Complex Number

Discussion in 'Math' started by xiahbaby, Apr 14, 2009.

  1. xiahbaby

    Thread Starter New Member

    Feb 4, 2009
    9
    0
    Find all cube roots of -3(surd of 3) + 3j in polar form.

    Can anyone show me the working?:confused:
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Hopefully the cube root of the magnitude doesn't pose any problems for you.

    I 'm guessing you are more concerned with the phasor angles and how many solutions there might be.

    -3√3+3j has a polar form angle of 150°.

    So the cube root's simplest phasor angle would be 50° (using 150°/3).

    But clearly other solutions will exist. There must be a recursive or indexed offset which will rotate the cubed phasor's position back to 150° but indexed by 360° multiples. To index the resultant cubed value by one or more increments of ±360°, one must have a either a positive or negative index of 120°.

    So there would be an infinite number of solutions with phase angles given by

    50°± N x 120° where N=0, 1, 2, 3, ....etc.

    For N=0, angle = 50° --> 3 x 50° --> 150°
    For N=+1, angle = 170° --> 3 x 170° --> 510° --> 150° + 360°
    For N=-1, angle = -70° --> 3 x -70° --> -210° --> 150° - 360°

    and so on.

    The magnitude would be constant for each phasor.
     
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