$V_m=$Homogeneous polynomials in complex variable with total degree $m$,
Let $U\in SU(2)$ is just a linear map on $\mathbb{C}^2$, Define a Linear Transformation $\Pi_m:V_m\rightarrow V_m$ given by $[\Pi_m(U)f](z)=f(U^{-1}z)$ where $f(z)=a_0z_1^m+a_1z_1^{m-1}z_2+\dots +a_mz_2^m$, $z=(z_1,z_2)\in\mathbb{C}^2$
The representation is the map $\Pi_m: SU(2) \to GL(V_m)$ where for $U \in SU(2)$, $\Pi_m(U)$ is the transformation that takes $f \in V_m$ to $f \circ U^{-1}$, i.e. $ (\Pi_m(U)(f))(z) = f(U^{-1}(z))$. Note that if $f$ is a homogeneous polynomial if $f(z) = z_1^{m_1} z_2^{m_2}$ and $U^{-1} = \pmatrix{a & b\cr c & d\cr}$, $f \circ U^{-1} = (a z_1 + b z_2)^{m_1} (c z_1 + d z_2)^{m_2}$.
Now I want to compute the corresponding Lie Algebra Representation $\pi_m$, According to the definition it can be computed as $\pi_m(X)=\frac{d}{dt}\Pi_m(e^{tX})|_{t=0}$ where $X=\begin{pmatrix}X_{11}&X_{12}\\X_{21}&X_{22}\end{pmatrix}$
is some matrix. So $(\pi_m(X)f)(z)=\frac{d}{dt}f(e^{-tXz})|_{t=0}$, I took a curve $z(t)\in\mathbb{C}^2$ as $z(t)=e^{-tX}z$, $z(t)=(z_1(t),z_2(t))$ by Chainrule $\pi_m(X)f=\partial f/\partial z_1 dz_1/dt+\partial f/\partial z_2 dz_2/dt|_{t=0}$ How ever $dz/dt|_{t=0}=-Xz$, so could any one tell me what will be next few steps so that I can get the representation of this Lie algebra? Thank you for the help!