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The spectrum of an element $f$ in a Banach algebra is defined as the set $$\sigma(f) := \left\{ \lambda \in \mathbb{C} \space\ \big| \space\ f - \lambda \space\ \text{is not invertible} \right\}$$

I have figured out the non-trivial idempotents (satisfies $f^2 = f$) have spectrum $\left\{ 0 , 1 \right\}$, and the nilpotents (satisfies $f^n = 0$ for some $n$) have spectrum $\left\{ 0 \right\}$. But I am unsure what can be said about the spectrum of a cyclic element. An element is said to be cyclic when $f^n = 1$ for some $n$.

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    $\begingroup$ If $f$ is a normal element, you can use the spectral mapping thoerem. This might give some inspiration. $\endgroup$ Apr 5, 2017 at 14:12

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Let $e$ be the unit for your algebra. Then $$ (x-\lambda e)(x^{n-1}+\lambda x^{n-2}+\cdots+\lambda^{n-2}x+\lambda^{n-1}e)=(1-\lambda^n)e. $$ From this it follows that $x-\lambda e$ is invertible if $\lambda^n \ne 1$, with inverse $$ (x-\lambda e)^{-1}=\frac{1}{1-\lambda^n}(x^{n-1}+\lambda x^{n-2}+\cdots+\lambda^{n-2}x+\lambda^{n-1}e). $$ So the spectrum of $x$ is a subset of $\{ \lambda : \lambda^n =1 \}$, but it may be a proper subset, which you can demonstrate with a diagonalize matrix in $\mathbb{C}^{N}$.

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