Question

Let $z_1$ be a complex root of the equation $a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_{1}z+a_0=3(|z|<1)$, where $|a_i|<2$ for $i=0,1,2,\dots ,n$. Then

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

Answer 1

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

$a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_{1}z+a_0=3$
Putting $z_1$ in the equation,
$|a_n{z}_1^n+a_{n-1}{z}_1^{n-1}+\cdots +a_1{z}_1+a_0|=|3|$
$\Rightarrow |a_n||z_1{|}^n+|a_{n-1}||z_1{|}^{n-1}+\cdots +|a_1||z_1|+|a_0|$
$\ge |a_n{z}_1^n+a_{n-1}{z}_1^{n-1}+\cdots +a_1{z}_1+a_0|$
$=3$
$\Rightarrow |a_n||z_1{|}^n+|a_{n-1}||z_1{|}^{n-1}+\cdots +|a_1||z_1|+|a_0|\ge 3$
$\Rightarrow 2(|z_1{|}^n+|z_1{|}^{n-1}+\cdots +|z_1|+1)>3[\because |a_i|<2]$
$\Rightarrow |z_1{|}^n+|z_1{|}^{n-1}+\cdots +|z_1|+1>\frac{3}{2}$
$\Rightarrow 1+|z_1|+\cdots +|z_1{|}^{n-1}+|z_1{|}^n>\frac{3}{2}$

Now, $1+|z_1|+\cdots +|z_1{|}^{n-1}+|z_1{|}^n=\frac{1-|z_1{|}^{n+1}}{1-|z_1|}$
So, $\frac{1-|z_1{|}^{n+1}}{1-|z_1|}>\frac{3}{2}$
$\Rightarrow 2-2|z_1{|}^{n+1}>3-3|z_1|(\because |z_1|<1\Rightarrow 1-|z_1|>0)$
$\Rightarrow 0<2|z_1{|}^{n+1}<3|z_1|-1$
$\Rightarrow 3|z_1|-1>0$
$\Rightarrow |z_1|>\frac{1}{3}$