$\omega$ is a complex cube root of unity and if $Z$ is a complex number satisfying $|\mathrm{Z}-1| \leq 2$…
$\omega$ is a complex cube root of unity and if $Z$ is a complex number satisfying $|\mathrm{Z}-1| \leq 2$ and $\left|\omega^2 Z-1-\omega\right|=a$, then the set of possible values of $a$ is
$0 \leq a \leq 2$
$|\omega| \leq a \leq \frac{\sqrt{3}}{2}+2$
$\frac{1}{2} \leq a \leq \frac{\sqrt{3}}{2}$
$0 \leq a \leq 4$
Solution
$\begin{aligned}
& \text {Given, }|Z-1| \leq 2 \text { and }\left|\omega^2 Z-1-\omega\right|=a \\
& \Rightarrow\left|\omega^2 Z+\omega^2\right|=a \qquad \left(\because 1+\omega+\omega^2=0\right) \\
& \Rightarrow\left|\omega^2\right||Z+1|=a \Rightarrow|Z-1+2|=a \\
& \Rightarrow|Z-1|+2 \geq a \Rightarrow 2+2 \geq a \\
& \Rightarrow 4 \geq a \text { and } a \geq 0
\end{aligned}$
So, $0 \leq a \leq 4$.