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

  1. Feb 11, 2014 #1
    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. jcsd
  3. Feb 11, 2014 #2

    Dick

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    ##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?
     
  4. Feb 11, 2014 #3
    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.
     
  5. Feb 11, 2014 #4

    Dick

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    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
  6. Feb 11, 2014 #5
    So you mean every point on that hyperbola excluding where each axis is 1,-1?
     
  7. Feb 11, 2014 #6

    Dick

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    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.
     
  8. Feb 11, 2014 #7
    Like this?
     

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  9. Feb 11, 2014 #8

    Dick

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    Good job. Like that. What's the interior?
     
  10. Feb 11, 2014 #9
    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.
     
  11. Feb 11, 2014 #10

    Dick

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    Good enough, so now what's the boundary and what's the closure?
     
  12. Feb 11, 2014 #11
    The boundary is {z ε: Re(z^2)=1} and there is no closure since it's an open set?
     
  13. Feb 11, 2014 #12

    Dick

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