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Openset / closedset / connected

  1. May 3, 2006 #1
    {z^2: z = x+iy, x>0, y>0}

    i am a lil confused about the notation to represent the set ...

    i'm used to seeing {z: z = x+iy, x>0, y>0}
    but what effect does squaring z have?

    i thought the set was open simply because x>0 and y>0 ... but apprently i was wrong ... (or maybe not?) ... i dunno ... i need to know what squaring that z means
     
  2. jcsd
  3. May 4, 2006 #2

    AKG

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    It's the set of squares of complex numbers with positive real and imaginary parts. Another way to write it would be:

    [tex]\{z\ :\ \exists x>0,\, y>0\ s.t.\ z = (x+iy)^2\}[/tex]
     
  4. May 4, 2006 #3
    in that case ... wouldnt it be an open set?
    and it will be above real axis? (meaning the boundary is upper plane or lower plane? getting confused with terminology a little)
     
  5. May 4, 2006 #4
    well you've got strict inequalities everywhere...
     
  6. May 4, 2006 #5

    AKG

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    The boundary is just the real line. Note it's usually good to distinguish "strict upper half plane" and "non-strict upper half plane" so you don't confuse yourself or others.
     
  7. May 6, 2006 #6
    As for the openness/closedness, [tex]z \mapsto z^2[/tex] is a holomorphic mapping, and its domain is open and connected, so...
     
  8. May 9, 2006 #7
    thats what i thought .... i wrote down "open" as my answer and the prof circles it and I dont think I got any points for it ... yes, i didnt write connected, but I should at least get half the points or something. oh well maybe he didnt gimme any credit, because I didnt explain why I think it's open set
     
  9. May 10, 2006 #8
    Yes, well the crucial point here is that you are applying the open mapping theorem, which works only when a number of conditions are satisfied. The open mapping theorem is very nontrivial and even counterintuitive, so you should properly document its application.
     
  10. May 17, 2006 #9
    I think you'll have better luck looking at the function [tex]z \mapsto \sqrt{z}[/tex]. The preimage of an open set under a continuous function is open.
     
    Last edited: May 17, 2006
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