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.
