Sketch the following regions and state the interior and the closure:
a) |z-2+i|≤1
b) Im(z)1

Question

Sketch the following regions and state the interior and the closure:
a) |z-2+i|≤1
b) Im(z)>1

anonymous · Answer

a) z=x+iy so |x+iy-2+i|-> |(x-2)+i(y+1)|≤1 
So (x-2)2+(y+1)2≤1 

So it would just be a circle on the real plane? And the interior would be the equation with < instead of ≤ right? I'm not sure how to write the closure though.

b)Im(z)= y so it would be a straight line at y=1 on the real plane and there is no interior or closure since it is an open set right?

anonymous · Answer

@satellite73 @phi @UnkleRhaukus

anonymous · Answer

The closure of b) is the set 
$$
\{  z \in C |  Im(z) \ge 1\}
$$
So it contains the line y=1 and anything above it in the xy-plane

anonymous · Answer

@eliassaab isn't that still open though? If it's everything above y=1 then there is no set closure??

unklerhaukus · Answer

right, the solution set for b)   is not closed

for example    z_1 = 2i , 
                      z_2 = 3i .      

z-1-z_3 = -i     which is not in the solution set

anonymous · Answer

I'm a bit confused as to what it does have. Does that mean the interior is Im(z)>1? @UnkleRhaukus

unklerhaukus · Answer

i think so, but i dont really understand the concept of interior