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Contour integral property

  1. Feb 6, 2005 #1
    Hi, I'm having a bit of trouble with this question.

    Use the property |integral over c of f(z)dz|<=ML
    to show |integral over c of 1/(z^2-i) dz|<=3pi/4

    where c is the circle |z|=3 traversed once counterclockwise


    thanks in advance for any tips.
     
  2. jcsd
  3. Feb 6, 2005 #2

    Janitor

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    I'm not sure what "ML" means, though presumably it is a constant. Is the "L" a subscript by chance?

    At any rate, a good start would be to factor the denominator of the integrand, and determine the poles. Once you have done that, ask yourself if the poles lie inside the circle |z|=3. That should get you well on the way to proving what you are asked to prove.
     
  4. Feb 6, 2005 #3
    I think the ML
    is M*L two different constants

    I geuss I could do the pole thing, thanks for the reply.
     
  5. Feb 7, 2005 #4

    Galileo

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    I`m sure M means the upper bound of |f(z)| on C and L is the arc length of the contour C.

    [tex]\left|\int_C f(z)dz\right|=\left|\int_a^bf(z(t))\frac{dz(t)}{dt}dt\right|[/tex].

    Now use the fact that:
    [tex]\left|\int_a^b f(t)dt\right|\leq \int_a^b \left|f(t)\right|dt[/tex]
     
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