## anonymous one year ago Someone please help me with this problem, I need to show my work and I have no idea how or where to start : Find the cube roots of 27(cos 330° + i sin 330°).

1. IrishBoy123

you know $$e^{i \theta} = cos \theta + i \ sin \theta$$??

2. anonymous

I've seen it before in my lessons but I have never actually applied it to an problems

3. IrishBoy123

it's is very useful to know you can write $$27(cos 330° + i sin 330°)$$ as $$\large 27 \ e^{i \ \frac{11}{6}\pi}$$ so you want $$\large \sqrt[3] { 27 \ e^{i \ \frac{11}{6}\pi}}$$

4. anonymous

That looks very confusing to solve Im uncertain as to how im supposed to take the cube root of that number

5. IrishBoy123

there's something related called deMoivre, which might be easier for you right now $$(cos \theta + i \ sin \theta)^n = cos \ n \ \theta + i \ sin \ n \ \theta$$

6. anonymous

Yes, I've heard of DeMoivre's Theorem

7. IrishBoy123

for you $$n = \frac{1}{3}$$ and don't forget the $$\sqrt[3] 27$$ !

8. anonymous

So $(\cos \theta+i \sin \theta)^{\frac{ 1 }{ 3 }}=\cos (\frac{ 1 }{ 3 })\theta+i \sin (\frac{ 1 }{ 3 })\theta ?$

9. IrishBoy123

not that simple! you will get roots following this pattern $$27^{1/3} \ cis(\frac{2π \ m+\theta }{3})$$ for m = 0, 1, 2

10. IrishBoy123

ie 3 roots for a cubic

11. anonymous

What is $cis(\frac{ 2pim+\theta }{ 3 }) ???$ Im confused. I know there will be 3 answers to this because its asking for cube roots.

12. IrishBoy123

sorry to confuse you, cis is just a shorthand for $$cos + i sin$$ so stuff that argument into both of them

13. anonymous

What does the p stand for in that above formula?

14. anonymous

I believe he meant pi times m, it just put pim instead of writing the letter for pi

15. IrishBoy123

so you will have $$3(cos \frac{\theta}{3} + i sin \frac{\theta }{3})$$ $$3(cos \frac{\theta + 2 \pi}{3} + i sin \frac{\theta + 2 \pi}{3})$$ $$3(cos \frac{\theta + 4 \pi}{3} + i sin \frac{\theta + 4 \pi}{3})$$

16. anonymous

$\pi m$ not pim

17. anonymous

is what he meant

18. IrishBoy123

from deMoivre, i think you have already seen how to get the first root ie when you wrote : $$(cosθ+isinθ)^{1/3}=cos(1/3)θ+isin(1/3)θ$$ the other roots come up only because the sinusoid repeats itself with period $$2 \pi$$ and you need to look at all solutions for a cube root, the roots will repeat themselves after you have the first 3. for, say, sixth root, you will get six distinct roots and then they will repeat.... and don;t forget the $$\sqrt[3]{27}$$ ....