Yes, just the same. You can easily check what the completion of $k(1/t)=k(t)$ is. Best to look first at the $(t)$-adic completion of the ring $k[t]$: this is the ring $k[[t]]$ of formal power series in $t$ over $k$. Then the associated complete ring is $k((t))$, the (finite-tailed) formal Laurent series over $k$. When $k$ is finite, your field is a member of the small but tremendously interesting class of locally compact but not discrete topological fields.