Let us fix $k_{1},\cdots,k_{n}\in \mathbb{Z}$, $T^n=\big\{(z_{1},\cdots,z_{n})\mid |z_{k}|=1,k\in \{1,\cdots,k\}\big\}$ and $w=\frac{i}{2}\sum dz_{i}\wedge d\overline{z_{i}}=\sum dx_{i}\wedge dy_{i}= \sum r_{i}dr_{i}\wedge d\theta_{i}$ the symplectic form on $\mathbb{C}^n$.
I am trying to prove that $\varphi\colon T^n\longrightarrow \text{Diff}(\mathbb{C}^n)$ where $$\varphi_{(t_{1},\cdots,t_{n})}(z_{1},\cdots,z_{n})=(t_{1}^{k_{1}}z_{1},\cdots,t_{n}^{k_{n}}z_{n})$$ is a hamiltonian action with moment map given by $\mu \colon \mathbb{C}^n\longrightarrow (t^n)^*\cong\mathbb{R}^n$, $$\mu(z_{1},\cdots,z_{n})=-\frac{1}{2}(k_{1}|z_{1}|^2,\cdots, k_{n}|z_{n}|^2).$$
I have already proven the equivariance, so I need to show that for all $X\in t^n\cong \mathbb{R}^n$, $$d\mu^x=\iota_{X^\#}\omega.$$
Suppose that $X=(a_{1},\cdots,a_{n})\in \mathbb{R}^n$. Then, $\mu^{X}((z_{1},\cdots,z_{n}))=-\frac{a_{1}k_{1}}{2}|z_{1}|^2-\cdots-\frac{a_{n}k_{n}}{2}|z_{n}|^2$. Then, $$d\mu^X= -\frac{a_{1}k_{1}}{2}\overline{z_{1}}\frac{\partial}{\partial z_{1}}-\cdots-\frac{a_{n}k_{n}}{2}\overline{z_{n}}\frac{\partial}{\partial z_{n}}-\frac{a_{1}k_{1}}{2}z_{1}\frac{\partial}{\partial \overline{z_{1}}}-\frac{a_{n}k_{n}}{2}z_{n}\frac{\partial}{\partial \overline{z_{n}}}.$$
Now, I have problems when computing $X_{(z_{1},\cdots,z_{n})}^{\#}$ for $(z_{1},\cdots,z_{n})\in \mathbb{C}^{n}.$ I should take a curve in $T^n$ that crosses the identity at $0$ and such that the derivative at $0$ is $(a_{1},\cdots,a_{n})$. I have taken $(e^{a_{1}ti},\cdots,e^{a_{n}ti})$. Then, I have to compute the derivative at 0 of $\varphi_{(e^{a_{1}ti},\cdots,e^{a_{n}ti})}(z_{1},\cdots,z_{n})$ which is $(k_{1}a_{1}z_{1}i,\cdots,k_{n}a_{n}z_{n}i$).
I do not know how to continue... I think that the problem is that I do not understand how is that written is terms of the basis of $T_{(z_{1},\cdots,z_{n})}\mathbb{C}^n\cong \mathbb{R}^{2n}$.
Can anyone help me, please?
$\mathbf{EDIT}$: I believe that it is not correct to take the curve $(e^{a_{1}ti},\cdots,e^{a_{n}ti})$ because its derivative at $0$ is $i(a_{1},\cdots,a_{n})$.