1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Openset / closedset / connected
  1. Connected Sets (Replies: 7)

  2. Interval => Connected (Replies: 3)

  3. Connected Sets (Replies: 6)

  4. Connected sets (Replies: 3)

  5. Connected Set (Replies: 4)