Results 1 to 4 of 4

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

    My attempt : First I parametrize C by \gamma (t)=\cos t +i \sin t.
    Second, I try to find f'(z). The only method I know is to use the definition : 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

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

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

    \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 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:

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

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

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


    <br />
\begin{aligned}<br />
\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} \\<br />
&=e^{2k\pi i(-1+i)}\left(\frac{1}{i}\;z^{i}\Biggr|_{z_0}^{z_1}\r  ight) \\<br />
&=-ie^{2k\pi i(-1+i)}\left(z_1^i-z_0^i\right) \\<br />
&=-ie^{2k\pi i(-1+i)}\left(e^{-arg(z_1)}-e^{-arg(z_0)}\right),\quad 0<arg(z)<2\pi<br />
\end{aligned}<br />

    So that

    \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; August 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
    7
    Quote Originally Posted by arbolis View Post
    I must find the value of \int _{C} z^{-1+i} dz, where the integrand is such that 0<\arg z<2\pi and C is the positively oriented circle of radius 1.

    My attempt : First I parametrize C by \gamma (t)=\cos t +i \sin t.
    That's a good start, but I would keep to complex exponentials and write the parametrisation as z=e^{it}. Then dz = ie^{it}dt and z^{-1+i} = \bigl(e^{it}\bigr)^{-1+i} = e^{-t}e^{-it}. The integral then becomes \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: September 13th 2011, 07:16 AM
  2. Complex variable line integral...
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 4th 2009, 05:14 AM
  3. Complex line integral
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: September 1st 2009, 06:32 PM
  4. Proof regarding a complex line integral equality
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 30th 2009, 01:03 AM
  5. Replies: 2
    Last Post: February 7th 2009, 06:12 PM

Search Tags


/mathhelpforum @mathhelpforum