= Solution
{c}
A real or complex <sequence> $(a_n)$ is a <Cauchy sequence> if
$$
\forall\varepsilon>0\ \exists N\in\mathbb N\ \forall m,n\geq N:\quad |a_m-a_n|<\varepsilon.
$$
For $z_n=x_n+iy_n$, the <complex modulus> gives
$$
|x_m-x_n|\leq|z_m-z_n|,\qquad |y_m-y_n|\leq|z_m-z_n|.
$$
Thus the same $N$ works for both coordinate <Cauchy sequences>. Conversely, if both real coordinate <Cauchy sequences> converge to $x$ and $y$, then $|z_n-(x+iy)|\leq|x_n-x|+|y_n-y|\to0$. This proves the needed completeness for <complex numbers> from the assumed <completeness of the real numbers>.
Now suppose $\sum_{n\geq0}|z_n|$ converges. Its real <partial sums> form a <Cauchy sequence>. Given $\varepsilon>0$, sufficiently late tails therefore satisfy $\sum_{k=n+1}^m|z_k|<\varepsilon$. For $s_m=\sum_{k=0}^mz_k$, the <triangle inequality> yields
$$
|s_m-s_n|\leq\sum_{k=n+1}^m|z_k|<\varepsilon\qquad(m>n\geq N).
$$
The complex <partial sums> are a <Cauchy sequence>, hence converge by the coordinate argument. Therefore \b[<absolute convergence> implies convergence] for a complex <series>.
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