Charles' and Pete's answer are (almost) the same: First there is a map
$\mathrm{dlog}\colon \mathcal{O}_X^\ast \rightarrow \Omega^1_X$ taking $f$ to
$df/f$ (just to show that it also makes algebraic sense) which indeed induces a
group homomorphism $H^1(X,\mathcal{O}_X^\ast)\rightarrow H^1(X,\Omega^1_X)$
giving one version of the Chern class. In the other version we have an exact
sequence $0\rightarrow 2\pi i\mathbb Z\rightarrow \mathcal O_X\rightarrow
\mathcal{O}_X^\ast\rightarrow0$ which gives a map $H^1(X,\Omega^1_X) \rightarrow
H^2(X,2\pi i\mathbb Z)$. Combined with the inclusion $2\pi i\mathbb
Z\subseteq\mathbb C$ and the projection on the $(1,1)$-part it gives the
previous Chern class. Of course the sheaf $2\pi i\mathbb Z$ is isomorphic to
$\mathbb Z$ but using the latter forces one to use the map $\mathbb Z
\rightarrow \mathbb C$ taking $1$ to $2\pi i$. It is better to use the sheaf
$2\pi i\mathbb Z$. One other reason for that is to keep track of complex
conjugation. If $X$ comes from a real algebraic variety so that it has an
antiholomorphic involution $\overline{(-)}$. Then we have
$\overline{c_1(L)}=c_1(\overline L)$ when we let complex conjugation do what it
should do on $2\pi i\mathbb Z$ (if one uses $\mathbb Z$ one has to throw in a
sign). This is completely analogous to the case of étale cohomology where the
first Chern class takes value in $H^2_{et}(X,\mathbb Z_\ell(1))$, where $\mathbb
Z_\ell(1)$ is the inverse limit of $\{\mu_{\ell^n}\}$. Similarly the $n$'th
Chern class lies most naturally in cohomology of $(2\pi i)^n\mathbb Z=(2\pi
i\mathbb Z)^{\otimes n}$ resp. $\mathbb Z_\ell(n):=(\mathbb Z_\ell(1))^{\otimes n}$.