Question

If $|z-1|\le 2$ and $|\omega z-1-{\omega }^2|=a$ (where $\omega$ is a cube root of unity) then complete set of values of $a$ is

• A. $0\le a\le 2$
• B. $\frac{1}{2}\le a\le \frac{\sqrt{3}}{2}$
• C. $\frac{\sqrt{3}}{2}-\frac{1}{2}\le a\le \frac{1}{2}+\frac{\sqrt{3}}{2}$
• D. $0\le a\le 4$

Answer 1

The correct option is D $0\le a\le 4$

$|\omega z-1-{\omega }^2|=a$
$\Rightarrow |z+1|=a\Rightarrow |z-1+2|=a$
$\Rightarrow |z-1|+2\ge a$
$\Rightarrow 0\le a\le 4$