The behaviour of a vector field $v^z$ at $z=\infty$ can be described by first moving to coordinates that are well-defined in that coordinate patch. Defining $w = \frac{1}{z}$, we find
$$
c^w = \frac{dw}{dz} c^z = - z^{-2} c^z
$$
Now, $c^w$ must be well-defined at $w=0$. This implies that $z^{-2} c^z$ must be well-defined as $z \to \infty$. Thus $c^z$ cannot go faster than $z^2$ as $z \to \infty$.

Performing a similar analysis for $b_{zz}$, we find
$$
b_{ww} = \left( \frac{dw}{dz} \right)^{-2} b_{zz} = z^4 b_{zz}
$$
Now, since $z^4 b_{zz}$ has to be well-defined as $z\to\infty$, it must be that $b_{zz} \sim z^{-4}$ at large $z$.

This post imported from StackExchange Physics at 2014-04-14 16:22 (UCT), posted by SE-user Prahar