If -z-1- - 2 and - z -1 - -2- - a where - is a cube root of unity- then complete set of values of a is

The correct option is C 0 ≤ a ≤ 4 | z 1 | ≤ 2 Since w is the cube root of unity. ∴ w ^ 3 =1 & 1+w+ w ^ 2 =0 | z 1 | ≤ 2 ...

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Sum of Coefficients of All Terms

If |z-1| ≤ ...

Question

If $| z - 1 | \leq 2$ and $∣ ∣ ω z - 1 - ω^{2} ∣ ∣ = a$ (where $ω$ 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}}{2}

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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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Similar questions

| z - 1 | \leq 2

| ω z - 1 - ω^{2} | = a

ω

a

| z - 1 | \leq 2

| w z - 1 - w^{2} | = a

ω

(1 + \frac{1}{ω}) (1 + \frac{1}{ω^{2}}) + (1 + \frac{2}{ω}) (1 + \frac{2}{ω^{2}}) + (1 + \frac{3}{ω}) (1 + \frac{3}{ω^{2}}) + . . . . . + (1 + \frac{n}{ω}) (1 + \frac{n}{ω^{2}})

| z - 1 | \leq 2

| ω z - 1 - ω^{2} | = a

ω

a

z_{1} + z_{2} + z_{3} = A, z_{1} + z_{2} ω + z_{3} ω^{2} = B

z_{1} + z_{2} ω^{2} + z_{3} ω = C

1, ω, ω^{2}

{| A |}^{2} + {| B |}^{2} + {| C |}^{2} =

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