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?
 A: Using $z \overline z = |z|^2$, the integral becomes
$$
  I = \int_l (z^2 + 1) \, dz \, .
$$
The integrand is a holomophic function in the simply-connected domain $\Bbb C$, therefore the integral
has the same value for any path connecting $1$ with $-1$.
Choosing a straight line gives
$$
 I = \int_1^{-1} (x^2 + 1)\, dx
$$
which is now easy to evaluate.
A: Parametrize the curve $l$: $z = e^{i\theta}$, $0<\theta<\pi$. 
Then calculate the integral by the definition.
$$
\int_{l}(z^2+\bar{z}z)dz=\int_{0}^{\pi}(e^{2i\theta}+1)ie^{i\theta}d\theta =
i\int_{0}^{\pi}(e^{3i\theta}+e^{i\theta})d\theta
= i((1/3i)e^{3i\pi}+(1/i)e^{i\pi}) - i((1/3i)+(1/i)) = 
i((1/3i)(-1)+(1/i)(-1)) - i((1/3i)+(1/i)) = 
 -(8/3).
$$
A: 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.)
A: Hint : $z \bar z =|z|^2$ $$\Rightarrow \int_{l} z^2dz+|z|^2\int_{l}dz$$
