Show all roots of $P(z)=z^n+a_{n-1}z^{n-1}+...+a_0$ lie within $|z|<\sqrt{1+|a_{n-1}|^2+...+|a_0|^2}$. Show all roots of $P(z)=z^n+a_{n-1}z^{n-1}+...+a_0$ lie within $|z|<\sqrt{1+|a_{n-1}|^2+...+|a_0|^2}$. I saw similar questions, but I saw no relation between $\sqrt{1+|a_{n-1}|^2+...+|a_0|^2}$ and any domain or circle(there was the usage of Cauchy-Schwartz Inequality but it has been quite remiss in making a statement about the set in which it all could take place, from what I did see.)
I do know that $|z^n|>|P(z)-z^n|$ which means, following Rouché's theorem, that both $z^n$ and $P(z)$ have the same number of roots in the domain in which the inequality holds(even though it isn't, for a moment, quite clear what roots $z^n$ has besides $z=0$). I could really use your help here.  
 A: Let $|z|>1,$ and $P(z)=z^n+\sum_{j=0}^{n-1}a_jz^j$
By Cauchy-Schwartz inequality, we have
\begin{align}
|P(z)|&=|z|^n\left|1+ \sum_{j=0}^{n-1}a_j\frac{1}{z^{n-j}}\right|\\&\geq |z|^n\left\{1-\left| \sum_{j=0}^{n-1}a_j\frac{1}{z^{n-j}} \right|\right\}\\&\geq |z|^n\left\{1-\left(\sum_{j=0}^{n-1}\left|a_j\right|^2\right)^{1/2}\left( \sum_{j=0}^{n-1}\frac{1}{|z^{n-j}|^2} \right)^{1/2}\right\}\\&= |z|^n\left\{1-\left(\sum_{j=0}^{n-1}\left|a_j\right|^2\right)^{1/2}\left( \frac{1}{|z|^2} +\frac{1}{|z|^4}+\ldots+\frac{1}{|z|^{2n}}\right)^{1/2}\right\}\\&> |z|^n\left\{1-\left(\sum_{j=0}^{n-1}\left|a_j\right|^2\right)^{1/2}\left( \frac{1}{|z|^2} +\frac{1}{|z|^4}+\ldots\right)^{1/2}\right\}\\&= |z|^n\left\{1-\left(\sum_{j=0}^{n-1}\left|a_j\right|^2\right)^{1/2}\left( \frac{1}{|z|^2-1}\right)^{1/2}\right\}>0
\end{align}
If $1-\left(\sum_{j=0}^{n-1}\left|a_j\right|^2\right)^{1/2}\left( \frac{1}{|z|^2-1}\right)^{1/2}>0$ or $|z|>\left(1+\sum_{j=0}^{n-1}\left|a_j\right|^2\right)^{1/2}.$
Thus,  all the zeros of $P(z)$ whose modulus is greater than $1$ lie in $|z|\leq \left(1+\sum_{j=0}^{n-1}\left|a_j\right|^2\right)^{1/2}.$ Those zeros whose modulus is less than or equal to $1$ already satisfy this inequality. 
