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Homework Help: Complex circle

  1. May 13, 2008 #1
    1. The problem statement, all variables and given/known data

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


    2. Relevant equations

    z=a+bi

    3. 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]
     
    Last edited by a moderator: Apr 23, 2017
  2. jcsd
  3. May 13, 2008 #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].
     
  4. May 13, 2008 #3

    rock.freak667

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    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.
     
  5. May 14, 2008 #4
    So [itex]|(x,y)-(1,1)| \geq \sqrt{2}[/itex].

    But what are (x,y)?

    z(x,y)
     
  6. May 14, 2008 #5

    rock.freak667

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    You just put [itex]|z-(-1-i)| \geq \sqrt{2}[/itex]
     
  7. May 14, 2008 #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...
     
  8. May 14, 2008 #7

    rock.freak667

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    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.
    |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]
     
  9. May 14, 2008 #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)
     
  10. May 14, 2008 #9

    rock.freak667

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    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]
     
  11. May 14, 2008 #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..
     
  12. May 14, 2008 #11

    rock.freak667

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    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.
     
  13. May 14, 2008 #12
    but what about x=-1 and y=-1, you think it will be correct? I don't think so...
     
  14. May 14, 2008 #13
    Isn't the center of the circle at 1+i?
     
  15. May 14, 2008 #14
    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!
     
  16. May 14, 2008 #15
    David can you please tell me your opinion?
     
  17. May 14, 2008 #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)|.
     
  18. May 14, 2008 #17

    rock.freak667

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    Seems today I keep confusing my + and - signs...... but I was hoping that the general method was correct. Seems it was wrong, sorry OP.
     
  19. May 15, 2008 #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?
     
  20. May 15, 2008 #19
    Yes, that looks right. So what would you have for c)?
     
  21. May 15, 2008 #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...
     
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