# Greens theorem and cauchy theorem help

1. ### honghong322

5
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

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2. ### HallsofIvy

40,225
Staff Emeritus
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
$$\int_C F(z,z*)dz= 2i \int_R \frac{\partial}{\partial z*}F(z,z*) dxdy$$
where $F(z,z*)= P(x,y)+ iQ(x,y)$ and I presume you know that with z= x+ iy, z*= x- iy.

Green's Theorem says that
$$\int_C (Ldx+ Mdy)= \int_R\int \left(\frac{\partial M}{\partial x}- \frac{\partial L}{\partial y}\right)dxdy$$
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 $$\frac{\partial}{\partial z*}F(z,z*)$$ 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)"
$$\int_C \frac{g(z)}{(z-z_0)^3} dz$$
where C is a closed path enclosing z0 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
$$\int_C f(z)dz= 0$$ 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)/(z-z0)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)(z-z0)+ (g"(z0)/2)(z-z0)(2+ .... Dividing that by (z-z0)2 gives the "Laurent" series
$$\frac{g(z)}{(z-z_0)^2}= \frac{g(z_0)}{(z-z_0)^2}+ \frac{g'(z_0)}{(z-z_0)}+ \frac{g"(z_0)}{2}+ ...$$
You should be able to show that the integral of a constant time (z- z0)n
around a circle centered on z0 (use the formula z-z0= Reit with $0\le t\le 2\pi$ on the circle of radius R, center z0) 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 z0. From your contour draw a straight line to distance R from z0, go in a circle around z0, then a straight line back to the contour. Those, patched together, give you a contour that does NOT enclose z0 and so, by Cauchy's integral theorem, has integral 0. Now, move the two straight contours together so they cancel out.

3. ### honghong322

5
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.

Last edited: Sep 21, 2007
4. ### honghong322

5
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?

5. ### HallsofIvy

40,225
Staff Emeritus
That was supposed to be
$$\frac{g(z_0)}{(z-z_0)^2}= \frac{g(z_0)}{(z-z_0)^2}+ \frac{g'(z_0)}{(z-z_0)}+ \frac{g"(z_0)}{2}+ ...$$
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)/(z-z_0)^2+ g'(z_0)/(z- z_0)+ g"(z_0)+ g'''(z_0)(z-z_0)+ ...
The Taylor's series for g, divided by (z-z_0)^2.

?? Because the problem asks you to show that one integral is equal to the integral of that! The left side is an integral around the boundary and the right side is an integral over the area. That's exactly what Green's theorem is about but you have to be know the functions being integrated in order to show that Green's theorem applies!

Last edited: Sep 23, 2007