Can you prove cauchy's integral formula based on the assumptions in Conway book, operator theory, VII, 4.2?
4.2. Cauchy's Integral Formula If $\mathcal X$ is a Banach space, $G$ is an open subset of $\mathbb C$, $f : G \to \mathcal X $ is analytic, $\gamma$ is a closed rectifiable curve in $G$ such that $n(\gamma;a)=0$ for every $a \in \mathbb C\setminus G$, and $\lambda \in G \setminus \{\gamma\}$, then for every integer $k\geq 0$, $$ n(\gamma;\lambda)f^{(k)}(\lambda) = \frac{k!}{2\pi i} \int_\gamma (z-\lambda)^{-(k+1)}f(z)dz $$
This is immediate from the definitions, plus the ordinary scalar-valued Cauchy's Integral Formula.
Suppose $\Lambda\in X^*$, and let $g=\Lambda\circ f$. So $g$ is analytic (by definition or not, depending on which definition of "analytic" you took). CIF shows that $$\Lambda(n(\gamma;\lambda)f(\lambda))=n(\gamma;\lambda)g(\lambda)=\frac1{2\pi i}\int_\gamma\frac{g(z)\,dz}{z-\lambda}=\frac1{2\pi i}\int_\gamma\frac{\Lambda\circ f(z)\,dz}{z-\lambda},$$and by definition this shows that $$n(\gamma;\lambda)f(\lambda))=\frac1{2\pi i}\int_\gamma\frac{f(z)\,dz}{z-\lambda},$$which is what you want for $k=0$.
For $k>0$ it's the same, after you prove by induction that $$g^{(k)}=\Lambda\circ f^{(k)}.$$