When is k[x,y]/I complete for the (x,y)-adic topology? Let $k$ be a field. If necessary, add assumptions on $k$ or just take $k=\mathbb{C}$.
It is easy to classify the ideals $I \subseteq k[x]$ such that $k[x]/I \to k[[x]]/(I)$ is an isomorphism, namely $I=(x^n)$ for some $n \in \mathbb{N}$.
Is it also possible to classify those ideals $I \subseteq k[x,y]$ such that $k[x,y]/I \to k[[x,y]]/(I)$ is an isomorphism? Clearly, $I=(x,y)^n$ is an example.
 A: Let $I$ be such an ideal. If $I\not\subset (x,y)$, then $(I)=k[[x,y]]$, so $k[x,y]/I=0$ and $I=k[x,y]$. 
Suppose now $I\subseteq (x,y)$. As $k[[x,y]]/I$ is a local ring, being quotient of a local ring, $k[x,y]/I$ is also local. So $V(I)$ is one point and is necessarily equal to $(0,0)$ as $(0,0)\in V(I)$. Therefore by Hilbert's Nullstellensatz, $\sqrt{I}=(x,y)$. Conversely, if $\sqrt{I}=(x,y)$, then $I$ contains a power of $(x,y)$, and $k[x,y]/I$ is Artian and noetherian, hence complete.
Conclusion, $k[x,y]/I\to k[[x,y]]/(I)$ is an isomorphism if and only if $\sqrt{I}\supseteq (x,y)$. 
Edit 2: More generally, let $A$ be a (commutative) noetherian ring, $\hat{A}$ the completion of $A$ with respect to a maximal ideal $ \mathfrak m$, let $I$ be an ideal of $A$. Then the canonical homomorphism $A/I\to \hat{A}/I\hat{A}$ is an isomorphism if $\sqrt{I}\supseteq \mathfrak m$. The proof is exactly the same. On the other hand, the isomorphism implies that either $I=A$ or $\mathfrak m$ is the only maximal ideal containing $I$. 
