Results 1 to 2 of 2

Math Help - Solved Complex Analysis

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    18

    Solved Complex Analysis

    \int\frac{sin(\pi.z)}{z^{2}(z-2)}dz

    This was the the problem i was give and it asks to integrate around the contours center a=-2, a=0, a=2 and radius 1.

    i) a=-2 gives and answer of 0 because it is holomorphic everywhere

    ii) a=0 gives and answer of \pi i

    iii) a=2 gives and answer of 0 as \int \frac{f(z)}{z-2} dz, when f(z)=\frac{sin(\pi.z)}{z^{2}}, = 2\pi.i \frac{sin(\pi.2)}{2^{2}}=0

    could some please check my working for me as am a bit confused about sin(\pi.z) around the contours.

    thanks bobisback
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    The essential point is that the integral can be written as...

    \int_{\gamma}\frac{\sin (\pi z)}{z^{2}\cdot (z-2)}\cdot dz= \pi \int_{\gamma} \frac{sinc (z)}{z\cdot (z-2)}\cdot dz (1)

    ... where sinc (z) = \frac{\sin (\pi z)}{\pi z} is an entire function [i.e. a fuction which is analytic in the whole complex plane...] , so that the function to be integrate has only single poles in z=0 and z=2. In this case we can apply the Cauchy's integral formula...

    \int_{\gamma} f(z)\cdot dz = 2\pi j \sum_{n} r_{n} (2)

    ... where the r_{n} are the residues of all single poles inside \gamma. Now we have three possibilities...

    a) if \gamma is the unit circle centered in a=-2 there are no poles inside \gamma and is \int_{\gamma} f(z)\cdot dz =0...

    b) if \gamma is the unit circle centered in a=0 there is a pole in z=0 inside \gamma and its residue is \lim_{z \rightarrow 0} \pi \frac{sinc (z)}{z-2} = - \frac{\pi}{2} so that is \int_{\gamma} f(z)\cdot dz = - j \pi^{2}...

    c) if \gamma is the unit circle centered in a=2 there is a pole in z=2 inside \gamma and its residue is \lim_{z \rightarrow 2} \pi \frac{sinc (z)}{z} = 0 so that is \int_{\gamma} f(z)\cdot dz = 0...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 12
    Last Post: June 2nd 2010, 03:30 PM
  2. [SOLVED] Complex Analysis: Limits!!!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 15th 2009, 06:42 AM
  3. [SOLVED] Complex Analysis Proof
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 6th 2009, 10:28 AM
  4. [SOLVED] Complex Analysis
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 29th 2009, 09:27 PM
  5. [SOLVED] complex analysis help...
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 28th 2009, 04:52 PM

Search Tags


/mathhelpforum @mathhelpforum