Let $$\pi \; : \; \mathbb{C}^{n+1} \backslash \{ 0 \} \longrightarrow \mathbb{P}^n \; : \; (z^0, \ldots z^n ) \longmapsto (z^0 : \cdots : z^n )$$ be the map onto homogeneous coordinates of the complex projective space $\mathbb{P}^n$. Denote by $\pi_{\ast}$ the push-forward of $\pi$ to the corresponding tangent spaces and denote the pullback by $\pi^{\ast}$.
Within my lecture on complex manifolds, my professor stated that
$$ \pi_{\ast} \sum_{i = 0}^n z^i \frac{\partial}{\partial z^i} \enspace = \enspace 0 $$
However, he did not prove it, he just vaguely outlined an idea for a proof by stating that when computing
$$ \pi^{\ast} \; d \bigg( \frac{z^i}{z^0} \bigg)$$
then
$$\pi_{\ast} \frac{\partial}{\partial z^i}$$
should somehow follow.
I am interested in how this proof is carried out in more detail, because I don't see where his outlined idea is going.
EDIT 1:
Thank you for your answers so far. I think I understood the approaches in the comments . However, I am also interested in the way my professor is suggesting. If I am not mistaken, the following equations hold:
$$ \pi_{\ast} \bigg( \sum_{i = 0}^n z^i \frac{\partial}{\partial z^i} \bigg) (f) \enspace = \enspace \sum_{i = 0}^n z^i \, \pi_{\ast} \bigg( \frac{\partial}{\partial z^i} \bigg) (f) \enspace = \enspace \sum_{i = 0}^n z^i \frac{\partial}{\partial z^i} \, ( \pi^{\ast} f ) $$
where $f$ is an arbitrary function. How do I proceed and where do I need $\pi^{\ast} d \big( \tfrac{z^i}{z^0} \big)$? Did I already go astray?
The functions $u^i: \mathbb{P}^n \rightarrow \mathbb{C}$, given by $$u_i(z^0:\cdots:z^n) = \frac{z^i}{z^0},\ 1 \le i \le n, $$ are well-defined. Moreover, their differentials $du^1, \dots, du^n$ are a basis of $T_{(z^0:\cdots:z^n)}^*\mathbb{P}^n$. Using this fact, you can show that to prove $$ \pi_*\left(\sum_{j=0}^n z^j\frac{\partial}{\partial z^j}\right) = 0, $$ it suffices to show that, for each $1 \le i \le n$, $$ \left\langle\sum_{j=0}^n z^j\frac{\partial}{\partial z^j},\pi^* du^i\right\rangle = 0\text{ on }\mathbb{C}^{n+1}\backslash\{0\}. $$ This is now a straightforward calculation.