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Sketching the closure of a set

  1. Aug 31, 2008 #1
    1. The problem statement, all variables and given/known data

    Sketch the closure of the set:Re(1/z)=< 1/2


    [ b]2. Relevant equations[/b]



    3. The attempt at a solution

    Re(1/z)=Re(1/(x+iy)

    Re(1/(x+iy))=< 1/2. Not really sure how to sketch 1/2 on a complex plane. Maybe 1/2 can be written in a complex form: 1/2= (1/2)+(0)*i=1/2 and therefore , 1/2 only lies exclusively on the x-axis.

    Not even entirely sure how to sketch the Re(1/z) on a complex plane. Generally, Re(z)=x and therefore would Re(1/z)=1/x?
     
  2. jcsd
  3. Aug 31, 2008 #2

    tiny-tim

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    Hi Benzoate! :smile:

    ok … now you need to simplify this fraction so that you can just read off the real part.

    The standard trick is to multiply top and bottom by (x - iy). :smile:
     
  4. Aug 31, 2008 #3
    so Re (1/z) = Re(1/(x+iy))=Re(1/(x+iy)*(x-iy)/(x-iy))=Re(x-iy)/(x^2-y^2)). How will this new expression make it easier for me to mapped Re(1/z) in a complex plane. In addition, how would I mapped 1/2 on a complex plane? Does 1/2=1/2+0*i
     
  5. Aug 31, 2008 #4

    tiny-tim

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    First … it's Re(1/z) = Re(x-iy)/(x^2 + y^2), not x^2-y^2 :wink:

    second … Re(x + iy) < 1/2 if … ? :smile:
     
  6. Aug 31, 2008 #5
    Only if x less than or equal to 1/2 and y is less than or equal to zero

    Wouldn't Re(x+iy/(x^2+y^2)) Be sketched only on the x-axis since all real numbers lie only on the x-axis. Would it be correct to say that 1/2 lies only only the real part of the complex plane since 1/2=1/2+i*0?
     
  7. Aug 31, 2008 #6

    tiny-tim

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    No … the value of y doesn't matter …

    Re(x+iy) = x …

    so Re(x+iy) < 1/2 just means x < 1/2 …

    so Re(x+iy) < 1/2 for x < 1/2 and for all y. :smile:
    Forget 1/2.

    And you're not sketching Re(x+iy/(x^2+y^2)).

    You're sketching x + iy such that Re(x+iy/(x^2+y^2)) ≤ 1/2.
     
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