greens theorem and cauchy theorem help
1 Attachment(s)
I'm doing these in order to prepare for my quiz in a week. I have no clue where to get started or the first step in attempting problem 3 and problem 4. Please do not solve it, I just want a guide and a direction... thanks
if you guys don't mind, please download and have a look! 
You should be aware that many people will not open "word" files! (I probably shouldn't myself but I have very strong virus protection.)
Problem 3 asks you to use Green's Theorem to show that [tex]\int_C F(z,z*)dz= 2i \int_R \frac{\partial}{\partial z*}F(z,z*) dxdy[/tex] where [itex]F(z,z*)= P(x,y)+ iQ(x,y)[/itex] and I presume you know that with z= x+ iy, z*= x iy. Green's Theorem says that [tex]\int_C (Ldx+ Mdy)= \int_R\int \left(\frac{\partial M}{\partial x} \frac{\partial L}{\partial y}\right)dxdy[/tex] Again, the left integral is over a closed path while the right integral is over the area inside the path. Looks to me like you need to determine what [tex]\frac{\partial}{\partial z*}F(z,z*)[/tex] looks like in terms of partial derivatives with respect to x and y. With z= x iy, that's an exercise in using the chain rule. Problem 4 asks you to integrate, using "Cauchy's integral theorem (NOT Cauchy's integral formula)" [tex]\int_C \frac{g(z)}{(zz_0)^3} dz[/tex] where C is a closed path enclosing z_{0} and g(z) is analytic and single valued inside and on C. Of course, "Cauchy's integral formula" would give the result trivially. I presume this is an exercise in proving Cauchy's integral formula. Cauchy's integral theorem says that [tex]\int_C f(z)dz= 0[/tex] where f(z) is analytic and single valued at every point inside and on C. You can't use it directly because, of course, g(z)/(zz[sub]0_{)2 is NOT analytic at z0. You might try this: since g(z) is analytic at z0, it is equal to its Taylor series there: g(z)= g(z0)+ g'(z0)(zz0)+ (g"(z0)/2)(zz0)(2+ .... Dividing that by (zz0)2 gives the "Laurent" series [tex]\frac{g(z)}{(zz_0)^2}= \frac{g(z_0)}{(zz_0)^2}+ \frac{g'(z_0)}{(zz_0)}+ \frac{g"(z_0)}{2}+ ...[/tex] You should be able to show that the integral of a constant time (z z0)[sup]n} around a circle centered on z_{0} (use the formula zz_{0}= Re^{it} with [itex]0\le t\le 2\pi[/itex] on the circle of radius R, center z_{0}) is 0 for every n except 1. And you should be able to get a specific value for that case. You will need to use Cauchy's integral theorem to argue that integral around any such contour C is equal to the integral around a small circle with center z_{0}. From your contour draw a straight line to distance R from z_{0}, go in a circle around z_{0}, then a straight line back to the contour. Those, patched together, give you a contour that does NOT enclose z_{0} and so, by Cauchy's integral theorem, has integral 0. Now, move the two straight contours together so they cancel out. 
oh wow, this is going to take me a couple hours to soak in. Thanks for the pointers and I'll see where I can get from this. Does that last formula not show up ? I cannot see it and it's just a big red X thanks.

I'm still not understanding number 3 correctly... you told me to find d/dz^* F(Z,Z^*)
why do I have to find this? 
Quote:
[tex]\frac{g(z_0)}{(zz_0)^2}= \frac{g(z_0)}{(zz_0)^2}+ \frac{g'(z_0)}{(zz_0)}+ \frac{g"(z_0)}{2}+ ...[/tex] I don't know why it didn't show up. I've checked the LaTex and can't find anything wrong in it. It still won't show! It is g(z_0)/(zz_0)^2+ g'(z_0)/(z z_0)+ g"(z_0)+ g'''(z_0)(zz_0)+ ... The Taylor's series for g, divided by (zz_0)^2. Quote:

All times are GMT 5. The time now is 07:51 AM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums