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! 

Note that since $\text{Im}(z)=\frac1{2i}(z-\bar z)$, that 
$$z\text{Im}(z)=\frac1{2i}(z^2-|z|^2)$$
Since $z^2$ is analytic, we have
$$\begin{align}
\oint_{|z-\frac13 |=3}z\text{Im}(z)\,dz&=\frac i2\oint_{|z-\frac13 |=3}|z|^2\,dz\\\\
&=-\frac {3}2 \int_0^{2\pi} \left|\frac13 +3e^{i\phi}\right|^2 e^{i\phi}\,d\phi\\\\
&=-\frac {3}2 \int_0^{2\pi} \left(\frac{10}9 +e^{i\phi}+e^{-i\phi}\right)e^{i\phi}\,d\phi\\\\
&=-3\pi 
\end{align}$$
