Results 1 to 4 of 4

Math Help - complex analysis

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    2

    complex analysis

    hello. I would like some direction in how to approach these problems

    1.Calculate the line integral
    <br />
\int\limits_{-ipi}^{ipi}({e^z-z^2}) dz<br />
calculated along circle at origin & radius r=pi

    2.Find the sum of the fourier series of f(x)= 1 -x on [-pi, pi]

    the 1st i tried substituting x+ iy but it looks strange and the parametrization of the line, should I change to polar coordinates? Im trying to use z(t)= a.t + b

    the 2nd i set f(x)= 1-sinx which makes it an odd function. but Im not sure how to solve it. like how to get values of n in sin(nt)dt.
    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 function f(z)= e^{z} - z^{2} is analyitic in the whole complex plane, so that is...

    \int_{\gamma} f(z)\cdot dz =0 (1)

    ... for any closed path \gamma. That means that the integral of f(*) from -i \pi to + i \pi doesn't depend from the path connecting the limits of integration, so that we can choose the direct path along the i \omega axis and is...

    \int_{- i \pi}^ {+ i \pi} (e^{z} - z^{2})\cdot dz = i \int_{-\pi}^{+ \pi} (e^{i \omega} + \omega^{2}) \cdot d\omega = i \int_{-\pi}^{+ \pi} (cos \omega + \omega^{2}) \cdot d\omega = \frac{2}{3} i \pi^{3} (2)

    Kind regards

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

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Second question: is...

    f(x)= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} a_{n} \cos n x + b_{n} \sin n x (1)

    ... where...

    a_{0} = \frac{1}{\pi}\cdot \int_{-\pi} ^{ + \pi} (1-x)\cdot dx = 1

    a_{n} = \frac{1}{\pi}\cdot \int_{-\pi} ^{ + \pi} (1-x)\cdot \cos nx \cdot dx= 0

    b_{n} = \frac{1}{\pi}\cdot \int_{-\pi} ^{ + \pi} (1-x)\cdot \sin nx \cdot dx= 2\cdot \frac {(-1)^{n}}{n} (2)

    ... so that is...

    f(x) = 1 - 2 \sin x + \sin 2x - \frac{2}{3} \sin 3x + \dots (3)

    Kind regards

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

  4. #4
    Newbie
    Joined
    May 2010
    Posts
    2
    Thanks for your help. Could you give me an idea of what steps you used or did you just plug in formulas?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 4th 2011, 06:30 AM
  2. Replies: 6
    Last Post: September 13th 2011, 08:16 AM
  3. Replies: 1
    Last Post: October 2nd 2010, 02:54 PM
  4. Replies: 12
    Last Post: June 2nd 2010, 03:30 PM
  5. Replies: 1
    Last Post: March 3rd 2008, 08:17 AM

Search Tags


/mathhelpforum @mathhelpforum