i argued as follows: by fundamental theorem of algebra i get that every non constant complex polynomial is surjective. Moreover, every polynomial is obviously continous. By general topology, we can characterize dense subsets of a topological space as those subsets which intersect non trivially any non empty open subset. Now, if $f:X\rightarrow Y$ is a continous surjective map of topological space, then it maps dense subsets of $X$ to dense subsets of $Y$, since: take $E$ dense in $X$,take $B$ open non-empty in $Y$, then $f^{-1}(B)$ is open (by continuity)and non-empty (by surjectivity) in $X$, hence $f^{-1}(B)\cap E$ is non-empty by the density of $E$ in $X$, hence $B\cap f(E)$ is non-empty. Applying this to $f=P$, the polynomyal, i get that $(P\circ f)(D(z_0,\epsilon)-\{z_0\})$ is dense in $\mathbb C$. Now, i know that if $z_0$ is essential, then $(P\circ f)(D(z_0,\epsilon)-\{z_0\})$ is dense in $\mathbb C$, but now i should prove the reverse. I have that $(P\circ f)(D(z_0,\epsilon)-\{z_0\})$ is dense in $\mathbb C$ and i must prove that $z_0$ is essential. It's definitely clear that $z_0$ cannot be removable, and intuitively i can understand it is not a pole, but i can't exclude this second case with a formal proof.