You say the integral is "boundedby |z- 1|= 2" but this is a path integral. Do you mean that |z- 1|= 2isthe path?

If so, the only point at which is not analytic is at z= 2 which is inside the curve. That means that the integral around any simple closed path thatalsoencloses 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 with t going from 0 to . Now, , so the integral becomes

Can you see that that is 0?

(More generally, if f(z) has a "pole" at then the integral around will be times theresidueof f at . However, the "residue" of function f at is the coefficient of the in the Laurent series for f around . Here the entire Laurent series for f around the z= 2isthat single term . There is NO term. Its coefficient is 0 so the residue is 0 so the integral is 0.)