Consider the complex space $\mathbb{C}^{n+1}$ with the usual hermitian inner product $\langle \cdot,\cdot\rangle$. We can construct the projective space $\mathbb{C}P^{n}$ by the equivalence relationship $\phi \sim \psi \iff \phi=a\psi$ for some $a\in\mathbb{C}$ and taking the quotient: $$ \mathbb{C}P^{n} = \mathbb{C}^{n+1}/\sim. $$ I want to prove that the tangent space $T_{[z]}\mathbb{C}P^{n}$ at a point $[z] \in \mathbb{C}P^n$ ($[z]$ denotes the equivalency class of $z$) is isomorphic to the space $z^{\perp}$. I've been trying to follow the steps described in the answer to this related question but I haven't had much success. My issue is that, given the canonical projection $$ \pi:\mathbb{C}^{n+1}\backslash \{0\}\to \mathbb{C}P^n\\ z\mapsto[z], $$ how does one compute the explicit expression for the pushforward $\pi_*$? Given that the pushforward is defined by $$ \pi_*:T_z\mathbb{C}^{n+1}\to T_{[z]}\mathbb{C}P^n, $$ once i have the expression for $\pi_*$, how can I show that $T_{[z]}\mathbb{C}P^n \cong z^{\perp}$?
I'm sorry if this is an obvious question or if it doesn't make sense, I've only recently started to study projective spaces and geometry, so any help is appreciated!