Sketch Complex Regions

1. Jan 26, 2014

alexcc11

1. The problem statement, all variables and given/known data
Sketch the following regions and state the interior and the closure:
a) |z-2+i|≤1
b) Im(z)>1

2. Relevant equations
z=x+iy

3. The attempt at a solution
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?

2. Jan 26, 2014

SammyS

Staff Emeritus
What is the definition of the closure of a set?

For (b):

y = 1 is the solution to Im(z) = 1, not Im(z) > 1 .

3. Jan 26, 2014

alexcc11

My book says that a set is closed if it contains all boundary points, but Im(z)>1 is open, so there is no closure. Do you know if the other parts are correct, referring to graphing them?

4. Jan 26, 2014

SammyS

Staff Emeritus
You are confusing a closed set with the closure of a set.

5. Jan 26, 2014

alexcc11

How do you show the closure of a set if it's an open set? For part a, would the showing the closure just be writing the equation out since it's a closed set?

6. Jan 26, 2014

SammyS

Staff Emeritus
First of all, does your book give the definition of the closure of a set?

Secondly, your answer to (a) was correct, since the closure of a closed set is the set itself.

7. Jan 26, 2014

alexcc11

The book just says Ω is closed if Ω <=> C'Ω where Ω in C'Ω has a line above it and = {zεC: each neighborhood = D_ε(z) intersects Ω

An example is: E={zεC: .5≤|z|<1}

E_1= {zεC :|z|=1/2}
E_2= {zεC :|z|=1}
E_3= {zεC : .5<|z|<1}

The closure is:E_1 $\bigcup$E_2 $\bigcup$E_3

but I don't know who to write that with these problems.

8. Jan 26, 2014

SammyS

Staff Emeritus
Nothing there gives a definition of the closure of a set.

The given example can be helpful.

The set E itself is neither closed nor open.

Notice that $\ E_3\$ is open and is the interior of $\ E\ .$

Furthermore, $\ E_1\cup E_2\$ is the boundary of $\ E\ .$

One can also say that $\ E\cup E_2\$ is closed. It's also the closure of $\ E\ .$

The closure of a set $\ X\$ can be thought of as the "smallest" closed set which contains $\ X\ .$

9. Jan 26, 2014

alexcc11

that really helps! Thanks a lot!