Let $G$ be a complex linear algebraic group acting linearly on finite-dimensional complex vector space $V$.
If an orbit $G\cdot x$ is closed in the standard topology on $V$, is it also Zariski-closed?
Sometimes it is easier to see that an orbit is closed in the usual topology rather than showing that it satisfies some polynomial equations. But I am not sure if this is sufficient to infer that the orbit is Zariski-closed. I couldn't find any counterexample though.
Every Zariski-closed set is also closed in the Euclidean topology, but the converse is false in general, because the Zariski topology is much coarser than the Euclidean topology. However, for complex algebraic group actions, the orbit is closed in Zariski topology iff it is closed in Euclidean topology because orbits are constructible sets. For a proof see Borel's book on linear algebraic groups.
Borel's result: If $G$ is a complex algebraic group and $X$ is a complex algebraic variety with regular action, then each orbit $G(x)$, $x\in X$ is a smooth algebraic variety, open in its closure $\overline{G(x)}$. Its boundary $\overline{G(x)}\setminus G(x)$ is a union of orbits of strictly lower dimension. Each orbit $G(x)$ is a constructible set, hence $\overline{G(x)}$ coincides with the closure $\overline{G(x)}^d$ in the standard Euclidean topology.