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The cube roots of unity when represented...

The cube roots of unity when represented on Argand diagram form the vertices of an equilateral triangle.

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The correct Answer is:
True
Since cube roote of unity are 1, `omega , omega^2` given by
`A(1,0),B(-1/2,sqrt(3)/2),C(-1/2,-sqrt(3)/2)`
`rArr AB= BC=CA= sqrt(3)` ltb rgt Hence , cube roots of untiy from an equilateral triangle .
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Knowledge Check

  • Which of the following points are the vertices of an equilateral triangle ?

    A
    `(a,a),(-a,-a),(2a,a)`
    B
    `(a,a),(-a,-a),(-a sqrt(3),a sqrt(3))`
    C
    `(sqrt(2)a,-a),(a,sqrt(2)a),(a-a)`
    D
    `(0,0),(a,-a),(a,sqrt(2)a)`

  • If the origin and the non - real roots of the equation 3z^(2)+3z+lambda=0, AA lambda in R are the vertices of an equilateral triangle in the argand plane, then sqrt3 times the length of the triangle is

    A
    2 units
    B
    1 units
    C
    3 units
    D
    4 units

  • If z _(1) , z _(2) and z _(3) represent the vertices of an equilateral triangle such that

    A
    `z _(1) + z _(2) = z _(3)`
    B
    `z _(1) + z _(2) + z _(3) =0`
    C
    `z _(1) z _(2) = (1)/( z _(3))`
    D
    `z _(1) - z _(2) = z _(3) - z _(2)`

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