Is the Set {z^2: z = x+iy, x>0, y>0} Open or Closed?

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{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 don't know ... i need to know what squaring that z means
 
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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:

\{z\ :\ \exists x>0,\, y>0\ s.t.\ z = (x+iy)^2\}
 
in that case ... wouldn't 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)
 
sweetvirgogirl said:
in that case ... wouldn't 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)

well you've got strict inequalities everywhere...
 
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.
 
As for the openness/closedness, z \mapsto z^2 is a holomorphic mapping, and its domain is open and connected, so...
 
Tantoblin said:
As for the openness/closedness, z \mapsto z^2 is a holomorphic mapping, and its domain is open and connected, so...
thats what i thought ... i wrote down "open" as my answer and the prof circles it and I don't 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
 
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.
 
I think you'll have better luck looking at the function z \mapsto \sqrt{z}. The preimage of an open set under a continuous function is open.
 
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