Consider the Banach space $A:=\ell^{1}(\mathbb{Z})$ together with a multiplication defined by $$(a\ast b)_{k}:=\sum_{n\in\mathbb{Z}}a_{n-k}b_{k}.$$ Then $A$ is a commutative Banach algebra (Although I don't see why $\|a\ast b\|_{1}\leq\|a\|_{1}\|b\|_{1}$, suggestions?). Given a complex number $z$ with unit length, how do I show that $$h\colon A\to\mathbb{C},\qquad h(a):=\sum_{k\in\mathbb{Z}}a_{k}z^{k}$$ is an algebra homomorphism that preserves the identity?
Unwinding the definitions, I find that $$h(a\ast b)=\sum_{k\in\mathbb{Z}}\bigg(\sum_{n\in\mathbb{Z}}a_{n-k}b_{k}\bigg)z^{k}.$$ But then I got stuck immediately.
Norm inequality:
$$\|a\ast b\| = \sum_k |(a\ast b)_k| = \sum_k \left|\sum_n a_{n-k}b_k\right| \le \sum_k \sum_n |a_{n-k}||b_k| = \sum_k |b_k|\sum_n |a_{n-k}|. $$
For a fixed $k$, the inner sum runs through all of $\Bbb Z$ so you get $\|a\|$ for that, and then you're left with $\|b\|$ for the other sum.
It's a bit easier to work backwards in my opinion for showing the homomorphism property.
$$h(a)h(b) = \left(\sum_k a_k z^k\right)\left(\sum_j b_j z^j\right) = \sum_k\sum_j a_k b_j z^{j+k}.$$
For fixed $k$, let's let $n = j+k$, then $n$ also sweeps out all of $\Bbb Z$ and also $j = n-k$, giving
$$h(a)h(b) = \sum_k\sum_n a_k b_{n-k} z^k.$$
A simple sum swap and reindex gives you $h(a\ast b)$. Note that you need the series convergence (first part) to swap sums around like this.