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I don't understand this estimation lemma example

  1. Dec 2, 2011 #1
    I don't understand this estimation lemma example :(

    We are given the 'curve',

    SUtPh.png

    And part of the example is showing that the contour integral over the top semicircle [itex]C_R[/itex] tends to zero.

    Apparently we use the estimation lemma and the fact that, [itex] |z^2 +1| \geq |z|^2 - 1 [/itex] to show,

    [itex]\left| \int_{C_R} \frac{e^{iz}}{z^2 + 1} dz \right| \leq \int_0^\pi \frac{e^{-Rsin(t)}}{R^2 - 1} dt \leq \frac{2\pi R}{R^2 - 1} \to 0[/itex] as [itex] R \to \infty [/itex]

    However I don't understand the part where [itex] e^{iz} [/itex] 'goes to' [itex] e^{-Rsin(t)} [/itex], is this some sort of parameterisation on the curve [itex] C_R [/itex]?

    Help would be much appreciated!

    Thanks
     
  2. jcsd
  3. Dec 2, 2011 #2

    micromass

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    Re: I don't understand this estimation lemma example :(

    Note that

    [itex]|e^{a+bi}|=|e^a||e^{bi}|=e^a[/itex]

    Now write out

    [tex]e^{iz}=e^{iR(\cos(t)+i\sin(t))}[/tex]
     
  4. Dec 2, 2011 #3
    Re: I don't understand this estimation lemma example :(

    Ah thank you that makes sense, but then for the next '[itex] \leq [/itex]' step surely as the arclength of [itex] C_R [/itex] is [itex] \pi R [/itex] and as [itex] e^{-Rsin(t)} [/itex] is always less than 1 for [itex] t \in [0,\pi] [/itex], then by the estimation lemma it should be [itex] \frac{\pi R}{R^2 - 1} [/itex] instead of [itex] \frac{2\pi R}{R^2 - 1} [/itex]?
     
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