# $(a-\lambda)^{-1}$ 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.

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 $$C^*(a)$$. $$\checkmark$$