Find the positive integer $n$ such that $p(z)=z^5-30z^2+1$ has exactly $3$ zeros(counting multiplicity) in $\{z \in \mathbb{C}:n<|z|Could anyone advise me on how to find the positive integer $n$ such that $p(z)=z^5-30z^2+1$ has exactly $3$ zeros(counting multiplicity) in $\{z \in \mathbb{C}:n<|z|<n+1\} \ ?$
Hints will suffice, thank you. 

Use Rouché's theorem to count the zeros of $p$ in the disk $\lvert z\rvert < n$ for $n = 1,2,\dotsc$.
For $n = 1$, we can choose $g(z) = - 30z^2$, since that is the term of $p$ with the largest modulus on $\lvert z\rvert = 1$. Thus $\lvert p(z) - g(z)\rvert = \lvert z^5+1\rvert \leqslant 2 < 30 = \lvert g(z)\rvert$ for $\lvert z\rvert = 1$, and we see that $p$ has as many zeros as $g$ in the disk $\lvert z\rvert = 1$, namely two. For sufficiently larger $n$, we must choose a different $g$, I'll leave it to you to find out which $g$ is appropriate for which $n$.
