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

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.
 
  • #10
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?
 
  • #11
The boundary is {z ε: Re(z^2)=1} and there is no closure since it's an open set?
 
  • #12
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.
 
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