Can we ignore the ``holomorphic trivialisation'' in the definition of a holomorphic vector bundle?

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I have learnt two definitions about holomorphic vector bundles over a complex manifold $M$.

  1. $E\to M$ is a smooth complex vector bundle with a trivialisation such that the transition functions are holomorphic

  2. $E\to M$ is a smooth complex vector bundle and $E$ is a complex manifold such that the map $E\to M$ is holomorphic (from Wikipedia)

It is easy to see that the first definition implies the second and I wanted to show that the second also implies the first.

In Wikipedia, they say

Specifically, one requires that the trivialization maps \begin{equation} \phi_U\colon \pi^{-1}(U)\to U\times\mathbb C^k \end{equation} are biholomorphic maps. This is equivalent to requiring that transition functions \begin{equation} t_{UV}\colon U\cap V\to > \text{GL}_k(\mathbb C) \end{equation} are holomorphic maps.

The problem happens to say ``the trivialisation maps $\phi_U$'s are biholomorphic'', because I think if we only assume $E\to M$ satisfies the second definition, we can only get smooth trivialisations. Even $E$ has complex coordinate charts, I can not require the chart has that special form as $\phi_U$.