TITLE: What is the descriptive complexity of a set added by Cohen forcing?
QUESTION [9 upvotes]: I want to think of ZFC as not fully determining the powerset of the naturals, because you can add subsets with forcing and otherwise have a lot of control over the cardinality of the powerset of the naturals. But that suggests the question: what subsets does ZFC determine? There is also the related question: how complicated are the sets added by forcing?
Since Turing machines are absolute, models of ZFC should have the same hyperarithmetical sets. So $\Delta_1^1$ is absolute. 
Can Cohen forcing add a $\Pi_1^1$ set? If not, what is the descriptive complexity of sets added by Cohen forcing?
This isn't my field so I apologize if this question is ill formed or naive. References would also be appreciated.
Edit: I'm specifically interested in where a set/real $c$ added by forcing fits into the lightface hierarchy.

REPLY [8 votes]: Jensen proved in this paper that, beginning with $V = L$, it is possible to add a real $a$ by forcing such that $a$ is $\Delta^1_3$ in $L[a]$.
This result is the best possible, in the sense that one can never add $\Sigma^1_2$ or $\Pi^1_2$ reals by forcing. This follows from Shoenfield's Absoluteness Theorem (mentioned already by Asaf in the comments). In fact, Shoenfield's theorem implies that all $\Sigma^1_2$ and $\Pi^1_2$ reals are constructible; so they can't be added by forcing because they're already in the ground model. (For a proof, see Theorem 25.20 and Corollary 25.21 in Jech's book).
I do not know whether anyone has improved on Jensen's result to show that a $\Delta^1_3$ real can be added to any ground model by forcing.
Jensen's forcing is a bit difficult to understand. For an easier-to-understand example of a non-constructible $\Delta^1_3$ real, there is $0^\sharp$. Even if $0^\sharp$ does exist (which is not provable in ZFC -- it is a large cardinal axiom), it cannot be added by any set-sized notion of forcing. So $0^\sharp$ does not answer your question, but it seems related, so I thought I'd share.