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Math Help - evaluating complex integrals (probably cauchy formula)

  1. #1
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    evaluating complex integrals (probably cauchy formula)

    1.Evaluate the following integrals
    {\int _{\partial B_2(i)}}\frac{e^{i\zeta}}{\zeta ^{4} - 4\zeta ^2} d\zeta

    {\int _{\partial B_3(-1)}}\frac{\zeta^{3}}{\zeta ^{2} + \zeta^2 -2} d\zeta

    {\int _{\partial B_1(1)}}\frac{sin(\pi\zeta)}{\zeta ^{2} - 1} d\zeta


    6. let z_0  \epsilon B_r(z_0), let  r>o and let  f:B_r[z_0] \to \mathbb{C} be continous such that f|_{B_r(z_0)} is holomorphic. Show that
    f(z) = \frac{1}{2\pi i} \int _{\partial B_r(z_0)} \frac{f(\zeta)}{\zeta - z}d\zeta

    for all  z \epsilon B_r(z_0)
    (hint: for  \theta \epsilon (0,1), apply the cauchy integral formula to B_{\frac{r}{\theta}}(z_0) \ni z \mapsto f(\theta(z-z_0)+z_0); then let \theta \to 1.)

    wow this was a lesson in the use of these symbols and latex or w/e
    anyways, thats whats left of my assignment due this time tomorrow
    if someone could help me get on track I'd appreciate it
    #6 looks like the original cauchy statement but i think it's saying that the border may not be holomorphic
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  2. #2
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    I got #6 myself
    so only need 1, a/b/c now
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  3. #3
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    Maybe some are unfamiliar with your notation like I was. Ok, let's assume it's what I think it is: \partial B_2(i) means the boundary of a circular disc with center at i and radius 2. Just never seen that notation before. If so then I'll write it as:

    \mathop\oint\limits_{D(i,2)}\frac{e^{iz}}{z^4-4z^4}dz

    (using \zeta makes it just unnecessarily cluttered)

    So that contour encloses only the pole at the origin and you want to solve it using Cauchy's Integral Formula:

    f^{(n)}(w)=\frac{n!}{2\pi i}\mathop\oint\frac{f(z)}{(z-w)^{(n+1)}}dz

    where f(z) is analytic in an on the contour. Then:

    \mathop\oint\limits_{D(i,2)}\frac{e^{iz}}{z^4-4z^4}dz=\oint\frac{\frac{e^{iz}}{(z^2-4)}}{(z-0)^2}dz=2\pi i\frac{d}{dz}\left(\frac{e^{iz}}{z^2-4}\right)\Bigg|_{z=0}=\pi/2

    Here's a check in Mathematica:

    Code:
    In[3]:=
    NIntegrate[(Exp[I*z]/(z^4 - 4*z^2))*2*I*
        Exp[I*t] /. z -> I + 2*Exp[I*t], 
      {t, 0, 2*Pi}]
    N[2*Pi*I*D[Exp[I*z]/(z^2 - 4), z] /. z -> 0]
    
    Out[3]=
    1.5707963267949334 - 8.459899447643693*^-14*I
    
    Out[4]=
    1.5707963267948966
    Do the other ones just like that but I bet you figured it out already.
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  4. #4
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    ok, I kept getting confused at the denominator being 0 but I guess that only matters on the border?
    and when I do part b) I get two different solutions depending on what I take as f
    example:

    2\pi i f(1) = \frac{2}{3}\pi i for f(z) = \frac{z^3}{z+2}
    and
    2\pi i f(-2) = \frac{16}{3}\pi i for f(z) = \frac{z^3}{z-1}

    is there a reason why I wouldn't use one or the other? or can I not use this approach for some reason
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  5. #5
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    Hey, the denominator is never zero. It's like asking if the denominator is zero for \int_1^2 \frac{1}{x}dx. Never gets ther right? Same dif with the integration path for the first one: D(i,2)=\{z : z=i+2e^{it},\;0\leq t \leq 2\pi\}. That never gets to the point 0 or \pm 2. For the second one:

    \mathop\oint\limits_{D(-1,3)}\frac{z^3}{2z^2-2}dz=\frac{1}{2}\left[\oint\frac{\frac{z^3}{z-1}}{z+1}dz+\oint\frac{\frac{z^3}{z+1}}{z-1}dz\right]

    Where the contours on the right side are now only going around each respective pole in accordance with the requirements of Cauchy's Integral Theorem (we can split up D(-1,3) that way without problems right? Now:

    =\frac{1}{2}\left[2\pi i\left(\frac{z^3}{z-1}\Bigg|_{z=-1}+\frac{z^3}{z+1}\Bigg|_{z=1}\right)\right]

    =\pi i(1)=\pi i

    Also . . . until you get good with these, check them numerically:

    Code:
    In[1]:=
    NIntegrate[(z^3/(2*z^2 - 2))*3*I*Exp[I*t] /. 
       z -> -1 + 3*Exp[I*t], {t, 0, 2*Pi}]
    
    Out[1]=
    -2.220446049250313*^-16 + 3.1415926306916244*I
    If the numerical results don't agree with the symbolic results, and you're pretty sure you've set up the numerical calculations correctly, look for an error in the symbolic results.
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  6. #6
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    \mathop\oint\limits_{D(-1,3)}\frac{z^3}{z^2 + z-2}dz

    I screwed up on the original the denominator only one of them is squared
    I'll try a few things though to get this
    and unfortunately I have no way of checking these numerically since in pure math we don't use mapl or matlab or mathematica or anything so I don't have those
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