Let $f$ be holomorphic on the annulus $\{z:\;1-\epsilon<|z|<1+\epsilon\}$. Define: $$\phi:\;D\to\mathbb{C};\quad \phi(w)=\frac{1}{2\pi i}\int_C \frac{f(z)}{z-w}\,\mathrm{d}z$$ where $C$ is the unit circle, $D$ is the open unit disk. I want to show that $\phi$ is the restriction of a holomorphic function defined on $\{z:\;|z|<1+\epsilon\}$
Work:
I have proved that $\phi$ is indeed analytic on $D$ via a series expansion & series/integral swap using uniform convergence. I also know that $f$ must have a Laurent series on compact subsets of the annulus, so write: $$f(w)=\sum_{-\infty}^{\infty} c_k w^k\;\text{where}\;c_k=\frac{1}{2\pi i}\int_C\frac{f(z)}{z^{k+1}}\mathrm{d}z$$ Then reversing the series expansion I did, I get for $w\in D:$ $$f(w)=\phi(w)+f_{\text{principal}}(w)$$ I'm not really confident about what I'm doing. Am I along the right lines?
Hint: For $1-\epsilon<r<1+\epsilon$ define $\phi_r(w)$ for $|w|<r$ by $$\phi_r(w)=\frac1{2\pi i}\int_{C_r}\frac{f(z)}{z-w}dz,$$where $C_r$ is the circle $|z|=r$. Show that if $|w|<r_1<r_2$ then $\phi_{r_1}(w)=\phi_{r_2}(w)$.