Question

Show that the roots of the equation $a_1{z}^3+a_2{z}^2+a_{3}z+z_4=3$, where $|a_i|\le 1,i=1,2,3,4$ lie outside the circle with centre origin and radius 2/3.

Answer 1

Give that
$a_1{z}^3+a_2{z}^2+a_{3}z+a_4=3$
we have
$\Rightarrow |3|=|a_1{z}^3+a_2{z}^2+a_{3}z+a_4|$
$\Rightarrow 3\le |a_1{z}^3|+|a_2{z}^2|+|a_{3}z|+|a_4|$
$\Rightarrow 3\le |a_1||z^3|+|a_2||z^2|+|a_3||z|+|a_4|$
$\Rightarrow 3\le |a_1||z{|}^3+|a_2||z{|}^2+|a_3||z|+|a_4|$
$\Rightarrow 3\le |z{|}^3+|z{|}^2+|z|+1$ $(\because |a_i|\le 1|)$
$\Rightarrow 3\le 1+|z|+|z{|}^2+|z{|}^3<1+|z|+|z{|}^2+|z{|}^3+......\infty$
$\Rightarrow 3<1+|z|+|z{|}^2+|z{|}^3+....\infty$
$\Rightarrow 3<\frac{1}{1-|z|}$ $(\because |z|<1)$
$\Rightarrow 1-|z|<\frac{1}{3}$
$\Rightarrow \frac{2}{3}-|z|<0$
$\Rightarrow |z|>2/3$
Hence $\frac{2}{3}<|z|<1$
Ans: 1