1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Simple complex integration
Loading...