Can the complement of a non-recursive language be RE
From your question I believe there is some confusion of the terms being used here. A recognizable language L is one such that a turing machine exists accepting it,  halting on every yes instance. There are also co-recognizable languages, those are languages whose complements are recognizable. Now, if a language happens to be both recognizable and co-recognizable, it is recursive, or decidable (why?). So a language can be non-recursive, but it may still be recognizable or co-recognizable. What can you say about the emptiness problem for turing machines $E_{TM} = \{\left &lt; M \right &gt;\ :\ \mathcal L(M) = \varnothing\}$ ?