$\oint_{C}(A-\lambda I)^{-1}\,d\lambda=0$ implies interior of $C$ is in the resolvent. Suppose that $A$ is a bounded linear operator on a complex Banach space $X$ with resolvent set $\rho(A)$. If $C$ is a simple closed smooth curve in $\rho(A)$ such that
$$
                  \oint_{C}(\lambda I -A)^{-1}\,d\lambda =0,
$$
then show that the the interior $\mbox{Int}(C)$ of $C$ is contained in the resolvent set $\rho(A)$. No assumptions are made about the spectrum of $A$ or about $A$.
NOTE: I discovered a proof, but it's very contorted, and it seems to me that there should be a short proof. I feel that it should be almost obvious from the analytic functional calculus. This is not something I found in a book or that was part of homework.
 A: This is a corollary from Riesz functional calculus and spectral mapping theorem and the proof below is standard( as you can find it in many textbooks on functional analysis ).
Setup 1:
Let $\sigma(A)$ be the spectrum of $A$. We can write $\sigma(A)=K_1\cup K_2$, where $K_1$ is contained in $Int(C)$ and $K_2$ is outside the closure of $Int(C)$. 
Setup 2:
Since the spectrum of $A$ is a compact set, whence $K_1$ and $K_2$ are compact. Thus we can find open sets $U_1, U_2$ s.t. 
$K_1\subset  Int(C)\subset U_1$ and $K_2\subset U_2\subset \text{exterior of } \bar{U}_1$. 
Moreover, we pick up two simple closed curves $\Gamma_i\subset U_i$, s.t. $K_i$ is enclosed in $\Gamma_i (i=1,2)$.
 Also we define $\Gamma=\Gamma_1\cup \Gamma_2.$
Then  $\sigma(A)\subset Int(\Gamma)\subset U=U_1\cup U_2$  and we can do functional calculus with analytic functions on $U$.
Proof:
We define $f(\lambda)=\chi_{U_1}(\lambda)$, namely, the characteristic function on $U_1$. Then we see that $f$ is analytic on $U$ because it is constant on $U_i(i=1,2)$ and $U_1, U_2$ are apart, i.e., $dist(U_1, U_2)>0$. 
Since  $f(\lambda)$ vanishes outside $U_1$ and $\Gamma_1\subset U_1$,  we see that
$$f(A)=\int_{\Gamma} \chi_{U_1}(\lambda)(\lambda I-A)^{-1}d\lambda
=\int_{\Gamma_1} (\lambda I-A)^{-1}d\lambda= \int_C (\lambda I-A)^{-1}d\lambda = 0,$$
where the last equality is from your assumption.
On the other hand, since $f\in H(U)$, from spectral mapping theorem, we have
$$\sigma(f(A)) = f(\sigma(A)) = \chi_{U_1}(\sigma(A)) = \chi_{U_1}(K_1)\cup\{0\}.$$ But $f(A)=0$ (whence $\sigma(f(A))=\{0\}$) and $K_1\subset U_1$, so we must have $K_1$ equal to empty. 
Therefore, $Int(C)\cap\sigma(A)\subset U_1\cap\sigma(A)=K_1$ is empty,
i.e., $Int(C)\subset \rho(A).$
