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.