Context
Let $V$ be a 2-dimensional evaluation representation of the quantum loop algebra $\mathcal{U}_{q}\left(\mathcal{L}\mathfrak{sl}_{2}\right)$ with $a=q$. Also, for $m\in\mathbb{Z}$, the Drinfeld generator $x_{m}^{+}$ ($x^{-}_{m}$) acts by $k^{m}X$ ($k^{m}Y$) on $V$.
We recall that $\mathcal{U}_{q}\left(\mathcal{L}\mathfrak{sl}_{2}\right)$ contains the Chevalley generators $X_{1}=x^{+}_{0}$, $X_{0}=k^{-1}x^{+}_{1}$, $Y_{1}=x^{-}_{0}$, $Y_{0}=x^{-}_{-1}k$ with the coproduct $$ \triangle\left(X_{0}\right)=X_{0}\otimes1+k^{-1}X_{0},\triangle\left(X_{1}\right)=X_{1}\otimes1+k\otimes X_{1},\triangle\left(k\right)=k\otimes k $$ $$ \triangle\left(Y_{0}\right)=Y_{0}\otimes k+1\otimes Y_{0},\triangle\left(Y_{1}\right)=Y_{1}\otimes k^{-1}+1\otimes Y_{1} $$ and for $m,m^\prime\ge0$ such that $m+m^\prime\ne0$: $$ \left[x_{-m}^{+},x_{-m^{\prime}}^{-}\right]=\phi_{-m-m^{\prime}}/\left(q^{-1}-q\right), $$ where $$ \phi^{-}\left(z\right)=k^{-1}\exp\left(\left(q^{-1}-q\right)\sum_{m>0}h_{-m}z^{-m}\right)=\sum_{m\ge0}\phi_{-m}z^{-m} $$
Problem
Consider $$ T\left(z\right)=\exp\left(\sum_{m>0}\frac{h_{-m}\left(q-q^{-1}\right)z^{-m}}{\left(q^{2m}-q^{-2m}\right)}\right) $$
Let $g\left(z\right)$ be the eigenvalue of $T\left(z\right)$ over $\left(V\right)_{q}$.
How to show that the eigenvalues of $\left(g\left(z\right)\right)^{-1}T\left(z\right)$ on $V\left[\left[z\right]\right]$ are $1$ and $1-zq^{-1}$?
Along with that, I would like to understand better how $T\left(z\right)$ acts on $V\left[\left[z\right]\right]$.
My thoughts so far
The $1$ eigenvalue is obvious, with $\left(V\right)_{q}$ being an eigenspace. I think that $\left(V\right)_{q}z$ is probably an eigenspace corresponding to eigenvalue $1-zq^{-1}$. However, I'm not sure how to get the result. Perhaps it can be done using Drinfeld polynomials or considering eigenvalues of $h_{-m}$ and noticing something about $T\left(z\right)$.