Sketch Complex Regions: |z-2+i|≤1 and Im(z)>1

[alexcc11]

Homework Statement

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

Homework Equations

z=x+iy

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?

 

[SammyS]

Homework Statement

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

Homework Equations

z=x+iy

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?

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 .

 

[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?

 

[SammyS]

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?

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

 

[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?

 

[SammyS]

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?

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.

 

[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.

 

[SammyS]

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.

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 $\ X\ .$

 

[alexcc11]

that really helps! Thanks a lot!