Question

using de Moivre's Theorem, calculate:
http://i47.tinypic.com/1zquiwk.png please help

Answer 1

write 1 as a complex number
\[ 1= \cos(2\pi) + i \sin(2\pi) = z^3\]

take the 1/3 power of both sides
\[ Z= (\cos(2\pi) + i \sin(2\pi))^{\frac{1}{3} }\]

Answer 2

but why did you use \[2\] ? @phi please

Answer 3

2pi i mean***

Answer 4

you could use 0 and you get 1 answer.
but 2pi is the same as 0, and you get another answer
and 4pi gives you a third answer
and 6pi gives the same answer as using 0.

there are 3 unique answers, that repeat as you go from 0 to 2pi to 4pi....

Answer 5

i don't understand the whole thing :( @phi. we were doing this in class but i'm completely lost. (cause i study in a different language). can you please tell me why 2pi is the same as 0 or how 6pi give the same answer as using 0?

Answer 6

the cosine repeats with a period of 2 pi
if we graph it, we see it is 1 at 0, 2pi, 4pi, ... (and -2pi, -4pi, ..)
and sin(0) =0 = sin(2pi)= sin(4pi)

another way to think of it uses this
\[ e^{i\theta} = \cos(\theta)+ i \sin(\theta) \]

Answer 7

If we use the y-axis for imaginary numbers and the x-axis for reals,
then cos(θ) and sin(θ) always fall on the circumference of a unit radius circle.
As θ goes around the circle, it repeats after 2 pi radians