Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - Contour within Contour

  1. #1
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9

    Contour within Contour

    Let f(z) be a complex function defined on an open non-empty set S. Let \gamma (t) be a contour wholly inside S. Does there exist another contour \zeta (t) wholly inside S with the property (abusing notation) that \partial \gamma \subseteq I(\zeta) (meaning such that this contour lie strictly within this new contour).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    I have an idea how to do this, but I am a noob in topology.

    Given a contour \gamma^*. Can a circle be "shrunk" to be strictly contained in the interior \gamma^*?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,853
    Thanks
    321
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    I have an idea how to do this, but I am a noob in topology.

    Given a contour \gamma^*. Can a circle be "shrunk" to be strictly contained in the interior \gamma^*?
    Only if the region you are "shrinking" the contour over is connected. (Someone check me on this, I don't think so but it might have to be path connected.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by topsquark View Post
    Only if the region you are "shrinking" the contour over is connected. (Someone check me on this, I don't think so but it might have to be path connected.)

    -Dan
    Can you give an example of a contour whose interior is not connected?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    Can you give an example of a contour whose interior is not connected?
    By 'contour' do you mean the path is a simple closed curve?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,853
    Thanks
    321
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    Can you give an example of a contour whose interior is not connected?
    Simple. Consider \mathbb{R}^2 - \{ (x, y) |x^2 + y^2 = 1 \} and the contour which is the circle centered on the origin with radius 2. You can't shrink your contour to, say, a circle centered on the origin with radius 1/2.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Plato View Post
    By 'contour' do you mean the path is a simple closed curve?
    Yes, let \gamma (t): [a,b] \mapsto \mathbb{C} be so that:
    1) \gamma is a path, i.e. the components of \gamma (IM and RE) are continous.
    2) \gamma(a) = \gamma(b) i.e. it is closed.
    3) 0<|t_1-t_2|<b-a \to f(t_1)\not = f(t_2), i.e. it is simple.

    Quote Originally Posted by topsquark
    Simple.
    Is that supposed to be a pun! No, seriously I think you and I have a different understanding of what a contour is. See above.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Then look up the Jordan Curve Theorem.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Plato View Post
    Then look up the Jordan Curve Theorem.
    I am not looking it up because I will not understand it. But I am assuming you are saying "yes", i.e. given any contour we can draw a circle inside of it?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    The Jordan Curve Theorem states that any simple closed curve partitions the plane into three non-empty disjoint connected sets: the curve itself, a bounded set (the interior) and an unbounded set (the exterior). An arc from an interior point to an exterior point must contain a point of the curve.

    Moreover, to answer your particular question: yes about any interior point there is a closed disk (can be centered at the point) that is a subset of the interior.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,853
    Thanks
    321
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    No, seriously I think you and I have a different understanding of what a contour is.
    I think Plato ended up answering this, but to be clear I am defining a "contour" to be the same as a "curve." I believe the only difference in what we were talking about is that your contour was in \mathbb{C} and mine was in a subset of \mathbb{R}^2.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by topsquark View Post
    I think Plato ended up answering this, but to be clear I am defining a "contour" to be the same as a "curve."
    Actually I did not answer anything general about contours. I only answer what TPH was considering a contour, for him it is a simple closed curve. That is by no means standard. Most complex variables textbooks would consider a contour as the image of a continuous one-to-one mapping of [0,1]: a Jordan arc. If we do allow only f(0)=f(1) the we have a simple closed curve.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Plato View Post
    Most complex variables textbooks would consider a contour as the image of a continuous one-to-one mapping of [0,1]: a Jordan arc. If we do allow only f(0)=f(1) the we have a simple closed curve.
    You mean graduate textbooks in complex anaylis. I brought myself an undergraduate one which is why I have many questions about these contours.

    This is Mine 61th Post!!!
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    538
    Thanks
    11
    The original set must be open. If that is the case, then you can inscribe a circle inside all contours C contained in the set.

    To contruct it, well...

    - Jordan's theorem gives us that the interior of C is open, so there certainly are circles everywhere. Here, we might also drop connectedness (though topsq made an important point).

    On the other hand, that is some big theorem, and you don't normally call a bulldozer to strike a nail. There is a procedure that can produce a new curve inside this one, and more importantly - maintain the same winding number on the complement of S (as, in the important cases, S has holes where functions have poles). Take the curve to be C(t), t in [0,1]. Let n(t) be the normal to c(t), pointing inwards (since C is simple, this is always well defined). Now there is a s>0, such that sn(t) is wholly inside S (as this is open) and the contour
    C*(t)=C(t)+sn(t), t in [0,1]

    is a) inscribed in C and b) for a function f, I(C,f)=I(C*,f) on the compliment of S, so the Cauchy theorems apply.
    Last edited by Rebesques; July 8th 2007 at 02:30 PM.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Rebesques View Post
    - Jordan's theorem gives us that the interior of C is open, so there certainly are circles everywhere. Here, we might also drop connectedness (though topsq made an important point).
    I think I finally understand, thank you!

    I was discussing this with my professor by e-mail. He said that the book by Lang makes the following definition:

    Definition: Given a contour \gamma a point a is "inside" \gamma iff \oint_{\gamma} \frac{dz}{z-a} \not = 0. (I really like it).

    So we define I(\gamma) to be all points in \mathbb{C} such that the contour integral over Western Europe is non-zero.

    Now I think I understand, "open" means we can find a \delta >0 such that (the neighorbhood) \{z\in \mathbb{C} : |z - z_0|<\delta\} \subset \mathbb{I}(\gamma). So the circle |z-z_0| = \delta is what we seek. And this is possible by the Jordan Closed Curve theorem which Plato explains as saying that I(\gamma) is an open set.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. contour integral, limiting contour theorem with residue
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: May 23rd 2011, 10:00 PM
  2. Contour Integration?
    Posted in the Calculus Forum
    Replies: 0
    Last Post: May 4th 2010, 12:42 PM
  3. Contour Integral
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: December 7th 2009, 12:37 PM
  4. Contour Integration
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 30th 2009, 10:51 AM
  5. Contour Map
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 7th 2009, 05:01 PM

Search Tags


/mathhelpforum @mathhelpforum