1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Trigonometric Laurent Series and Complex Integration

  1. Dec 23, 2012 #1
    My task is to solve the integral [itex]\frac{1}{\cos 2z}[/itex] on the contour [itex]z=|1|[/itex] using a Laurent series.

    The easy part of this is the geometric part. I drew a picture of the problem. I see that there are two singularity points that occur within the contour region at [itex]\pm \frac{\pi}{4}[/itex]. I realize that one of the problems with a Taylor series expansion of sec z is that it is only convergent for [itex]\pm \frac{\pi}{2}[/itex].

    I have tried several methods of determining the Laurent series, but I am having difficulty with this problem. I have tried two methods, and I am not certain what might be a better strategy -- or if I am going down the wrong road completely. In one method (a), I create a Taylor series, and then construct a geometric series. In the other (b), I use a trigonometric identity and derive partial fractions.

    In method (a) I considered [itex]u=2z[/itex], [itex]\frac{1}{\cos{u}}[/itex]. I changed the cosine to a Taylor Series, and then recognized the geometric series, so that:

    [itex]\sum_{j=0}^\infty \left( \sum_{k=0}^\infty \frac{z^{(2k+2)}}{(2k+1)!} \right)^j[/itex]

    In method (b) I got
    [itex] \frac{1}{2} \left( \frac{1}{1-\sqrt{2} \sin z} - \frac{1}{1+\sqrt{2} \sin z}\right)[/itex]

    Is one of these methods better to keep following -- or am I on the wrong track altogether? Also, sorry to be not as thoroughly detailed... I wrote a lot, but the system logged me out during one of my previews and I lost everything.
  2. jcsd
  3. Dec 23, 2012 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The problem here is that you're expanding about z=0. You want to find the Laurent series about each pole. I'd start by writing ##\cos 2z = \cos 2[(z\pm\pi/4)\mp\pi/4]##.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook