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Rouché's theorem states that if $f,g$ are holomorphic on a neighborhood of a closed disc with boundary $C$ and $|g|<|f|$ on $C$, then $f$ and $f+g$ have the same number of zeros in the disc, counted with multiplicity. More generally the conclusion holds for a simple closed contour whose interior lies in their domain. To prove it, set $f_t=f+tg$ for $0\le t\le 1$. The strict boundary inequality ensures $f_t\ne 0$ on $C$. By the argument principle,is the integer number of zeros inside. The integrand is continuous in $t$ uniformly on $C$, since its denominator is bounded away from zero on the compact product $C\times [0,1]$. Thus $N(t)$ is continuous and integer-valued, so constant. This proves the theorem.

On $|z|=1$, the term $8z^6$ dominates the remaining terms because $3+1+2+1=7<8$. Thus the polynomial has six zeros inside the unit circle. On $|z|=2$, $3z^9$ dominates because its modulus is $1536$, whereas the sum of the other bounds is $512+32+16+1=561$. There are therefore nine zeros inside radius two. The strict inequalities exclude boundary zeros, so

For the fixed-degree polynomial limit, choose $d+1$ distinct complex nodes ${\zeta }_0,\dots ,{\zeta }_d$. Write $p_n(z)={\sum }_{j=0}^d{a}_{j,n}{z}^j$, inserting zero coefficients if necessary. The vector of values at the nodes equals the fixed Vandermonde matrix times the coefficient vector. Its determinant is ${\prod }_{i<j}({\zeta }_j-{\zeta }_i)\ne 0$. Convergence of the values therefore implies coefficient convergence, say $a_{j,n}\to a_j$. For $p(z)={\sum }_{j=0}^d{a}_j{z}^j$ and a compact set with $|z|\le R$,This proves that the given limit is a polynomial of degree at most $d$ without invoking any theorem about uniform limits of holomorphic functions.

When the limit has $d$ distinct roots, choose pairwise disjoint small closed discs around them, with no limit root on their boundaries. On each boundary $|p|$ has a positive minimum. Uniform convergence makes $|p_n-p|<|p|$ there for all sufficiently large $n$, so Rouché's theorem puts exactly one root $z_{i,n}$ of $p_n$ in each disc. For every smaller disc about $z_i$ the same argument applies eventually. The unique root in the original disc must then lie in the smaller one, provingThe literal demand for a root at every initial index needs this finite-index qualification: $p_1=1$, $p_n=z$ for $n\ge 2$ satisfies the hypotheses with limit $z$, but $p_1$ has no root. If every initial polynomial has a root, arbitrary choices at the finitely many exceptional indices extend the convergent sequences; otherwise the sequences necessarily start at an eventual index.