## samnatha Group Title 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 one year ago one year ago

1. phi Group Title

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

2. samnatha Group Title

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

3. phi Group Title

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

4. samnatha Group Title

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

5. samnatha Group Title

then it is 2 to the power of a third

6. phi Group Title

See http://en.wikipedia.org/wiki/De_Moivre's_formula#Applications you solve $\left( z^3 \right)^{\frac{1}{3}} = 8^{\frac{1}{3}}(\cos(2n\pi)+i \sin(2n\pi)^{\frac{1}{3}}$

7. phi Group Title

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

8. samnatha Group Title

yeah i kinda get thanks

9. samnatha Group Title

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

10. phi Group Title

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))