Let $\mathcal{A}$ be a Banach algebra with $1$, let $n \in \mathbb{N}$, and let $a\in \mathcal{A}$ such that $a^n=0$.
(a) Determine $\sigma(a)$.
(b) Let $f,g \in \mathrm{Hol}(a)$. Give a necessary and sufficient condition on $f$ and $g$ such that $f(a)=g(a)$.
For (a) I have the following:
\begin{align*} a^n=0&\implies a^{n+k}=a^na^k=0\cdot a^k=0\,\,\text{for all $k\ge1$}\\ &\implies\|a^{n+k}\|=0\,\,\text{for all $k\ge1$} \end{align*}
Thus, the spectral radius of $a$ is given by \begin{align*} r(a):&=\sup\{|\alpha|:\,\alpha\in\sigma(a)\}\\ &=\lim_{n\to\infty}\|a^n\|^{1/n}\\ &=0 \end{align*} Hence, we must have that $\sigma(a)=\{0\}$. Is this correct?
For (b) I know that $f,g\in\text{Hol(a)}$ means by definition that $f$ and $g$ are analytic in a neighborhood of $\sigma(a)$ and so, by part (a), $f$ and $g$ are analytic in a neighborhood of $0$. From here, I am thinking to use the spectral mapping theorem, this would give \begin{align*} &\sigma(f(a))=f(\sigma(a))=f(0)\\ &\sigma(g(a))=g(\sigma(a))=g(0). \end{align*} But I am not sure how I would go from here to get necessary and sufficient conditions to have that $f(a)=g(a)$. Any help here would be much appreciated.
