Results 1 to 3 of 3

Math Help - Almost Cauchy-Goursat theorem

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Almost Cauchy-Goursat theorem

    Let C be the positive oriented frontier of the semi disk 0\leq r \leq 1, 0\leq \theta \leq \pi and let f(z) be a continuous function defined in the semi disk, given by f(0)=0 and f(z)=z^{\frac{1}{2}}= \sqrt r e^{\frac{i\theta}{2}}, (r>0, -\frac{\pi}{2}< \theta < \frac{3\pi}{2}).
    Prove that \int _C f(z)dz=0.
    Is the C-G theorem applicable?
    ----------------------------------------------
    My attempt : I notice I can't apply C-G theorem because f(z) is not analytic in z=0 which is a point in C.

    I've absolutely no idea how to prove what they ask me.
    I'd like an idea...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    You can avoid the singularity at the origin by making the contour C take a small detour to avoid it. More precisely, replace the straight line section of C (going from 1 to +1) by three segments defined as follows. First, a straight line segment going from 1 to ε, then a semicircle given by \varepsilon e^{i\theta}, with θ going from π to 0, and finally a straight line segment from +ε to +1. The integral round this modified contour will be 0 (by Cauchy's theorem), and as \varepsilon\to0 the integral round the modified integral will tend to the original integral round C. The reason for this is that the function f(z) is small when z is close to 0, so the integrals round parts of the contours near the origin will also be small.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Opalg View Post
    You can avoid the singularity at the origin by making the contour C take a small detour to avoid it. More precisely, replace the straight line section of C (going from 1 to +1) by three segments defined as follows. First, a straight line segment going from 1 to ε, then a semicircle given by \varepsilon e^{i\theta}, with θ going from π to 0, and finally a straight line segment from +ε to +1. The integral round this modified contour will be 0 (by Cauchy's theorem), and as \varepsilon\to0 the integral round the modified integral will tend to the original integral round C. The reason for this is that the function f(z) is small when z is close to 0, so the integrals round parts of the contours near the origin will also be small.
    Ah I see what you mean! Thanks for the perfectly clear description.

    However I'm having difficulties to formally justify that the original integral is worth 0.
    I know that near the origin the function f(z) is small.

    Oh... maybe the integral is worth 0 because f(0)=0 which makes f being continuous in the disk. If f(0) would have been different from 0 then the integral wouldn't have been worth 0... or am I mistaking?
    And of course the integral over the whole region except the origin is worth 0 as you pointed out, by Cauchy-Goursat.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Applying Cauchy-Goursat theorem/ doing it the "hard way"
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: July 20th 2010, 12:01 PM
  2. Cauchy-Goursat
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 21st 2010, 12:48 AM
  3. cauchy goursat theorm etc
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: February 28th 2010, 07:15 PM
  4. Proof Cauchy-Goursat
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 12th 2009, 08:50 PM
  5. Cauchy-Goursat Theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 14th 2008, 08:59 AM

Search Tags


/mathhelpforum @mathhelpforum