Let $X$ be a holomorphic vector bundle over $Y$ (where $Y$ is an arbitrary complex manifold, not necessary projective). Does there exist an analytic subset $Z$ of $Y$ such that the restriction of $X$ to $Y \setminus Z$ is a trivial vector bundle?



non-trivial locus of  a holomorphic vector bundle
If your manifold is complex projective, then the answer is yes. Otherwise it is no. You can take a $K3$ surface without complex curves and just consider its tangent bundle. Of curse it will stay *holomorphically* non-trivial, if you throw away finite number of points from $K3$, since any holomorphic vector field on a $K3$ surface defined outside a finite set is zero.


On the other hand, I guess, in this example the bundle is topoligcally trivial, since its first Chern class is zero. 


**Added.** Let me sketch the proof of the statement, everything holds for compact complex aglebraic manifold. First every such a manifold admits a blow up that is projective. Pull back the complex bundle to the blow up. Then we get a holomorphic bundle over a complex projective manifold and such a bundle is algebraic. Hence it has meromorphic sections. Moreover we can chose meromorphic sections that a linearly independent at one point. It is clear that they trivialise the bundle over a complement to a complex analytic manifold.


