Let $E \to M$ be a complex vector bundle over a manifold $M$ with a connection $\nabla$. Denote by $\Omega$ the curvature matrix of $\nabla$. The $k$'th Chern class is then given by
$$ c_k(E)= [f_k(i/2\pi \Omega)] $$
where $f_k(A)$ is the coefficient of $\lambda^{n-k}$ in $\det(\lambda I + A)$.
I'm wondering why isn't $$c_1(E) \equiv 0?$$
Since $\Omega$ is defined locally in an framed open set $U$ with frame $e_1,\dots,e_n$ and $f_k$'s are invariant polynomials which means that the choice of frame shouldn't affect the cohomology class we can always choose the frame to be orthonormal in which case $\Omega$ is skew-symmetric and as $f_1(i/2\pi \Omega)$ is given by the trace of the matrix $i/2\pi \Omega$ this is identically $0$.
But these are complex bundles, so the Lie algebra of the unitary group consists of skew-Hermitian matrices. On the diagonal, you have purely imaginary entries. That is why you put in an $i/2\pi$ to get a real or even integral class.
It is true though if you do the real bundle, you get skew-symmetric matrices, so you get 0's on the diagonal. That is why Pontryagin classes only happen for half of the dimensions of the Chern classes.