Question

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°).

Answer 1

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

Answer 2

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

Answer 3

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 \(\large \sqrt[3] { 27 \ e^{i \ \frac{11}{6}\pi}}\)

Answer 4

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

Answer 5

there's something related called deMoivre, which might be easier for you right now

\((cos \theta + i \ sin \theta)^n = cos \ n \ \theta + i \ sin \ n \ \theta\)

Answer 6

Yes, I've heard of DeMoivre's Theorem

Answer 7

for you \(n = \frac{1}{3}\)

and don't forget the \(\sqrt[3] 27\) !

Answer 8

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

Answer 9

not that simple!

you will get roots following this pattern

\( 27^{1/3} \ cis(\frac{2Ï€ \ m+\theta }{3}) \)

for m = 0, 1, 2

Answer 10

ie 3 roots for a cubic

Answer 11

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

Answer 12

sorry to confuse you, cis is just a shorthand for \(cos + i sin\)

so stuff that argument into both of them

Answer 13

What does the p stand for in that above formula?

Answer 14

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

Answer 15

so you will have

\(3(cos \frac{\theta}{3} + i sin \frac{\theta }{3})\)
\(3(cos \frac{\theta + 2 \pi}{3} + i sin \frac{\theta + 2 \pi}{3})\)
\(3(cos \frac{\theta + 4 \pi}{3} + i sin \frac{\theta + 4 \pi}{3})\)

Answer 16

\[\pi m\]
not pim

Answer 17

is what he meant

Answer 18

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 \(\sqrt[3]{27}\) ....