Statement of the Problem: Let $a\in\mathbb{C}$ be a constant and consider the polynomial $$P(z)=z^{10}+a(z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1).$$ Given a radius $\rho>0$, use complex analysis to find an upper bound on $|a|$ of the form $|a|<\delta(\rho)$, $\delta(\rho)>0$, sufficient to guarantee that $P$ has all ten of its zeros in the disk $|z|<\rho$.
My work: Consider $f(z)=z^{10}$ and $g(z)=a\left(z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1\right)$. Then: $$\begin{align} |g(z)|&= |a \left(z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1\right)|\\ &\leq |a||z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1| \\ & \leq |a|\left(|z|^9+|z|^8+|z|^7+|z|^6+|z|^5+|z|^4+|z|^3+|z|^2+|z|+1\right)\end{align}$$ Now we consider the region $|z|=\rho$ where $\rho>1$ $$\begin{align} &\leq |a|\left(\rho^9+\rho^8+\rho^7+\rho^6+\rho^5+\rho^4+\rho^3+\rho^2+\rho+1\right) \\ &\leq |a|(10\rho^{10}) \end{align}$$ Here we choose $|a| \leq \frac{1}{10}$ $$\begin{align} &\leq 10|a|\rho^{10} \leq \rho^{10} = |f(z)| \end{align}$$ $$|g(z)|\leq|f(z)|$$ Thus $f$ and $f+g$ have the same number of zeros by Rouché in $|z|\leq\rho$.
Do I need to show work for the case where $0<\rho<1$? or is this sufficient to answer the question? Am I understanding the question and demonstrating knowledge of the subject in my answer? (PS, I am studying for qualifying exams so any commentary on clarifying my proof or making it simpler, or even a better, simpler solution would be much appreciated) thankyou
EDIT: Work for case where $0<\rho<1$: Say $0<\rho,1$, and $|z|<\rho$ then $$ \begin{align}|h(z)|& =|a\left(z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1 \right)| \\ & \leq |a|\left(|z|^9+|z|^8+|z|^7+|z|^6+|z|^5+|z|^4+|z|^3+|z|^2+|z|+1 \right)\\ & \leq |a| \left( \rho^9+\rho^8+\rho^7+\rho^6+\rho^5+\rho^4+\rho^3+\rho^2+\rho+1 \right) \\ & \leq |a| (10) \end{align}$$ Take $a\leq \frac{\rho^{10}}{10}$ $$ \leq \rho^{10} = |f(z)| \text{ for } |z|=\rho$$ Therefore we have shown that for the $0<\rho<1$ case, if we choose $a \leq \frac{\rho^{10}}{10}$ then $$|g(z)| \leq |f(z)|$$
Putting both cases together we have that for $a\leq\min\{\frac{1}{10},\frac{\rho^{10}}{10}\}$, $|g(z)|\leq|f(z)|$.
Thus $f$ and $f+g$ have the same number of zeros (10) in $|z|\leq\rho$
Edit from comment by user reuns
Note that $z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1=\frac{z^{10}-1}{z-1}$.
Using $g(z)=z^{10}$ and $f(z)=z^{10}+a\frac{z^{10}-1}{z-1}$
Then for $a<0$ we have that $|f(z)|<|g(z)|$ all z
Using $z=ρw$, the normalized equation for $w$ is $$ w^{10}+a(ρ^{-1}w^9+...+ρ^{-9}w+ρ^{-10})=0. $$ The Lagrange bound for the size of the roots is $$ R=\max(1,|a|(ρ^{-1}+...+ρ^{-9}+ρ^{-10}) $$ As we want $|z|\leρ$, we need $R\le 1$, thus $$ |a|\le \frac{1}{ρ^{-1}+...+ρ^{-9}+ρ^{-10}}=\frac{ρ^{10}}{1+ρ+..+ρ^9}. $$