Question: My function is $z^{2} \hat z^{3}$
I don't know if this can be:
Taking the radio circle 2
\begin{equation}
\int z^{2} \hat z^{3} dx = \int \frac{64}{z} dx 
\end{equation}
This because $z \hat z= |z|^2 = 4$ but this happens twice, so $z \hat z z \hat z= |z|^2 = 16$ y then there is a $\hat z$ without $z$ so $z \hat z = 4 \rightarrow \hat z = \frac{4}{z}$ 
but I do not know if this is valid

Answer: 
First, $z^2\bar z^3=(z\bar z)^2\bar z=|z|^4\bar z$ so using the unit circle $\gamma=e^{it}$ gives
$$ \int\_{\gamma}|z|^4\bar z dz=\int\_0^{2\pi} e^{-it}ie^{it}dt=2\pi i$$
In your case the integrand is

$$(\rho e^{i\theta})^2(\rho e^{-i\theta})^3de^{i\theta}=2^5e^{-i\theta}de^{i\theta}$$ giving the integral

$$\left.32i\right|\_0^{2\pi}.$$