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. $\mathbb Z_\ell(n):=(\mathbb Z_\ell(1))^{\otimes n}$.