re(z)[itex]\geq[/itex]0 , Im(z)[itex]\geq[/itex]0, |z|[itex]\leq[/itex]2 ?

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

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

and for |z|[itex]\leq[/itex]2

[itex]x^2+y^2[/itex]=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 [itex]\in (0,2)[/itex]?

Since |z|[itex]\leq[/itex]4
 
  • #11
Redbelly98
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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
[itex]
x^2+y^2=9
[/itex]

then what is troubling you about
[itex]
x^2+y^2 \leq 4
[/itex]

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

x,y [itex]
\in (0,2)
[/itex]

?
I understand the answers... Just I need confirmation, that's alll....
 
  • #13
HallsofIvy
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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 [itex]x^2+y^2 \leq 4[/itex] ?

Is it x [itex]
\in (0,+\propto)
[/itex]

Is it
[itex]
\in (0,+\propto)
[/itex]
 
Last edited:
  • #15
dx
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None of those are correct. Are you aware of the interpretation of complex numbers as points in the complex plane?
 
  • #16
dx
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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
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?
 
  • #19
I asked my professor and he told me that I should draw a circle and present the results...
 
  • #20
dx
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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
Redbelly98
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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
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
dx
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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 [tex] e^{i\theta} = cos(\theta) + i sin(\theta) [/tex]?
 
  • #24
Redbelly98
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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
Redbelly98
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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)?

And what about Re(z)[itex]\geq[/itex]0 , Im(z)[itex]\geq[/itex]0, |z|[itex]\leq[/itex]2 ?

for Re(z), x [itex]\in (0,+\propto)[/itex]

for Im(z) y [itex]\in (0,+\propto)[/itex]

and for |z|[itex]\leq[/itex]2

[itex]x^2+y^2[/itex]=4

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

See response #9.
 
  • #27
Redbelly98
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Halls, I was trying to ask the OP that question.
 
  • #28
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  • #29
Redbelly98
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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:

[tex]x \geq 0[/tex] and [tex]y \geq 0[/tex]

So it won't be a full circle.
 
  • #31
dx
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No it doesnt.
 
  • #33
dx
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You said these were the conditions :

1. [tex] Re(z) \geq 0 [/tex]
2. [tex] Im(z) \geq 0 [/tex]
3. [tex] |z| \leq 2 [/tex]

Part of your picture is to the left of the vertical axis, so it doesnt satisfy (1). Part of your picture is below the horizontal axis, so it doesnt satisfy (2). And part of it is further away from the origin than 2, so it doesnt satisfy (3).
 
  • #34
Redbelly98
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Physicsissuef, I see you included points where x < 0 as well as y < 0. Does that not violate two of the requirements?
 

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