Question

The distances of the roots of the equation $|\sin {\theta }_1|z^3+|\sin {\theta }_2|z^2+|\sin {\theta }_3|z+|\sin {\theta }_4|=3$, from $z=0$, are

• A. Greater than $\frac{2}{3}$
• B. less than $\frac{2}{3}$
• C. greater than $|\sin {\theta }_1|+|\sin {\theta }_2|+|\sin {\theta }_3|+|\sin {\theta }_4|$
• D. les than $|\sin {\theta }_1|+|\sin {\theta }_2|+|\sin {\theta }_3|+|\sin {\theta }_4|$

Answer 1

The correct option is A

Greater than $\frac{2}{3}$

$\because |\sin {\theta }_1|z^3+|\sin {\theta }_2|z^2+|\sin {\theta }_3|z+|\sin {\theta }_4|=3|\sin {\theta }_1|z^3+|\sin {\theta }_2|z^2+|\sin {\theta }_3|z+|\sin {\theta }_4||\le |\sin {\theta }_1||z{|}^3+|\sin {\theta }_2||z{|}^2+|\sin {\theta }_3||z+|\sin {\theta }_4|\le |z{|}^3+|z{|}^2+|z|+1<1+|z|+|z{|}^2+\dots \infty =\frac{1}{1-|z|}\therefore 3<\frac{1}{1-|z|}\Rightarrow \frac{1}{3}>1-|z|\Rightarrow |z|>\frac{2}{3}\therefore |z-0|>\frac{2}{3}$