Let $f$ be an analytic function on a neighborhood $\{|z|\leq 1\}$ and $\gamma$ is a unit circle oriented counterclockwise. Show that for $0<|z|<1$ $$2\pi i f(z)=\int_\gamma \frac{f(w)}{w-z}~dw-\int_\gamma \frac{f(w)}{w-\frac{1}{\bar z}}~dw$$
My idea was to use chauchy's integral formula (we had the following version):
Let $\Omega\subset \Bbb{C}$ be an open subset. Let $f:\Omega\rightarrow \Bbb{C}$ be analytic. Suppose $\gamma:[a,b]\rightarrow \Omega$ is a closed curve or a cycle and take $z\in \Omega\setminus \gamma([a,b])$ Suppose that $\forall a\in \Bbb{C}\setminus \Omega $ $$n(\gamma,a)=0$$ where $n(\gamma, a)$ is the winding number of $\gamma$ around $a$. Then $$\frac{1}{2\pi i}\int_\gamma \frac{f(w)}{w-z}~dw=f(z)n(\gamma,z)$$
The only problem I have is that in the theorem we need $\Omega$ to be open but $\{|z|\leq 1\}$ is not open. Can I still apply the theorem?
P.s. we also had a remark that the Cauchy formula holds if $\Omega$ is simply connected. Using this remark it would work right?
Thanks for your help.
“Let $f$ be an analytic function on a neighborhood $\{|z|\leq 1\}$” means that $f$ is holomorphic in some open set $\Omega$ containing $ \{|z|\leq 1\}$. Then $$ \{|z|\leq 1\} \subset B_r(0) \subset \Omega $$ for some $r > 1$ and you can apply Cauchy's integral formula to $B_r(0)$, which is simply-connected.