@DanielFischer I managed it without using essential singularity .. :)
my idea is $f(z)^n + \sum\limits_{k=0}^{n-1} a_k(z)f(z)^k = 0$, where $a_k(z) = \frac{p_k(z)}{q_k(z)} \in \mathbb{C}(x)$
I divide it with $z^{nN}$, where $N > max_k (deg (p_k))$
Then writing $F(z)=\frac{f(z)}{z^N}$, the expression becomes $F(z)^n + \sum\limits_{k=0}^{n-1} \frac{a_k(z)}{z^{N(n-k)}}F(z)^k = 0$
Now, since $\lim\limits_{z \to \infty} \frac{a_k(z)}{z^{N(n-k)}} = 0$, we can get a $R>0$ such that for all $|z| >R$ we have $|\frac{a_k(z)}{z^{N(n-k)}}| < 1$.