$(a-\lambda)^{-1}$ belongs to the $C^{*}$-subalgebra generated by $a$? Let $\mathcal{A}$ be a $C^{*}$-algebra and $a \in \mathcal{A}$ be fixed. The $C^{*}$-subalgebra generated by $a$ is the closure (with respect to the norm of $\mathcal{A}$) of the set of all polynomials in $a$ and $a^{*}$.
Suppose $\lambda \in \mathbb{C}$ is an element of the resolvent of $a$, so that $(a-\lambda)^{-1}$ exists. I want to prove that this element belongs to the $C^{*}$-subalgebra generated by $a$.
I honestly don't know exactly how to start. My guess was to formally expand $(a-\lambda)^{-1}$ as a Laurent series in $a$ and argue that it is an element of the desired algebra because it is the limit of polynomials in $a$. However, this Laurent series is only convergent if $\lambda > \|a\|$, whereas the statement holds for every $\lambda$ in the resolvent of $a$, so this is probably not the best approach.
 A: I assume that $A$ is unital. The $C^*$-subalgebra generated by $a$ is denoted by $C^*(a)$. Since $ C^*(a) = C^*(a - \lambda)$, it is enough to consider the case $\lambda = 0$.  That is, we need to show:

If $a \in A$ is invertible, then $a^{-1} \in C^*(a)$.

First, let's consider the special case that $a$ is normal. Then $C^*(a)$ is commutative and we have isomorphism of $C^*$-algebras
$$C^*(a) \cong C(\mathrm{Spec}(a))$$ which maps $a$ to the identity function $t$. Since $a$ is invertible, $0 \notin \mathrm{Spec}(a)$. It follows that $1/t$ is a well-defined element in $C(\mathrm{Spec}(a))$, which is clearly inverse to $t$. It follows that $a$ is invertible in $C^*(a)$, and we are done.
In the general case, we consider the element $a \cdot a^*$. It is self-adjoint and hence normal. Also, it is invertible. Thus, the special case shows that $(a \cdot a^*)^{-1} = (a^*)^{-1} \cdot a^{-1}$ is contained in $C^*(a \cdot a^*)$. We have $ C^*(a \cdot a^*) \subseteq C^*(a)$ and $a^* \in C^*(a)$. Thus, $a^{-1} = a^* \cdot \bigl((a^*)^{-1} \cdot a^{-1}\bigr)$ is contained in $C^*(a)$. $\checkmark$
