I know the notion of analytic as following
The Operator $A$ is analytic if $A=\sum_{k=0}^{\infty} \epsilon^k B_k$ where $B_k \in L(\mathcal{H})$
However, I can't understand why $$R_z (A(\epsilon))=R_z (A_0) {[I+R(\epsilon)+(z-z_0 )R(\epsilon)R_z (A_0 )]}^{-1} $$ implies $R_z (A(\epsilon))$ is analytic in $\epsilon$ uniformly with respect to $z\in \Gamma$ where $R_z(A)$ is resolvent operator, $R(\epsilon)=G(\epsilon){(A_0 - z_0 I)}^{-1}$, $A(\epsilon)=A_0+G(\epsilon),$ $\Gamma$ is contour.
I don't know how do I state but it is in "Lecture Notes on Schrodinger Equations" by Alexander Pankov, page 64, Theorem 5.4.1
Please help me.