Conformal weights in Laurent expansions Currently I am taking an introductory class in conformal field theory and ran into some confusion. Everywhere I run into Laurent expansions of several important quantities, and (mostly) all of them boil down to the following: 
Suppose $\phi (z)$ is a primary field with conformal dimension $h$, then it may be Laurent expanded as:
$$\phi(z) = \sum_{n \in \mathbb{Z}} c_n z^{-n-h}$$
With $c_n$ the Laurent modes (constants). For a particular reference of this, see for example di Francesco's book eqn. (6.7) or Blumenhagen eqn. (2.42) in which it is claimed (without motivation beforehand):
$$T(z) = \sum_{n \in \mathbb{Z}} z^{-n-2} L_n$$
Is there any mathematical or physical reason why we include these conformal weights of the objects in the Laurent expansion (why not simply $\sum_{n \in \mathbb{Z}} z^{n} c_n$) ? Or is it simply convenient in calculations?
Bonus (related) question: in Blumenhagen's book (page 13) the following expression is stated:
$$z'=z + \epsilon(z) = z + \sum_{n \in \mathbb{Z}}\epsilon_n (-z^{n+1})$$
Where $\epsilon(z)$ is assumed to be meromorphic and $\epsilon_n$ constant. Is this particular expansion (so with $z^{n+1}$ instead of $z^n$) similarly done simply for convenience?
 A: Let me use the book by di Francesco et al. Including the conformal weight in the power of $z$ in the Laurent expansion, as in equation (6.8), allows you to have the nice property (6.10), namely $\phi_{-m} = \phi^{\dagger}_m$. For the energy momentum tensor, it gives $L_{-m} = L^{\dagger}_m$. 
It is just a nice convention. 
A: The conformal weights are included because they come from the conformal transformation which maps the cylinder to the radial plane. Start with the cylinder where $x \sim x + 2\pi$. Define complex coordinates $w=t+ix$ so $w \sim w + 2\pi i$. Then, all operators admit the expansion
$$
\Phi(w,{\bar w}) = \sum_{m,n} \phi_{m,n} e^{-m w} e^{-n {\bar w} } , \qquad m,n \in {\mathbb Z}
$$
Now, we perform the conformal transformation which moves us to the radial plane. In particular, we define
$$
z = e^{ w} = e^{ i x} e^{t} , \qquad {\bar z} = e^{ {\bar w}}.
$$
Then, note that $t \to -\infty$ maps to the origin and $t \to \infty$ maps to infinity. Now, under conformal transformations a primary field transforms as
\begin{align}
\Phi(z,{\bar z}) &= (\partial_z w)^h(\partial_{\bar z} {\bar w})^{{\bar h}}\Phi (w,{\bar w}) =  \sum_{m,n} \phi_{m,n} z^{-m-h} {\bar z}^{-n-{\bar h}} 
\end{align}

ASIDE
In Euclidean signature, the adjoint property is
$$
\Phi(t,x)^\dagger = \Phi(-t,x) ~~\implies~~ \Phi(w,{\bar w})^\dagger = \Phi(-{\bar w},-w) ~~\implies~~  \phi_{m,n}^\dagger  = \phi_{-m,-n} .
$$
This is because time evolution is defined by $\Phi(t,x) = e^{H t} \Phi(0,x) e^{-H t}$. We assume here that $\Phi(0,x)$ is Hermitian. On the radial plane, the adjoint property then reads
\begin{align}
\Phi(z,{\bar z})^\dagger &= \sum_{m,n} \phi^\dagger_{m,n}{\bar z}^{-m-h} z^{-n-{\bar h}}\\
&= {\bar z}^{-2h} z^{ - 2 {\bar h} }\sum_{m,n} \phi_{m,n} (1/{\bar z})^{-m-h} (1/z)^{-n-{\bar h}} \\
&= {\bar z}^{-2h} z^{ - 2 {\bar h} } \Phi(1/{\bar z},1/z). 
\end{align}
A: This is simply a convenience, and in the end you should be able to work with any convention. In fact, in the mathematical literature on vertex operator algebras you will often encounter,
$$ Y(\omega,z) = \sum \omega_n z^{-n-1} = \sum L_n z^{-n-2}$$
where $Y$ maps a state to a field, i.e. the stress-energy tensor has modes expanded as if $h=1$ in your notation. This might seem odd but it is actually useful to have a consistent convention to use for any state, regardless of conformal dimension, when you have operators in your algebra of varying conformal dimension.
