 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 $\epsilon >0$, for sufficiently large $n,k$ every coordinate difference is at most $\epsilon$. Letting $k\to \infty$ gives ${\max }_j|a_j^{(n)}-a_j|\le \epsilon$. Thus ${\mathbb{C}}^m$ with its maximum norm is complete.

Let $F(a)=\sum a_j{z}_j$. The triangle inequality givesTake $\delta =\sum \Vert z_j\Vert$. This estimate also makes $a\mapsto \Vert F(a)\Vert$ 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 givesA 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 every finite-dimensional normed complex space is a Banach space.