Let $M$ be a smooth real manifold and $B$ an Hermitian vector bundle over it. Then one can define Chern classes as
$$c(B)=\sum c_i(B)t^i=\det \left( I+\frac{it\Omega}{2\pi} \right) \in H_{DR}^*(M),$$
where $\Omega$ is the curvature form of any connection $\nabla$ on $B$.
I am wondering why it is necessary for $B$ to be Hermitian. My ideas are:
1) Otherwise it will depend on the choice of a connection. But one can consider $(M, g)$ Riemannian and take as $\nabla$ Levi-Civita connection, then functoriality and Whitney sum formula seem to still work.
2) There is something complex in the idea of curvature. It seems strange that the curvature is purely imaginary (because $i\Omega$ should lie in real de-Rham cohomology). Maybe, connection should be compatible with Hermitian metric?
Let $B \longrightarrow M$ be a rank $k$ complex vector bundle. Note that a choice of Hermitian metric on $B \longrightarrow M$ reduces the structure group from $\mathrm{GL}(k,\Bbb C)$ to $\mathrm{U}(k)$. Any other choice of Hermitian metric will give an isomorphic reduction. Hence the choice of Hermitian metric does not make any difference in our study of $B \longrightarrow M$ as a complex vector bundle.
To define the Chern polynomial as a polynomial with real coefficients, we need to assume $\nabla$ is a Hermitian connection with respect to our chosen Hermitan metric. The reason for this is as follows. An arbitrary connection $\nabla$ will have curvature $\Omega$, which will be a $\mathfrak{gl}(B)$-valued $2$-form. Here $\mathfrak{gl}(B)$ denotes the space of bundle endomorphisms of $B \longrightarrow M$. On the other hand, if $\nabla$ is a Hermitian connection, then the curvature $\Omega$ will be a $\mathfrak{u}(B)$-valued $2$-form, where $\mathfrak{u}(B)$ denotes the space of skew-Hermitian bundle endomorphisms of $B \longrightarrow M$. The eigenvalues of skew-Hermitian endomorphisms are all purely imaginary, so that for Hermitian $\nabla$, the Chern polynomial has real coefficients.