Given:

$\displaystyle \int_C z(\overline{z}+1)^2 dz$

where $\displaystyle C$ is the circle $\displaystyle |z+1|=3$ in a clockwise direction.

I'm thinking that the answer is simple 0 since it's composite function seems to be analytic everywhere, but it's continuous nowhere, since:

$\displaystyle (x+iy)(x-iy+1)^2 = (x+iy)(x^2-2iyx-y^2+2x-2iy+1)$$\displaystyle = x^3-iyx^2 +y^2x-iy^3+2x^2+2y^2+x+iy$ which clearly won't satisfy the Cauchy-Riemann equations