can somebody explain to me why ∫(1/((z-2)^3)dz=0(integral is bounded by |z-1|=2)?????? z is complex number consder
You say the integral is "bounded by |z- 1|= 2" but this is a path integral. Do you mean that |z- 1|= 2 is the path?
If so, the only point at which $\displaystyle 1/(z- 2)^3$ is not analytic is at z= 2 which is inside the curve. That means that the integral around any simple closed path that also encloses z= 2 will be the same. In particular, the integral around, say |z- 2|= 1, the circle of radius 1 with center at 2, must have the same integral. But on the path, we can use the parameterization $\displaystyle z= 2+ e^{it}$ with t going from 0 to $\displaystyle 2\pi$. Now, $\displaystyle z-2= e^{it}$, $\displaystyle dz= ie^{it}dt$ so the integral becomes $\displaystyle \int_0^{2\pi} (e^{it})^{-3}(ie^{it}dt)= i\int_0^{2\pi} e^{-2it}dt$
Can you see that that is 0?
(More generally, if f(z) has a "pole" at $\displaystyle z= z_0$ then the integral around $\displaystyle z_0$ will be $\displaystyle \frac{1}{2\pi i}$ times the residue of f at $\displaystyle z_0$. However, the "residue" of function f at $\displaystyle z= z_0$ is the coefficient of the $\displaystyle (z- z_0)^{-1}$ in the Laurent series for f around $\displaystyle z_0$. Here the entire Laurent series for f around the z= 2 is that single term $\displaystyle \frac{1}{(z- 2)^3}= (z- 2)^{-3}$. There is NO $\displaystyle (z- 2)^{-1}$ term. Its coefficient is 0 so the residue is 0 so the integral is 0.)
OK,If i used the way to solve out the case ∫(1/(z-2))dz the path is |z|=1,then let z=exp(it) dz=iexp(it)dt so we get ∫(1/(z-2))dz = ∫(iexp(it)/(exp(it)-2) dt=ln|exp(it)-2|=0 (the integral from 0 to 2pi) is my argument right? i am not sure the step from ∫(iexp(it)/(exp(it)-2) dt to ln|exp(it)-2|here