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If arg(z - 1) = pi/6 and arg(z + 1) = 2 ...

If `arg(z - 1) = pi/6` and `arg(z + 1) = 2 pi/3` , then prove that `|z| = 1`.

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Knowledge Check

  • arg z= pi if and only if z is a

    A
    purely imaginary number
    B
    positive real number
    C
    negative real number
    D
    non negative real number

  • If arg((z-1)/(z+1))=(pi)/(3) then the of z is

    A
    circle
    B
    ellipse
    C
    triangle
    D
    straight line

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