# Homework Help: Determine the interior, the boundary and the closure of the set

1. Feb 11, 2014

### alexcc17

1. The problem statement, all variables and given/known data
Determine the interior, the boundary and the closure of the set {z ε: Re(z2>1}
Is the interior of the set path-connected?

2. Relevant equations
Re(z)=(z+z*)/2

3. The attempt at a solution
Alright so z2=(x+iy)(x+iy)=x2+2ixy-y2

so Re(x2+2ixy-y2)= x2-y2 >1

So would the image be a hyperbola that starts when each axis is >1 leaving a hole in the center?

Boundary: {z ε: Re(z2=1}
Interior: none
Closure: {z ε: Re(z2>1}

Since there is no interior the question of interior path connectedness is mout.

I'm not sure if this is right though

2. Feb 11, 2014

### Dick

$x^2-y^2=1$ is a hyperbola. But that's not what you have. You've got $x^2-y^2>1$. Can you figure out what that set is?

3. Feb 11, 2014

### alexcc17

The set is given {z ε: Re(z^2)>1} so x^2 -y^2=52 or anything larger than 1 which is still a hyperbola.

4. Feb 11, 2014

### Dick

You are thinking about it too hard. The boundary of your set is $x^2-y^2=1$, sketch a graph of that hyperbola and then figure out where all of the points that satisfy $x^2-y^2>1$ lie relative to that hyperbola. Then answer the interior question again.

Last edited: Feb 11, 2014
5. Feb 11, 2014

### alexcc17

So you mean every point on that hyperbola excluding where each axis is 1,-1?

6. Feb 11, 2014

### Dick

No, the set of points that satisfy $x^2-y^2>1$ includes lots of points that aren't on the hyperbola $x^2-y^2=1$. Rethink your answer about the interior.

7. Feb 11, 2014

### alexcc17

Like this?

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8. Feb 11, 2014

### Dick

Good job. Like that. What's the interior?

9. Feb 11, 2014

### alexcc17

Would the interior be: {z ε: Re(z^2)>1} so the set itself, since the boundary is what it approaches and isn't actually part of the set.

10. Feb 11, 2014

### Dick

Good enough, so now what's the boundary and what's the closure?

11. Feb 11, 2014

### alexcc17

The boundary is {z ε: Re(z^2)=1} and there is no closure since it's an open set?

12. Feb 11, 2014

### Dick

You'd better look up the definition of closure again.