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...$\displaystyle \displaystyle (\forall \varepsilon >0)(\exists \delta (\varepsilon )>0) \therefore |I-\varsigma _\pi | <\varepsilon$

"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 :

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