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Homework Help: Simple complex integration

  1. Aug 31, 2007 #1
    Integrate [tex] \int_C \sec^2 z dz \ \mbox{any path from } \ \frac{\pi}{4} \mbox{ to } \frac{\pi\iota}{4} \\ \sec^2 z = \frac{1}{\cos^2 z} \ \mbox{ which is equal to } \ \frac{1}{2(1+\cos 2z)} \\[/tex] Therefore [tex] \frac{2}{1+\cos 2z} = \frac{2}{1 + \cos2x\cosh2y -\iota \sin2x\sinh2y}\\[/tex]
    How do you split this fraction up to differentiate it w.r.t., [tex] u_x \mbox{,} v_y \\[/tex] i.e., the Cauchy - Riemann equations? It seems to me that [tex] sec^2 z [/tex] is not analytic.
    Therefore integrate as follows:[tex] z(t) = t + (\frac{\pi}{4} -t)\iota \ \frac{dz}{dt} = 1 - \iota \\[/tex] Therefore the integral is [tex] \int_0^{\frac{\pi}{4}} \sec^2(t+(\frac{\pi}{4}-t)\iota)(1 - \iota ) dt\\[/tex]. Am I correct so far? Is z substituted in [tex] \sec^2z \ [/tex] done correctly. Thanks for helping.
     
  2. jcsd
  3. Aug 31, 2007 #2

    Dick

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    If f(z) is analytic, then f(z)^2 is analytic and 1/f(z) is analytic (away from poles). Since cos(z) is analytic, there is no point in checking CR for sec(z)^2. They will be satisfied.
     
  4. Sep 1, 2007 #3
    Cycloid question

    The parametric equations of a cycloid are [tex] x=a(\theta - \sin\theta) \mbox{ and } y=a(1-\cos\theta) \\ [/tex] where a is a constant. Show that S^2=8ay, where s is the arc length measured from the point theta = 0.

    [tex] \frac{dx}{d\theta} = a(1 - \cos\theta) \ \frac{dy}{d\theta} = a\sin\theta \\ [/tex]. Then [tex] \int \sqrt{a^2(1-\cos\theta)^2 + a^2\sin^2 \theta} d\theta \\[/tex]. Now this gives [tex] \int \sqrt{2}a\sqrt{1 - \cos\theta} d\theta\\[/tex]. I could do with a hint as to, how to solve this integral. Can I simply say [tex] s^2=2a^2 \int (1 - \cos\theta)d\theta \\[/tex]. I certainly think not. Thanks for helping.
     
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