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Complex circle

  • #1

Homework Statement



What will be solutions of http://i26.tinypic.com/jj360y.jpg" complex circles.


Homework Equations



z=a+bi

The Attempt at a Solution




a)
[tex]1 \leq Re(z) \leq 2[/tex]

[tex]1 \leq Im(z) \leq 2[/tex]

What about z?

b)
[tex]-2 \leq Re(z) \leq 2[/tex]

[tex]-2 \leq Im(z) \leq 2[/tex]

What about z?

c) I don't know. I think |z|=[tex]\sqrt{2}[/tex]
 
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Answers and Replies

  • #2
You are using rectangles or circles for all of them, one is a rectangle, one is an annulus and one is a disk. Let me help you with the non-rectangular ones.

An annulus centered at [tex]z_0=x_0+iy_0[/tex] with inner and outer radii [tex]r_1,r_2[/tex] should be described by [tex]r_1 \leq |z-z_0| \leq r_2[/tex]. That should help you. A disk is just an annulus with [tex]r_1=0[/tex].
 
  • #3
rock.freak667
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c) I don't know. I think |z|=[tex]\sqrt{2}[/tex]
remember that the equation for a circle in complex form is |z-a|=r where a is a fixed complex number in the form [itex]x_0 +iy_0[/itex] and r is the radius. You are correct that [itex]r=\sqrt{2}[/itex]. Then centre of the circle is given by [itex](x_0,y_0)[/itex]. You should now be able to get the proper equation for the circle.

Now to deal with the shaded region. If [itex]|z-a| \geq r[/itex] then that means for the circle |z-a|=r, you would shade everything around the circle.
 
  • #4
So [itex]|(x,y)-(1,1)| \geq \sqrt{2}[/itex].

But what are (x,y)?

z(x,y)
 
  • #5
rock.freak667
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You just put [itex]|z-(-1-i)| \geq \sqrt{2}[/itex]
 
  • #6
1) Why (-1-i)?

2)Shouldn't it be |z-a|=[itex]\sqrt{2}[/itex], since as we can see on the picture, it can't be neither lower nor bigger than r...
 
  • #7
rock.freak667
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1) Why (-1-i)?
Because in the form |z-a|=r, a is a fixed complex number in the form, [itex]a=x_0+iy_0[/itex] where the centre of the circle is [itex](x_0,y_0)[/itex]. So to write the equation correctly, you must write a in that form.
2)Shouldn't it be |z-a|=[itex]\sqrt{2}[/itex], since as we can see on the picture, it can't be neither lower nor bigger than r...
|z-a|=r is just the circle alone with nothing shaded. Everything inside is the circle is shaded. So I think it would have to be such that [itex]|z-a| \leq r[/itex]
 
  • #8
I understand. But it should be [itex]|z-a| \geq r[/itex], right? So we will shade everything inside the circle... But in this case we can't define Re(z) and Im(z), since there are some negative values for Re(z) and Im(z)
 
  • #9
rock.freak667
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I understand. But it should be [itex]|z-a| \geq r[/itex], right? So we will shade everything inside the circle...
I think it would be [itex] \leq \sqrt{2}[/itex] since the distance from the centre to any point on the circumference is [itex]\sqrt{2}[/itex] if it was greater than or equal to [itex]\sqrt{2}[/itex] then you would shade where the distance from the centre to any point is greater than or equal to [itex]\sqrt{2}[/itex]

The inside of the circle is shaded. For all the points within that circle, the maximum distance is [itex]\sqrt{2}[/itex] i.e. the point is on the circumference. Therefore for the shaded part, I think it would be [itex]|z-a| \leq r[/itex]
 
  • #10
But if we say like that, what about Re(z) and Im(z), they can receive negative values if we don't define them..
 
  • #11
rock.freak667
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But if we say like that, what about Re(z) and Im(z), they can receive negative values if we don't define them..
Well I guess you could restrict Re(z) and Im(z) if you wanted, but I think that leaving it in the circle form would still be correct.
 
  • #12
but what about x=-1 and y=-1, you think it will be correct? I don't think so...
 
  • #13
737
0
Isn't the center of the circle at 1+i?
 
  • #14
Isn't the center of the circle at 1+i?
It's worse than that, way long ago they agreed that the region was the complement of a disk, when the picture clearly shows otherwise!
 
  • #15
David can you please tell me your opinion?
 
  • #16
HallsofIvy
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I don't know if rock.freak667 didn't understand the the question or was just giving you a hard time but the answer he gave is completely wrong.
The last picture is the interior, together with the circle itself, of a disk with center at (1,1) and radius [itex]\sqrt{2}[/itex]. Every point in the figure has distance from (1,1) less than or equal to [itex]\sqrt{2}[/itex]. Written as a complex number, (1, 1) is 1+ i (NOT -1-i) and the distance from z to 1+ i is |z- (1+ i)|.
 
  • #17
rock.freak667
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I don't know if rock.freak667 didn't understand the the question or was just giving you a hard time but the answer he gave is completely wrong.
The last picture is the interior, together with the circle itself, of a disk with center at (1,1) and radius [itex]\sqrt{2}[/itex]. Every point in the figure has distance from (1,1) less than or equal to [itex]\sqrt{2}[/itex]. Written as a complex number, (1, 1) is 1+ i (NOT -1-i) and the distance from z to 1+ i is |z- (1+ i)|.
Seems today I keep confusing my + and - signs...... but I was hoping that the general method was correct. Seems it was wrong, sorry OP.
 
  • #18
Look if |z|=r
r=[itex]\sqrt{2}[/tex]

[itex]|z|=r \leq \sqrt{2}[/itex]

out from [itex]r \leq \sqrt{2}[/itex]

and |z-a|=r

[itex]|z-a| \leq \sqrt{2} [/itex]

The interior part is [itex]r \leq \sqrt{2}[/itex]

Is this correct?
 
  • #19
737
0
Yes, that looks right. So what would you have for c)?
 
  • #20
z=(x,y)

a=(1,1)

[tex]|(x,y)-(1,1)| \leq \sqrt{2}[/tex]

[tex]|(x-1,y-1)| \leq \sqrt{2}[/tex]

[tex](x-1)^2+(y-1)^2 \leq 2[/tex]

But what about x and y? Should we define them? For ex. x=-1 and y=-1 is not the part that we look for...
 
  • #21
HallsofIvy
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Looking back at your original question, I see that I have the direction of the inequality wrong (and rock.freak667 was right): you want [itex]|z-(1+i)|\ge \sqrt{2}[/itex] not less than. x and y are variables any point (x,y) satisfying that inequality will be in the set shown.
 
  • #22
HallsofIvy you were right. It is [tex]r \leq \sqrt{2}[/tex]. Just look at the picture. The place which we should find (the circles) have radius less than [itex]\sqrt{2}[/itex]...
 
  • #23
All radius which are less then [itex]\sqrt{2}[/itex] are the radius of the shadowed place..

Sp the final answer is:



[tex]
(x-1)^2+(y-1)^2 \leq 2
[/tex]

right?
 
  • #24
HallsofIvy, somebody?
 
  • #25
There is a subtle point here that HallsofIvy mentioned but you overlooked-- you are not in R^2. You are in the complex plane. You should write your sets of points in terms of one complex number instead of two real numbers. Just z instead of x and y.
 

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