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

0 \leq a \leq 2

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\frac{1}{2} \leq a \leq \frac{\sqrt{3}}{a}

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\frac{\sqrt{3}}{2} - \frac{1}{2} \leq a \leq \frac{1}{2} + \frac{\sqrt{3}}{2}

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0 \leq a \leq 4

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Solution

The correct option is D $0 \leq a \leq 4$
$| w z - 1 - w^{2} | = a = | w z - 1 - (1 + w) | = a \Rightarrow | w z + w | = a = | w | | z + 1 | = a | z + 1 | = a$
$\Rightarrow | z - 1 + 2 | = a$
$| z - 1 | + 2 \geq a$
$0 \leq a \leq 4$

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