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.
You are right about the computation of the spectrum. Nevertheless, the Gelfand-Beurling formula is a bit of an overkill. A more elementary way to see this is the polynomial spectral theorem: if $p(z)$ is a polynomial, then we know that $\sigma(p(a))=\{p(z):z\in\sigma(a)\}$. Now take $p(z)=z^n$. Since $p(a)=a^n=0$, we have that $\{z^n:z\in\sigma(a)\}=\sigma(0)=\{0\}$, so $z^n=0$ for all $z\in\sigma(a)$, i.e. $z=0$, so $\sigma(a)=\{0\}$.
Now Cameron Williams is right about the second part. Since $f(a)=g(a)$ iff $(f-g)(a)=0$, it suffices to find a sufficient and necessary condition for $f(a)=0$ where $f\in\text{Hol}(a)$. Also, we now assume that $n$ is the least integer such that $a^n=0$ (that means that $a,a^2,\dots, a^{n-1}$ are all non-zero. You can always reduce to this case.
Since $f\in\text{Hol}(a)$, we have that $f$ is holomorphic at a disc around $0$, say $D(0,r)$. We can thus write $f$ as a power-series centered at $0$ for $z\in D(0,r)$, i.e. $$f(z)=\sum_{k\geq0}b_kz^k$$ for all $z\in D(0,r)$. Note that $f(a)$ is by definition $$f(a)=\sum_{k\geq0}b_ka^k=\sum_{k=0}^{n-1}b_ka^k$$ So, if $f(a)=0$, then $b_01+b_1a+\dots+b_{n-1}a^{n-1}=0$. Multiplying with $a^{n-1}$, we get that $b_0a^{n-1}=0$, hence $b_0=0$. So our initial equation becomes $b_1a+\dots+b_{n-1}a^{n-1}=0$. Multiplying with $a^{n-2}$ we get that $b_1a^{n-1}=0$, so $b_1=0$. Proceeding like this we see that $b_0=b_1=\dots=b_{n-1}=0$.
Conversely, if $b_0=\dots=b_{n-1}=0$, then it is evident that $f(a)=0$, by its definition.
To conclude: $f(a)=g(a)$ if-f the first $n$ terms in the power series of $f$ and $g$ around $0$ are equal, as Cameron Williams said in the comments.