Compute $\oint_{|z|=r}z^2 \sin(\bar z)dz$

Question

As mentioned above I am interested in the value of $$\oint_{|z|=r}z^2 \sin(\bar z)dz$$ where $r>0$ although I'm somewhat helpless at the moment. What I got so far: $$z^2 \sin(\bar z)=z^2 \sum_{n=0}^\infty(-1)^n\frac{(\bar z)^{2n+1}}{(2n+1)!}=|z|^2\overline{\sum_{n=0}^\infty(-1)^n\frac{z^{2n-1}}{(2n+1)!}}$$ and therefore we can write the integral as $$\overline{\sum_{n=0}^\infty\oint_{|z|=r}|z|^2(-1)^n\frac{z^{2n-1}}{(2n+1)!}dz}$$ but this seems to just make matters worse. Any good approach to this?

Answer 1

Let $z\bar z=r^2$ we have $$\oint_{|z|=r}z^2 \sin(\bar z)dz=\oint_{|z|=r}z^2 \sin(\frac{r^2}{z})dz$$ now youse $$z^2 \sin(\frac{r^2}{z})=z^2(\frac{r^2}{z}-\frac{r^6}{3!z^3}+\frac{r^{10}}{5!z^5}-\ldots)$$ we get that

$$\oint_{|z|=r}z^2 \sin(\bar z)dz=-2{\pi}i(\frac{r^6}{3!})$$