My professor recently made the claim that "if $f$ is holomorphic on the open disk $\mathbb{D}_{R}(0) = \{z \in \mathbb{C}: |z| < R\}$, then for $w = \frac{r^2}{\bar{z}}$, the integral of $\frac{f(\xi)}{\xi - w}$ vanishes over the circle of radius $0 < r < R$"
I don't see how this is possible, as we can parametrize this circle as $\xi = re^{it}$ for $0 < t < 2 \pi$, and then we get
$$\oint_{|z|=r} \frac{f(\xi)}{\xi - w}d\xi = \int_0^{2 \pi} \frac{f(re^{it})}{re^{it} - \frac{r^2}{re^{-it}}}ire^{it}dt$$ And the denominator in the argument of the latter integral is $0$.
Am I missing something here? Or perhaps perhaps I was misguided for treating $\bar{z}$ as $\bar{\xi}$ in the integral, and really $z$ is just some fixed point satisfying $|z|<r$ so that we can use Cauchy's theorem?
If $0<|z|<r$, then $|w|>r$. Choose $\rho$ such that $r<\rho< |w|$ and put $g(\xi):= \frac{f(\xi)}{\xi-w}$. The $g$ is holomorphic on $\mathbb{D}_{\rho}(0)$. Cauchy's theorem gives
$$\oint_{|z|=r} g(\xi)d\xi=0.$$