Every smooth manifold admits a complete Riemannian metric. In fact, every Riemannian metric is conformal to a complete Riemannian metric, see this note. What about in the Kähler case?
Does a Kähler manifold always admit a complete Kähler metric?
Of course, every metric on a compact manifold is complete, so the question is only of interest in the non-compact case.
One might hope that the proof in the aforementioned note will still be of use, but as is shown in this question, a metric conformal to a Kähler metric cannot be Kähler (with respect to the same complex structure) except in complex dimension one.
The only condition that I am aware of that ensures the existence of complete Kähler metrics is that the manifold is weakly pseuodoconvex (i.e. it admits a plurisubharmonic exhaustion function), see Demailly's Complex Algebraic and Analytic Geometry, Chapter VIII, Theorem 5.2.
Best Answer
Grauert proved that a relatively compact domain with real analytic boundary in C^n has a complete Kahler metric iff the boundary is pseudoconvex .For any relatively compact domain in C^n which is the interior of its closure ,with a complete Kahler metric, Diederich and Pflug showed that it is locally Stein. See the paper of Demailly in Annales ENS vol 15 1982 page 487 .