Find $\oint\limits_{|z-\frac{1}{3}|=3} z \text{Im}(z)\text{d}z$ In my test on complex analysis I encountered following problem:

Find $\oint\limits_{|z-\frac{1}{3}|=3} z \text{Im}(z)\text{d}z$

So first I observed that function $z\text{Im}(z)$ is not holomorphic at least on real axis. Therefore we have to intgrate using parametrization.
First, let's change variable $w = z - \frac{1}{3}$. So we got $\oint\limits_{|w|=3} (w+\frac{1}{3}) \text{Im}(w+\frac{1}{3})\text{d}w = \oint\limits_{|w|=3} (w+\frac{1}{3}) \text{Im}(w)\text{d}w = \frac{1}{2i}\oint\limits_{|w|=3} (w+\frac{1}{3}) (w-\bar w)\text{d}w$.
Then by letting $w=3e^{i \phi}$ we transform integral to the form 
$\frac{1}{2}\int\limits_{0}^{2\pi}(3e^{i \phi}+\frac{1}{3})(3e^{i \phi}-3e^{-i \phi})ie^{i \phi}\text{d}\phi = -\frac{1}{2}\int\limits_{0}^{2\pi}\text{d}\phi=-\pi$.
Is my reasoning correct? I don't quite sure about change of variable I made since function is not holomorphic at real axis. Is there any other way how this integral can be evaluated? Thanks! 

You can also use Stokes'theorem. Since $z\mathrm{Im}(z)=\frac{1}{2i}(z^2-z\overline{z})$ and $dx\wedge dy=\frac{1}{2i}d\overline{z}\wedge dz$ we have
$$
\oint_{|z-\frac{1}{3}|=3}\frac{z^2-z\overline{z}}{2i}dz=\iint_{|z-\frac{1}{3}|\le3}-z\frac{d\overline{z}\wedge dz}{2i}=\iint_{(x-\frac{1}{3})^2+y^2\le 9}(-x-iy)dx\wedge dy=\\
=-\iint_{(x-\frac{1}{3})^2+y^2\le 9}xdx\wedge dy-i\iint_{(x-\frac{1}{3})^2+y^2\le 9}ydx\wedge dy=\\
=-Average_D(x)\times Volume(D)-iAverage_D(y)\times Volume(D)=\\
=-\frac{1}{3}\times 9\pi-i\cdot 0\times 9\pi=-3\pi
$$
where $D=\{(x,y)\in\mathbb{R}|(x-\frac{1}{3})^2+y^2\le 9\}$
