# Openset / closedset / connected

## Main Question or Discussion Point

{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

AKG
Homework Helper
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 ... 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)

sweetvirgogirl said:
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)
well you've got strict inequalities everywhere...

AKG
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.

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 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

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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