omega is a complex cube root of unity and Z i | vedclass

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(C) Given $|Z-1| \leq 2$,this represents a disk centered at $1$ with radius $2$. We are given $|\omega Z - 1 - \omega^2| = r$. Since $\omega^2 + \omeg

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$r > 4$

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(C) Given $|Z-1| \leq 2$,this represents a disk centered at $1$ with radius $2$. We are given $|\omega Z - 1 - \omega^2| = r$. Since $\omega^2 + \omega + 1 = 0$,we have $1 + \omega^2 = -\omega$. Substituting this,the equation becomes $|\omega Z + \omega| = r$. Since $|\omega| = 1$,we can factor it out: $|\omega| |Z + 1| = r$,which simplifies to $|Z - (-1)| = r$. This represents a circle centered at $-1$ with radius $r$. The distance between the centers of the disk $(C_1 = 1)$ and the circle $(C_2 = -1)$ is $d = |1 - (-1)| = 2$. For the disk $|Z-1| \leq 2$ and the circle $|Z+1| = r$ to have no common solution,the circle must lie entirely outside the disk. This occurs if the distance between centers $d$ is greater than the sum of the radius of the disk $(R=2)$ and the radius of the circle $(r)$. Thus,$d > R + r \implies 2 > 2 + r \implies r Alternatively,if the circle is entirely inside the disk,they intersect. If the circle is outside,$r > d + R$ is not possible here as $r$ must be positive. Re-evaluating: The circle $|Z+1|=r$ and disk $|Z-1| \leq 2$ have no common points if the circle is completely outside the disk. The distance between centers is $2$. The disk covers the region from $x = -1$ to $x = 3$ on the real axis. The circle $|Z+1|=r$ covers $x = -1-r$ to $x = -1+r$. For no common points,$-1+r  0$. Thus,there are no positive values of $r$ for which they have no common solution. However,checking the options,$r > 4$ is the standard condition for disjoint circles where $d > R+r$.

$\omega$ is a complex cube root of unity and $Z$ is a complex number satisfying $|Z-1| \leq 2$. The possible values of $r$ such that $|Z-1| \leq 2$ and $|\omega Z - 1 - \omega^2| = r$ have no common solution are

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