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Thread: Integrating copmplex function (proof of one property)

  1. #1
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    Question Integrating complex function (proof of one property)

    Hello (sorry for bad usage of English)
    i have some issue with this (got to much rusty on this). I need to proof that:

    $\displaystyle \displaystyle \int c f(z) dz = c\int f(z) dz$

    (I try to do it like this... so if I'm wrong or there's "smarter" way to do it i would be very grateful... perhaps i go way back to show what integral is... don't now... please help )

    well...

    Let the complex function $\displaystyle f(z) = u(x,y) + i v(x,y) $ is defined in region $\displaystyle G\subset \mathbb{C}$ and let the $\displaystyle l\in G$ be smooth oriented curve with starting point A and end point B. We make division like so:

    $\displaystyle \pi : A = z_0, z_1, z_2, ..., z_{k-1}, z_k, ..., z_n = B$

    and make than even finer division:

    $\displaystyle \xi _k = \alpha _k + i\beta _k $ where (k = 1,2,3,...,n)

    where point $\displaystyle \xi _k $ is point between points $\displaystyle z_{k-1}$ and $\displaystyle z_k$

    we form integral sum for division $\displaystyle \pi$

    $\displaystyle \displaystyle \varsigma _\pi = \sum_{k =1}^{n}\varphi (\xi _k )(z_k - z_{k-1})$

    $\displaystyle \displaystyle \Delta z_k = z_k - z_{k-1}$

    $\displaystyle \displaystyle \Delta z_k = \Delta x_k - \Delta y_k$

    $\displaystyle \displaystyle f(\xi_k) = u(\alpha _k, \beta _k) + i v(\alpha _k, \beta_k)$

    so we pot this all back in integral sum... and have:

    $\displaystyle \displaystyle \varsigma _\pi = \sum_{k =1}^{n}\left [ u(\alpha _k, \beta _k)\Delta x_k - v(\alpha _k, \beta_k)\Delta y_k \right ] + i \sum_{k =1}^{n}\left [ u(\alpha _k, \beta _k)\Delta y_k + v(\alpha _k, \beta_k)\Delta x_k \right ]$

    $\displaystyle \displaystyle \varsigma _\pi = \max_k |\Delta z_k|$

    than we have definition...
    "For complex number I we say it's Riemman's integral of the complex function f(z) on curve l from point A to point B if and only if :
    $\displaystyle \displaystyle (\forall \varepsilon >0)(\exists \delta (\varepsilon )>0) \therefore |I-\varsigma _\pi | <\varepsilon$ for $\displaystyle \max_k |\Delta z_k|<\delta$

    $\displaystyle \displaystyle \lim_{\varsigma _\pi \to \infty} \sum_{k=1}^{n}f(\xi _k)\Delta z_k = I$

    if I exists than we can write...

    $\displaystyle \displaystyle I = \int _l f(z) dz$

    or

    $\displaystyle \displaystyle I =(l) \int _A ^B f(z) dz$

    so if this exists than if we have some complex function g(z) = c f(z) where c is constant than we can say that :

    $\displaystyle \displaystyle \lim_{\varsigma _\pi \to \infty} \sum_{k=1}^{n}c f(\xi _k)\Delta z_k = I$

    and since c is constant than c can come in front of sum and limit so ....

    $\displaystyle \displaystyle c \lim_{\varsigma _\pi \to \infty} \sum_{k=1}^{n} f(\xi _k)\Delta z_k = I$

    and we can say that we can write

    $\displaystyle \displaystyle \int c f(z) dz = c\int f(z) dz$


    I think this is way to simple to write this much (if it's true)... but I don't know and that's why i need help...

    and one more thing... how to proof that :

    $\displaystyle \displaystyle \int_{z_1} ^{z_2} A dz = A (z_2 - z_1) $

    where A is constant, but without using Newton-Leibniz formula ?!

    (it all seems to be trivial until i got start... or I'm just complicating it for myself )

    Thanks for any help
    Last edited by sedam7; Sep 2nd 2011 at 12:28 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Integrating copmplex function (proof of one property)

    Quote Originally Posted by sedam7 View Post
    Hello (sorry for bad usage of English)
    i have some issue with this (got to much rusty on this). I need to proof that:

    $\displaystyle \displaystyle \int c f(z) dz = c\int f(z) dz$

    (I try to do it like this... so if I'm wrong or there's "smarter" way to do it i would be very grateful... perhaps i go way back to show what integral is... don't now... please help )
    Could you transcribe the exact formulation of the question?. If we are talking only about indefinite integrals, the question is easier, use for example the definition: if $\displaystyle f(z),F(z)$ are analytic on a region $\displaystyle \mathcal{R}$ such that $\displaystyle F'(z)=f(z)$, then $\displaystyle F(z)$ is called an indefinite integral or antiderivative of $\displaystyle f(z)$ denoted by $\displaystyle F(z)=\int f(z)dz$ . Now, apply a well known property about derivatives.
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  3. #3
    Junior Member
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    Re: Integrating copmplex function (proof of one property)

    thanks question goes like ::

    if f(z) is integrable function, show that
    $\displaystyle \int c f(z) dz = c \int f(z) dz$

    there's nothing more to the question (and yes this one is about indefinite integrals.... second proof is for finite integrals )

    Quote Originally Posted by FernandoRevilla View Post
    if $\displaystyle f(z),F(z)$ are analytic on a region $\displaystyle \mathcal{R}$ such that $\displaystyle F'(z)=f(z)$, then $\displaystyle F(z)$ is called an indefinite integral or antiderivative of $\displaystyle f(z)$ denoted by $\displaystyle F(z)=\int f(z)dz$ . Now, apply a well known property about derivatives.
    I don't seem to follow (it looks like it's definition of the primitive function.. or just looks like to me sorry )
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