If $\left|a_k\right|<3,1\le k\le n$ , then all the complex numbers $z$  satisfying the equation $1+a_{1}z+a_2{z}^2+\mathrm{.....}+a_n{z}^n=0$ :

$\because \left|a_k\right|<3,1\le k\le n$ 

$\therefore \left|a_1\right|<3,\left|a_2\right|<3,\mathrm{.....},\left|a_n\right|<3$

$\because 1+a_{1}z+a_2{z}^2+\mathrm{.....}+a_n{z}^n=0$

$\Rightarrow$      $\left|a_{1}z+a_2{z}^2+\mathrm{.......}+a_n{z}^n\right|=\left|-1\right|$

$\Rightarrow$      $1=\left|a_{1}z+a_2{z}^2+\mathrm{...}+a_n{z}^n\right|$

      $1\le \left|a_1\right|\left|z\right|+{\left|\left|a_2\right|z\right|}^2+\mathrm{.....}+\left|a_n\right|{\left|z\right|}^n$

     $1<3\left(\left|z\right|+{\left|z\right|}^2+\mathrm{.....}+{\left|z\right|}^n\right)<3\left(\left|z\right|+{\left|z\right|}^2+\mathrm{.......}\infty \right)$

$\Rightarrow$        $1<\frac{3\left|z\right|}{1-\left|z\right|}\Rightarrow 1-\left|z\right|<3\left|z\right|\Rightarrow \left|z\right|>\frac{1}{4}$ .