We identify canonically $\mathbb{R}^{2n+2}$ with $\mathbb{C}^{n+1}$ which we equip with the standard hermitian product.
For each $z\in\mathbb{S}^{2n+1}$, we define $X_z:=iz$. It's easy to check using the hermitian product that $X$ defines a vector field on $\mathbb{S}^{2n+1}$.
We recall that the complex projective space is $\mathbb{C}\mathbb{P}^n:=\mathbb{C}^{n+1}/\mathbb{C}^*$ and let $p:\mathbb{S}^{2n+1}\rightarrow\mathbb{C}\mathbb{P}^n$ the application that maps each $z\in\mathbb{S}^{2n+1}$ to its class in $\mathbb{C}\mathbb{P}^n$.
I want to demonstrate that for all $z\in\mathbb{S}^{2n+1}$, $dp_z(X_z)=0$. I tried to work in local coordinates using the standard atlases of $\mathbb{S}^{2n+1}$ and $\mathbb{C}\mathbb{P}^n$, but the computations got too messy very quickly and I wonder if there is any easy way to demonstrate it.