Find all cube roots of -3(surd of 3) + 3j in polar form.

Can anyone show me the working?

 

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.