Evaluate:$\int_{l}(z^2+\bar{z}z)dz$ 
Evaluate:
  $\displaystyle=\int_{l}(z^2+\bar{z}z)dz\,\,\,\,\,\,\,\,\,\,l:|z|=1,0\leq\arg z\leq\pi$

My try:
$$\int_{l}(z^2+\bar{z}z)dz=\int_{0}^{\pi}(r^2e^{2i\theta}+re^{i\theta}???)dz$$
I'm stuck here, hints?

Since the integral equals $\int_l (z^2+1)\, dz,$ and this integrand has the antiderivative $z^3/3 + z$ (in all of $\mathbb C$), all you need to do is evaluate $z^3/3+z$ at the end points and subtract. (All we're using is the FTC of ordinary calculus here.)
