omega is a complex cube root of unity and if | vedclass

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(D) Given,$|Z-1| \leq 2$ and $|\omega^2 Z-1-\omega|=a$. Since $1+\omega+\omega^2=0$,we have $-1-\omega = \omega^2$. Substituting this into the second

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$0 \leq a \leq 4$

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(D) Given,$|Z-1| \leq 2$ and $|\omega^2 Z-1-\omega|=a$. Since $1+\omega+\omega^2=0$,we have $-1-\omega = \omega^2$. Substituting this into the second equation: $|\omega^2 Z + \omega^2| = a$. Since $|\omega^2| = 1$,this simplifies to $|Z+1| = a$. We can rewrite this as $|(Z-1)+2| = a$. By the triangle inequality,$|Z-1+2| \leq |Z-1| + |2|$. Given $|Z-1| \leq 2$,we have $a = |Z+1| \leq |Z-1| + 2 \leq 2 + 2 = 4$. Also,the modulus $a$ must be non-negative,so $a \geq 0$. Thus,the range of possible values for $a$ is $0 \leq a \leq 4$.

$\omega$ is a complex cube root of unity and if $Z$ is a complex number satisfying $|Z-1| \leq 2$ and $|\omega^2 Z-1-\omega|=a$,then the set of possible values of $a$ is

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