Assume that $f$ is analytic on a disk of radius $R$. Let $\gamma$ denote the positively oriented circle with center at $0$ and radius $r$, where $0 < r < R$. If $a$ is inside $\gamma$, prove that $$f(a) = \dfrac{1}{2\pi i} \int_\gamma f(z) \big(\dfrac{1}{z-a} - \dfrac{1}{z-\frac{r^2}{\overline{a}}}\big)$$
Attempt:
Let $g_1(z) = \dfrac{f(z)}{z-a}$ and $g_2(z) = \dfrac{f(z)}{z - \dfrac{r^2}{\overline{a}}}$
Then $g_1$ is analytic in the red-dotted region where $\gamma_2$ is the contour of a region centered around $a$ of radius $r_2$.
By Cauchy's theorem in the shaded region: Since $g_1$ has a singularity at $a$ : -
$2 \pi i~f(a) = \int_{\gamma_2} \dfrac{f(z)}{(z-a)}$ whereas in the case of $g_2 : z-\frac{r^2}{\overline{a}} \ne 0.$ For if, $ z-\frac{r^2}{\overline{a}} = 0 \implies z = \frac{r^2}{\overline{a}} \implies |z|~|a| = r^2.$ But $|z|,|a| < r,$ a contradiction. Thus:
$$ \int_{\gamma} \dfrac{f(z)}{z-\frac{r^2}{\overline{a}}} = 0$$
Therefore, by Cauchy's theorem: The integrals must be equal:
$\int_{\gamma} \dfrac{f(z)}{z-a} = \int_{\gamma_2} \dfrac{f(z)}{z-a}$
And thus, the result holds.
Could someone tell if I did this correctly? Thanks!