Let $\mathcal{A}$ be a unital Banach algebra. I want to prove that if $a \in \mathcal{A}$ and spectrum $\sigma(a) = \sigma_{1} \cup \sigma_{2}$ where $\sigma_{1} \cap \sigma_{2} = \emptyset$, $\sigma_{1},~\sigma_{2} \neq \emptyset$. Then there exists non-trivial idempotents $E_{1}$ and $E_{2}$. When $a_{k} = aE_{k}$ we have $\sigma(a_{k}) = \sigma_{k}.$
Proof:
Let $$g_{k}(z) = \begin{cases} & 1 ~~~~~~~~~~~~\text{ if } z \in \Omega_{k} \supseteq \sigma_{k},\\ &0 ~~~~~~~~~~~\text{if } z \in \Omega_{j} \supseteq \sigma_{j}~~\text{where }j \neq k. \end{cases} $$
where $\Omega_{k} \supseteq \sigma_{k}$ are open $\Omega_{1} \cap \Omega_{2} = \emptyset$ and where $k \in \{1,2 \}$. Then $g_{k} \in \text{Hol}(a)$. Further let $E_{k} = g_{k}(a) \in \mathcal{A}$ then $$E_{1} + E_{2} = (g_{1} + g_{2})(a) = 1 \in \mathcal{A}.$$
Let $a_{k} = E_{k}a$, if $f(z) = z$ on some $\Omega \supseteq \sigma(a)$, then $f \in \text{Hol}(a)$ and $$a_{k} = E_{k}a = (g_{k} \cdot f)(a) = (f\cdot g_{k})(a) = a E_{k}.$$
Hence $E_{k}$ commutes with $a$. Further, from the spectral mapping theorem, we have $\sigma(a_{k}) = \sigma((g_{k} \cdot f)(a)) = (g_{k} \cdot f)(\sigma(a))$.
But we have $$(g_{k} \cdot f)(z) = \begin{cases} & z ~~~~~~~~~~~~\text{ if } z \in \Omega_{k} \supseteq \sigma_{k},\\ &0 ~~~~~~~~~~~\text{if } z \in \Omega_{j} \supseteq \sigma_{j}~~\text{where }j \neq k. \end{cases} $$
This implies $(g_{k} \cdot f)(\sigma(a)) =\sigma_{k} \implies \sigma(a_{k}) = \sigma_{k}$.