Question

If $z_1$ is a root of the equation $a_0^n{z}^n+a_1{z}^{n-1}+....+a_{n-1^z}+a_n=3$, where $|a_i|<2$ for $i=0,1,....,n.$ Then,

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

Answer 1

Answer: (A)

$|z_1|>\frac{1}{3}$

According to question,

$l{a}_0{z}^n+a_1{z}^{n-1}+.............+a_{n-1}z+a_n=3\Rightarrow \left|a_0{z}^n+a_1{z}^{n-1}+.............+a_{n-1}z+a_n\right|=\left|3\right|\Rightarrow \left|a_0\right|{\left|z\right|}^n+\left|a_1\right|{\left|z\right|}^{n-1}+..........+\left|a_{n-1}\right|\left|z\right|+\left|a_n\right|\ge 3\Rightarrow 2({\left|z\right|}^n+{\left|z\right|}^{n-1}+...........\left|z\right|+1)>3\Rightarrow (1+\left|z\right|+{\left|z\right|}^2+...........+{\left|z\right|}^n)>\frac{3}{2}\Rightarrow \frac{1-{\left|z\right|}^{n+1}}{1-\left|z\right|}>\frac{3}{2}\Rightarrow 2-2{\left|z\right|}^{n+1}>3-3\left|z\right|\Rightarrow 2{\left|z\right|}^{n+1}<3\left|z\right|-1\Rightarrow 3\left|z\right|-1>0\therefore \left|z\right|>\frac{1}{3}sothecorrectoptionisA$