Your proposal does not work. For instance, the set $\{a-ai:a\in\mathbb{R}\}$ is bounded above but has no least upper bound. Its upper bounds are all numbers $x+iy$ with $x+y>0$, and there is no least such number (since, for instance, there is no least value that $x+y$ can take).
For a simple correct example, just pick any bijection $f:\mathbb{C}\to\mathbb{R}$, and define an ordering $\preceq$ on $\mathbb{C}$ by $a\preceq b$ iff $f(a)\leq f(b)$, where $\leq$ is the usual order on $\mathbb{R}$. Since the usual order on $\mathbb{R}$ is complete, so is $\preceq$.
complete ordered set
. – dxiv Mar 12 '17 at 08:13