Push-forward of vector fields of Heisenberg type group $\mathbb R \times \mathbb C$ I am learning some geometry by myself, so pleas be explicit as possible. My aim is to calculate the structure constants of the Lie algebra of $\mathbb R \times \mathbb C$ equipped with the low 
$$(x_0,x) \cdot (y_0,y) = (x_0+y_0+\frac{1}{2}Im(x\bar{y}), x+y)$$
I think that, with the trivial chart $(\mathbb R \times \mathbb C,Id_{\mathbb R \times \mathbb C}) $ this can be viewed as a real manifold. Moreover, it is a lie group.
Let $\ell_{(z_0,z)}$ be the left multiplication operator by $(z_0,z) \in \mathbb R \times \mathbb C$
I am confused with the combination of the real and complex variables in the definition of the group, for a basis of the tangent vectors at (0,0) I guess we should take something like $$\partial_{x_0}|_{(0,0)} , \partial_{x}|_{(0,0)} , \partial_{\bar{x}}|_{(0,0)}$$
but when I try to push-forward one of them by $\ell_{(z_0,z)}$ using the Jacobian matrix technique, which turns to be a $2\times 3$ matrix, I loose information about the last vector field (with the bar). I know I am probably doing a stupid error. I don't have time now to include my explicit computation, but I will add them later if it is needed.

Update : 
I want to add that I am interested in results without using matrix representation, for learning purpose and for my intention of doing computations in higher dimension and in other similar lie groups.
I tried to imitate the computation done in the following lecture note by M.L. Fels : An Introduction to Differential Geometry through Computation Page 61 § 4.2
$$ {\ell_{(z_0,z)}}_* \partial_{x_0}|_{(0,0)} = J|_{(0,0)} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} $$ 
$$ {\ell_{(z_0,z)}}_* \partial_{x}|_{(0,0)} = J|_{(0,0)} \cdot \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} $$ 
$$ {\ell_{(z_0,z)}}_* \partial_{\bar x}|_{(0,0)} = J|_{(0,0)} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} $$ 
where J|_{(0,0)}, is the Jacobian matrix ${\begin{pmatrix} 1 & -\frac{1}{4}\bar{z} & \frac{1}{4} z \\ 0 & 1 & 0 \end{pmatrix}}_{|(0,0)} = \begin{pmatrix} 1 & -\frac{1}{4}\bar{z} & \frac{1}{4} z \\ 0 & 1 & 0 \end{pmatrix}$ associated to the application $\ell_{(z_0,z)}$.
I am confusing the actual tangent vector at $\ell_{(z_0,z)}(0,0) = (z_0,z)$ with its coefficient in the base $\partial_{x_0}|_{(z_0,z)} , \partial_{x}|_{(z_0,z)} , \partial_{\bar{x}}|_{(z_0,z)}$.
The problem is that  these computations work well in  open sets of $\mathbb R^n$ but are not consistent when $\mathbb C$ is introduced. I guess I have to proceed otherwise? Probably by  rethinking about the chart and try to devise $\mathbb C$ into $\mathbb R \times \mathbb R$ ?
To illustrate the problem : if we do the computation for the following
$${\ell_{(z_0,z)}}_* \partial_{x}|_{(0,0)} =  \begin{pmatrix} -\frac{1}{4} \bar{z} \\ 1 \end{pmatrix}$$
the result is only two coefficients, but we need three to write the tangent vector in $\{\partial_{x_0}|_{(z_0,z)} , \partial_{x}|_{(z_0,z)} , \partial_{\bar{x}}|_{(z_0,z)}\}$
 A: Let an element of the Heisenberg group be
 $$g=\left(\begin{array}{ccc}
1 & a & b\\
0 & 1 & c\\
0 & 0 & 1
\end{array}\right).$$
Then you have
$$ L_{g}\left(X\right)=\left(\begin{array}{ccc}
1 & a & b\\
0 & 1 & c\\
0 & 0 & 1
\end{array}\right)\left(\begin{array}{ccc}
1 & x & y\\
0 & 1 & z\\
0 & 0 & 1
\end{array}\right)=\left(\begin{array}{ccc}
1 & x+a & y+az+b\\
0 & 1 & z+c\\
0 & 0 & 1
\end{array}\right),$$
Writing this in another way you have that $$L_{\left(a,b,c\right)}\left(x,y,z\right)=\left(x+a,\,y+az+b,\,z+c\right).$$
Now take the differential and you have the push forward
$$L_{g*}=\left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & a\\
0 & 0 & 1
\end{array}\right).$$
For example let's take the vectors
$$\left(E_{1}\right)_{\mathbb{1}}=\left(\begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right),\,\,\left(E_{2}\right)_{\mathbb{1}}=\left(\begin{array}{ccc}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right),\,\,\left(E_{3}\right)_{\mathbb{1}}=\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{array}\right),$$
Then for every point $g$ in the Heisenberg group you have a tangent base given by $$\left(E_{i}\right)_{g}=L_{g*}\left(E_{1}\right)_{\mathbb{1}},$$
  i.e.$$\left(E_{1}\right)_{g}=\left(\begin{array}{ccc}
0 & a & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right),\,\,\left(E_{2}\right)_{g}=\left(\begin{array}{ccc}
0 & 0 & a\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right),\,\,\left(E_{3}\right)_{g}=\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & a\\
0 & 0 & 0
\end{array}\right).$$
