Question

If $z_1$ is a root of the equation $a_0{z}^n+a_1{z}^{n-1}+.....+a_{n-1}z+a_n=4$, where $|a_i|<2$ for $i=0,1,2,....,n$, then

• A. $|z_1|>\frac{1}{3}$
• B. $|z_1|<\frac{1}{3}$
• C. $|z_1|>\frac{1}{4}$
• D. $|z_1|>\frac{1}{2}$

Answer 1

The correct option is A $|z_1|>\frac{1}{3}$

$a_0{z}^n+a_1{z}^{n-1}+a_2{z}^{n-2}+....+a_{n-1}z+a_n=3$

or $|3|=|a_0{z}^n+a_1{z}^{n-1}+....+a_{n-1}z+a_n|$

or $|3|\le |a_0||z{|}^n+|a_1||z{|}^{n-1}+....+|a_{n-1}||z|+|a_n|$

or $3<2(|z{|}^n+|z{|}^{n-1}|+....+|z|+1)$

or $1+|z|+|z{|}^2+....+|z{|}^n>\frac{3}{2}$

If $|z|\ge 1$, the inequality is clearly satisfied. For $|z|<1$, we must have,

$\frac{1-|z{|}^{n+1}}{1-|z|}>\frac{3}{2}$

or $2-2|z{|}^{n+1}>3-x|z|$

or $2|z{|}^{n+1}<3|z|-1$

or $3|z|-1>0$

or $|z|>\frac{1}{3}$