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More Complex Analysis

  1. Mar 21, 2009 #1
    two questions here:

    (i) my notes say that [itex]\frac{1}{e^{\frac{1}{z}}-1}[/itex] has an isolated singularity at [itex]z=\frac{1}{2 \pi i n}, n \in \mathbb{Z} \backslash \{0\}[/itex]

    i can't see this though.....

    (ii) let [itex]b \in \mathbb{R}[/itex]. show

    [itex]\int_{-\infty}^{\infty} e^{-x^2} \cos{(2bx)} dx = e^{-b^2} \sqrt{\pi}[/itex]

    we take a rectangular contour [itex]\Gamma[/itex] with vertices (R,0) , (R,b) , (-R,b) , (-R,0)
    this integrand is holomorphic on the entire complex plane so Cauchy's Theorem tells us that

    [itex]\int_{\Gamma}e^{-z^2} \cos{(2bz)} dz=0[/latex

    now we define
    [itex]\gamma_1[/itex] between (-R,0) and (R,0)
    [itex]\gamma_2[/itex] between (R,0) and (R,b)
    [itex]\gamma_3[/itex] between (R,b) and (-R,b)
    [itex]\gamma_4[/itex] between (-R,b) and (-R,0)
    so that [itex]\Gamma= \gamma_1 \cup \gamma_2 \cup \gamma_3 \cup \gamma_4[/itex]

    now on [itex]\gamma_2[/itex] we use Jordan's lemma

    the length of the curve is b

    and we have [itex]|e^{-z^2} \cos{(2bz)}| \leq |e^{-z^2}|[/itex] as [itex]|\cos{(2bz)}| \leq 1[/itex]

    we parameterise [itex]\gamma_2[/itex] by [itex]z=R+it, t \in [0,b][/itex]

    so [itex]|e^{-z^2}| = |e^{-(R+it)^2}| \leq e^{t^2-R^2} \leq e^{b^2-R^2}[/itex]

    so [itex]\int_{\gamma_2} e^{-z^2} \cos{(2bz)} dz \leq be^{b^2-R^2} \rightarrow 0[/itex] as [itex]R \rightarrow \infty [/itex]

    and [itex]\gamma_4[/itex] vanishes simlarly

    my problem is how to treat [itex]\gamma_3[/itex]. is it meant to be rearranged so that we get the integral over [itex]\gamma_3[/itex] is proportional to the integral over [itex]\gamma_1[/itex]. i was going to say it ISN'T because we want it to be non-zero so that the integral over [itex]\gamma_1[/itex] doesn't end up as 0.

    this is what i've been trying:

    parameterise [itex]\gamma_3[/itex] by [itex]z=z+ib , x \in [-R,R] [/itex]

    [itex]\int_{\gamma_3} f(z) dz = \int_R^{-R} e^{-(x+ib)^2} \cos{(2b(x+ib))} dx[/itex]

    do i continue down this road or am i supposed to be using Jordan's lemma here:

    [itex]|e^{-z^2} \cos{(2bz)}| \leq |e^{-z^2}| = |e^{-x^2} e^{-2ixb} e^{b^2}| = |e^{b^2-x^2}| \leq e^{b^2}[/itex] as [itex]x \in [-R,R][/itex] and the length of this contour is b so

    we'd end up with our desired integral being equal to [itex]-be^{b^2}[/itex] which is a bit off

    any advice...........
     
  2. jcsd
  3. Mar 22, 2009 #2

    tiny-tim

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    Hi latentcorpse! :smile:

    When z = 1/2nπi, 1/z = 2nπ1, so e1/z = 1.
    Wouldn't it be easier to write cos(2bx) = Re(e2bxi), and then complete the square ? :wink:
     
  4. Mar 22, 2009 #3
    of course it would lol. i'm such a tool. i literally just finished reading that part of my notes before trying this question and completely ignored them!
    oh well ..... learn by doing examples eh?
     
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