= Solution
If $(a^{(n)})$ is Cauchy in the <maximum norm> on $\mathbb C^m$, every coordinate is Cauchy in $\mathbb C$ and has a limit $a_j$. Given $\varepsilon>0$, for sufficiently large $n,k$ every coordinate difference is at most $\varepsilon$. Letting $k\to\infty$ gives $\max_j|a_j^{(n)}-a_j|\leq\varepsilon$. Thus \b[$\mathbb C^m$ with its <maximum norm> is <complete>].
Let $F(a)=\sum a_jz_j$. The <triangle inequality> gives
$$
\|F(a)\|\leq\left(\sum_{j=1}^m\|z_j\|\right)\|a\|_\infty.
$$
Take $\delta=\sum\|z_j\|$. This estimate also makes $a\mapsto\|F(a)\|$ <continuous> on the maximum-<norm> unit sphere, which is <compact> by coordinatewise Bolzano-Weierstrass. <Linear independence> makes this function strictly positive there, so its minimum $\gamma$ is positive. <Homogeneity> then gives
$$
\boxed{\gamma\|a\|_\infty\leq\|F(a)\|\leq\delta\|a\|_\infty.}
$$
A <Cauchy sequence> in $X$ consequently has Cauchy coordinate vectors in $\mathbb C^m$. Their coordinate limit maps by the upper estimate to a limit in $X$. Therefore \b[every finite-dimensional normed complex space is a <Banach space>].
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