Let $p_1,\dots,$ $p_{n}$, $q_1,\dots,q_{n}$ be polynomials in the variables $z_1,\dots,z_m$ and $$ Y:=\{(z_1,\dots,z_m):\ q_1(z_1,\dots,z_m)\ne 0,\dots, q_{n}(z_1,\dots,z_m)\ne 0\}\subset \mathbb C^m. $$ We define the set $Z\subset\mathbb C^n$ as the image of $Y$ under the mapping \begin{equation*} %\begin{split} \phi:\ (z_1,\dots,z_m)\in Y\mapsto\left(\frac{p_1(z_1,\dots,z_m)}{q_1(z_1,\dots,z_m)},\dots,\frac{p_{n}(z_1,\dots,z_m)}{q_{n}(z_1,\dots,z_m)}\right)\in Z, %\end{split} \end{equation*} that is, $Z=\phi(Y)$. We define the varity $V$ as closure of $Z$ (we use the Zariski topology).
It is clear that if $Y_1$ is a closed subset of $Y$, then the closure of $\phi(Y_1)$ is a closed subset of $V$.
Question: Is it true that every closed subset of $V$ can be obtained in this way?