Question

please help me i have some idea of what to do but i need help !
use de moivre's theorem to solve each of these questions. plot your solutions on the complex plain.
(i) z^3 = 8

Answer 1

I think they want you to write 8 as
\[ 8=8(\cos(2 \pi) + i \sin(2 \pi))\]

Answer 2

well the theorm we have been using is diff cuz u have to add 2n(pi)

Answer 3

yes, that is even better.
8=8(cos(2nπ)+isin(2nπ))

Answer 4

like the way our teacher has been telling us is we have to get this \[\sqrt[3]{8?}\]

Answer 5

then it is 2
to the power of a third

Answer 6

See
http://en.wikipedia.org/wiki/De_Moivre's_formula#Applications

you solve
\[ \left( z^3 \right)^{\frac{1}{3}} = 8^{\frac{1}{3}}(\cos(2n\pi)+i \sin(2n\pi)^{\frac{1}{3}}\]

Answer 7

you get 3 different answers, for n=0,1,2. then they start repeating n=3 gives the same answer as n=0

Answer 8

yeah i kinda get thanks

Answer 9

wait usually ther is another angle you can't get 0 of 2n pi

Answer 10

in case it is not clear
8 = 8( cos(0) + i sin(0))
when we add in multiples of 2π we get
8 = 8( cos(2nπ ) + i sin(2nπ))