Supposed counterexample to Liouville's theorem I'm trying to understand Liouville's theorem, and I don't see why $f(z)=e^{-|z|^2}$ isn't a counterexample.  It's bounded ($0 < f(z) \leq 1$), so it must be somehow that it's not holomorphic.  Isn't it differentiable everywhere?
 A: Hint: $|z|^{2}=z\bar{z}$ but a complex function of complex variable $z\to\bar{z}$ is not differentiable (although its real and imaginary part are differentiable when considered as real functions).
A: No, it's nowhere differentiable (except at the origin). First off, the modulus function $f(z)=|z|$ is nowhere differentiable in $\mathbb{C}$, and since any point (other than the origin) is contained in some open set with a holomorphic branch of logarithm and square root defined, if $g(z) = e^{-|z|^2}$ were differentiable then so too would $(-\log(g(z)))^{1/2}=|z|$, which is a contradiction.
At the origin, $g(0)=1$ so: 
$$\frac{g(re^{i\theta})-g(0)}{re^i\theta}=\frac{e^{-r^2}-1}{re^{i\theta}} $$
So taking the limit as $r\rightarrow 0$ we get:
$$e^{-i\theta}\frac{d}{dx}\bigg(e^{-x^2}\bigg)\biggr\rvert_0$$
Where we take the real function $e^{-x^2}$, which is (real) differentiable at the origin. As pointed out in the comments below, this derivative is equal to zero so the function is differentiable at the origin with derivative $0$.
