Determine the interior, the boundary and the closure of the set

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alexcc17
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Homework Statement


Determine the interior, the boundary and the closure of the set {z ε: Re(z2>1}
Is the interior of the set path-connected?


Homework Equations


Re(z)=(z+z*)/2


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
 
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alexcc17 said:

Homework Statement


Determine the interior, the boundary and the closure of the set {z ε: Re(z2>1}
Is the interior of the set path-connected?


Homework Equations


Re(z)=(z+z*)/2


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

##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?
 
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.
 
alexcc17 said:
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.

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.
 
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So you mean every point on that hyperbola excluding where each axis is 1,-1?
 
alexcc17 said:
So you mean every point on that hyperbola excluding where each axis is 1,-1?

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.
 
Like this?
 
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alexcc17 said:
Like this?

Good job. Like that. What's the interior?
 
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.
 
alexcc17 said:
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.

Good enough, so now what's the boundary and what's the closure?
 
The boundary is {z ε: Re(z^2)=1} and there is no closure since it's an open set?
 
alexcc17 said:
The boundary is {z ε: Re(z^2)=1} and there is no closure since it's an open set?

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