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Complex analysis

  1. May 7, 2009 #1
    Hi all.

    I have some questions on complex analysis. They are very fundemental.

    1) Singularities of a complex functions are the points, where the functions fails to be analytic. Will a singularity then always be a point, where the numerator of the functions is zero?

    2) This question is on integration in the complex plane. If have a function f(z), then I have to specify the curve C (parametrized by z(t)), on which I wish to integrate f(z) along. If I just want to find the length of the curve, then which of the following integrals are correct?

    [tex]
    \int dz \quad \quad \text{or}\quad \quad \int z(t) dz.
    [/tex]

    I hope you can help. I would very much appreciate it.

    Best regards,
    Niles.
     
    Last edited: May 7, 2009
  2. jcsd
  3. May 8, 2009 #2
    Perhaps it would me more appropriate if this question was in the "Calculus and beyond"-homework forum. Would a moderator be kind enough to move the thread over there?
     
  4. May 8, 2009 #3
    are you looking at this stuff because of the born scattering section in griffiths? lol

    you are correct that a singularity of a function is a point where the function fails to be analytic but you are incorrect in saying that it's where the numerator is zero. for a rational function the singularities are where the denominator is zero. for other function it isn't necessarily so. for example [itex]y^2=x[/itex] has a singularity at (0,0) because the slope is infinite there. |x| also has a singularity at (0,0).

    for a line integral in complex space you need to specify a path z(t) where t is the parameter hence the path z is parameterized somehow by t.

    in the integral it looks like this:

    [tex] \int f(z)dz =\int f(z(t))\frac{dz}{dt}dt = \int f(z(t))z'(t)dt[/tex]

    you can think of dz as a line element and hence the arc length is just

    [tex] \int dz =\int z'(t)dt [/tex]

    edit

    actually that's wrong

    since the line integral over complex space is the integral over a sort of vector field the definition i wrote down is correct. finding the arc length though corresponds to a line integral over a scalar field hence to find the arc length it should look like this:

    [tex] \int |dz| =\int |z'(t)|dt =\int \sqrt{z'(t)*z'(t)^*}dt = \int \sqrt{u'(x(t),y(t))^2+v'(x(t),y(t))^2}dt[/tex]

    where the star in the exponent of the second z'(t) in the middle exression is complex conjugation.
     
    Last edited: May 8, 2009
  5. May 8, 2009 #4
    1) I actually meant the denominator (the lower part of the fraction), but I got the two terms mixed up.

    2) So if I want to find the arc length of the unit circle, then it is given by:

    [tex]
    \int_0^{2\pi} (\cos t + i\sin t)(-\sin t + i\cos t)dt ?
    [/tex]

    Thanks.
     
  6. May 8, 2009 #5
    if the parameterization you're using is [itex]z(t)=\cos(t)+i\sin(t)[/itex] then

    [tex] z'(t) = -\sin(t)+i\cos(t) [/tex]
    [tex] z'(t)*= -\sint(t)-i\cos(t) [/tex]
    [tex] z'(t)*z'(t)= \sin^2(t) +i\cos(t)\sin(t)-i\sin(t)\cos(t)+\cos^2(t) =\sin^2(t) +cos^2(t)[/tex]
    [tex] |dz| =\sqrt{z'(t)*z'(t)}dt =\sqrt{\sin^2(t) +\cos^2(t)}dt = dt[/tex]
     
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