Results 1 to 4 of 4

Thread: Complex line integral

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Complex line integral

    I must find the value of $\displaystyle \int _{C} z^{-1+i} dz$, where the integrand is such that $\displaystyle 0<\arg z<2\pi$ and $\displaystyle C$ is the positively oriented circle of radius 1.

    My attempt : First I parametrize $\displaystyle C$ by $\displaystyle \gamma (t)=\cos t +i \sin t$.
    Second, I try to find $\displaystyle f'(z)$. The only method I know is to use the definition : $\displaystyle f'(z)=\lim _{h \to 0}\frac{(z+h)^{-1+i}-z^{-1+i}}{h}$. I don't know how to calculate this limit... I'd like a hint.
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jul 2008
    From
    Sofia, Bulgaria
    Posts
    75
    Hi there!

    Let

    $\displaystyle \gamma : [a,b]\rightarrow C$
    $\displaystyle t\rightarrow\gamma(t)$

    be a rectifiable, smooth enough curve. Then we have:

    $\displaystyle \int_{\gamma}f(z)dz=\int^b_af(\gamma(t))\dot\gamma dt$

    You parametrisation is correct, try also applying to it the Euler formula.

    Find out about the Cauchy-Riemann differential equations (if you think you'll need a differentiation of the integrand function).

    For the integral, check out when a holomorphic function has an antiderivative and most importantly - WHERE.


    When you're ready, compare your resuld with the Cauchy's integral formula.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Nothing is being said about the multifunction-ness of the integrand. I'm kinda' rusty, but this is how it looks to me:

    The function $\displaystyle f(z)=z^{-1+i}$ is a multifunction with an infinite number of overlapping sheets, and the description above does not distinguish which sheet the integration is being performed over. I would write:

    $\displaystyle
    \begin{aligned}
    z^{-1+i}&=e^{[(-1+i)\left[\ln|z|+i(arg(z)+2k\pi)\right]} \\
    &=e^{[2k\pi i(-1+i)]}z^{-1+i}
    \end{aligned}
    $

    with $\displaystyle z^{-1+i}=e^{i\, arg(z)(-1+i)},\quad 0<arg(z)<2\pi$.

    For example, take $\displaystyle C$ as a section of the unit circle starting at $\displaystyle z_0=e^{\pi i/4}$ and ending at $\displaystyle z_1=e^{7\pi i/4}, $ then the integrand is analytic over any contiguous section of the multifunction over $\displaystyle C,$ so I can use antiderivatives:


    $\displaystyle
    \begin{aligned}
    \mathop\int\limits_{C} z^{-1+i}dz&=e^{2k\pi i(-1+i)}\mathop\int\limits_{C} z^{-1+i}dz,\quad 0<arg(z)<2\pi,\quad k\in\mathbb{Z} \\
    &=e^{2k\pi i(-1+i)}\left(\frac{1}{i}\;z^{i}\Biggr|_{z_0}^{z_1}\r ight) \\
    &=-ie^{2k\pi i(-1+i)}\left(z_1^i-z_0^i\right) \\
    &=-ie^{2k\pi i(-1+i)}\left(e^{-arg(z_1)}-e^{-arg(z_0)}\right),\quad 0<arg(z)<2\pi
    \end{aligned}
    $

    So that

    $\displaystyle \mathop\int\limits_{C}z^{-1+i}dz=-i e^{2k\pi i(-1+i)}\left(e^{-7\pi/4}-e^{-\pi/4}\right),\quad k\in\mathbb{Z}$
    Last edited by shawsend; Aug 30th 2009 at 11:05 AM. Reason: changes domain on k to Z
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by arbolis View Post
    I must find the value of $\displaystyle \int _{C} z^{-1+i} dz$, where the integrand is such that $\displaystyle 0<\arg z<2\pi$ and $\displaystyle C$ is the positively oriented circle of radius 1.

    My attempt : First I parametrize $\displaystyle C$ by $\displaystyle \gamma (t)=\cos t +i \sin t$.
    That's a good start, but I would keep to complex exponentials and write the parametrisation as $\displaystyle z=e^{it}$. Then $\displaystyle dz = ie^{it}dt$ and $\displaystyle z^{-1+i} = \bigl(e^{it}\bigr)^{-1+i} = e^{-t}e^{-it}$. The integral then becomes $\displaystyle \oint _{C} z^{-1+i} dz = \int_0^{2\pi}\!\!\!e^{-t}e^{-it}ie^{it}dt = i\!\!\int_0^{2\pi}\!\!\!e^{-t}dt$, which I guess should give you no trouble.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Sep 13th 2011, 07:16 AM
  2. Complex variable line integral...
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Sep 4th 2009, 05:14 AM
  3. Complex line integral
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Sep 1st 2009, 06:32 PM
  4. Proof regarding a complex line integral equality
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Aug 30th 2009, 01:03 AM
  5. Replies: 2
    Last Post: Feb 7th 2009, 06:12 PM

Search Tags


/mathhelpforum @mathhelpforum