TITLE: Holomorphic functions on a complex compact manifold are only constants
QUESTION [12 upvotes]: Is there a simple proof that every holomorphic function $M\to\mathbb{C}$ on a compact complex manifold $M$ is constant?

REPLY [5 votes]: proof. ‎$‎\left| f \right|:X \to R‎$‎ is a continuous function, ‎$X‎‎$ ‎is ‎compact, ‎‎$‎‎\Rightarrow‎ \left\{ {\left| {f\left( x \right)} \right|:x \in X} \right\}‎$‎ is
compact, hence bounded. Thus, there is an ‎$‎x_0  \in X$‎ such that ‎$‎‎\left| {f(x_0 )} \right| = M$ ‎is ‎maximal. ‎Let ‎‎$‎‎a = f(x_0 ) \in \mathbb{C}‎$ .‎ ‎‎Obviously, ‎$‎‎f^{ - 1} (a)$‎ is closed in ‎$X$‎ (it is a pre-image of a point); if
we are able to show that $‎‎f^{ - 1} (a)$ is open, too, then $‎‎f^{ - 1} (a) = X$, that implies ‎$‎‎f(x) = a~~~~\forall x \in X$‎. 
Let $x \in ‎‎f^{ - 1} (a)$ and ‎$ (U,z) $ be a chart with ‎$ z(x)=0 $‎. Then ‎$ F:foz^{ - 1} :z(U) \to ‎\mathbb{C}‎ $‎ is
holomorphic on the open subset ‎$ z(U) \subseteq C^n ;F(0)=f(x) = a $ ‎and‎ ‎$ \left| F \right| $‎ has a maximum in ‎$ z=0 $.‎
Let‎ ‎$ \varepsilon  > 0 $ ‎such ‎that‎ ‎$ B_\varepsilon  : = \left\{ {y \in C^n :\left\| y \right\| < \varepsilon } \right\} \subseteq z(U) $. ‎For‎ ‎$ y \in B_\varepsilon  $‎, the function ‎$ g(t): = F(ty) $‎ is holomorphic on ‎$ \left\{ {t \in C:\left\| {ty} \right\| < \varepsilon } \right\} $ ‎and‎ ‎$ 
\left| g \right| $‎ takes its maximum in ‎$ t=0 $.‎ By the "maximum principle", ‎$ g $‎ is constant ‎$  \Rightarrow a = g(0) = g(1) = F(y) $‎, that means ‎$ F(y) = a~~~ \forall y \in B_‎\varepsilon‎ $‎. Hence ‎$ f \equiv a $ ‎on‎ ‎$ z^{ - 1} (B_\varepsilon  ) $‎, an open subset of ‎$ X $‎ containing ‎$ x $‎. Hence ‎$ f^{ - 1} (a) $‎ is open.