SOLUTION: Find the sixth roots of -64. Enter the solutions in standard form, in any order. (Enter your answers separated by semicolons.) Thank you in advanced.

Algebra ->  Trigonometry-basics -> SOLUTION: Find the sixth roots of -64. Enter the solutions in standard form, in any order. (Enter your answers separated by semicolons.) Thank you in advanced.      Log On


   



Question 984942: Find the sixth roots of -64. Enter the solutions in standard form, in any order.
(Enter your answers separated by semicolons.)
Thank you in advanced.

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
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Notice that     -64 = 64(cos(180°) + i*sin(180°)).

The sixth roots of -64 are 6 complex numbers

1)  2(cos(30°) + i*sin(30°)) = 2%28sqrt%283%29%2F2+%2B+i%2A%281%2F2%29%29 = sqrt%283%29+%2B+i;                                                 (Notice that 30° = 180°/6)

2)  2(cos(30°+60°) + i*sin(30°+60°)) = 2(cos(90°) + i*sin(90°)) = 2%2A0+%2B+2%2Ai%29 = 2i;       (Notice that 60° = 360°/6)

3)  2(cos(30°+120°) + i*sin(30°+120°)) = 2(cos(150°) + i*sin(150°)) = 2%2A%28%28-sqrt%283%29%2F2%29+%2B+i%2A%281%2F2%29%29 = -sqrt%283%29+%2B+i%29;

4)  2(cos(30°+180°) + i*sin(30°+180°)) = 2(cos(210°) + i*sin(210°)) = 2%2A%28%28-sqrt%283%29%2F2%29+%2B+i%2A%28-1%2F2%29%29 = -sqrt%283%29+-+i%29;

5)  2(cos(30°+240°) + i*sin(30°+240°)) = 2(cos(270°) + i*sin(270°)) = 2%2A%280+-+i%29 = -2i%29;

6)  2(cos(30°+300°) + i*sin(30°+300°)) = 2(cos(330°) + i*sin(330°)) = 2%2A%28%28sqrt%283%29%2F2%29+%2B+i%2A%28-1%2F2%29%29 = sqrt%283%29+-+i%29.

If you want to see my lessons on complex numbers in this site, look in this
    - REVIEW of lessons on complex numbers
and especially in this one
    - How to take a root of a complex number.