Question

Find the cube roots of 8(cos 216° + i sin 216°).

Answer 1

Hey there Ms Elizabeth c:
So looks like we'll need to apply De'Moivre's Theorem, yes?

Answer 2

\[\Large\rm \left[8\cos216+\mathcal i \sin 216\right]^{1/3}\]So ummm...

Answer 3

Where you at girl? >.<
Remember De'Moivre's Theorem?

Answer 4

I guess we would want to write our angle like this:\[\large\rm \left[8\cos\left(216+360k\right)+\mathcal i \sin\left(216+360k\right)\right]^{1/3},~k=0,\pm1,\pm2,...\]This k represents any multiple of 360 degrees.
So this is telling us that we can spin around and around any number of times and land at the same spot if we spin in multiples of 360 degrees.

The `third root` will correspond to the `first three positive` values that k can take on.\[\large\rm \left[8\cos\left(216+360k\right)+\mathcal i \sin\left(216+360k\right)\right]^{1/3},~k=0,~1,~2\]Since we're counting from zero, 0, 1 and 2 will be the first three positive values for k.
After that, the k values would just end up giving us repeated values that we already got.

Answer 5

And then from there, you apply De'Moivre's Theorem, bringing the 1/3 inside.
Then to find your 3 roots:
Let k=0 and simplify. This will correspond to the first cubic root.
Then let k=1, simplify. This corresponds to the second cubic root.
And then once more with k=2.

Answer 6

thank you!!!!!! sorry I was getting ready for bed. Thank you so much, I'll solve for them now!

Answer 7

cool c:

Answer 8

quick question! should my calculator be in radian or degree mode to solve this problem?

Answer 9

We're currently working in degrees.
You could covert your 360k to 2k(pi)
but converting your 216 degrees to radians will be a little ugly I think.

So let's just stay in degrees. I think it'll work out ok.

Answer 10

ah can u possibly help me convert the 216 degrees to radians? I'm really bad with the unit circle etc

Answer 11

\[\Large\rm 216^o\]We want to multiply it by something that is equivalent to 1.
That way we don't change it's value, just the way it looks.
\(\Large\rm \pi\) is equivalent to \(\Large\rm 180^o\).
So we can multiply it by \(\Large\rm \dfrac{\pi}{180^o}\) without running into any problems.
The reason I put the 180 on the bottom is because the units are degrees on the 180.
So we'll get a nice cancellation with the degree units.\[\Large\rm 216^o\cdot \frac{\pi}{180^o}=\frac{216\pi}{180}\]Which can probably be simplified down a little bit....
I don't it's one of our special angles though.

Answer 12

oh right true okay thanks! I forgot that rule lol it's late. I'll check with u with the right answer in a second

Answer 13

so with the 216 degrees is it like 216π/180 *k?

Answer 14

No it would be like ummm...\[\Large\rm 216^o+360^ok\]
\[\Large\rm \frac{216\pi}{180}+\frac{360k \pi}{180}\]
Which simplifies to,\[\Large\rm \frac{6\pi}{5}+2k \pi\]Hopefully I got the simplification correct on the first term..
Yah I think it's ok.

Answer 15

for each root? so the simplification would be like, 6pi/5 + 2(0)pi for the first root, and so on for the others?

Answer 16

Well we have to apply De'Moivre's Theorem before you can plug your k values in.
Err no I guess we don't have to. Either order is fine.

But yes you have the right idea.
Plug in your k=0, simplify (apply De'Moivre's, bringing the 1/3 into the angle).
Then combine fractions if necessary (it won't be necessary for the first one).

Answer 17

okay I got these answers for the roots

1) 94.25
2) 100.5
3) 106.8

Answer 18

I'm not sure what you mean.
Your roots should consist of both a `real` and `imaginary` part.
I'm not sure what that single value is supposed to represent :o

Answer 19

ugh nevermind I'm sorry I'm not understanding this lmao

Answer 20

Ok ok ok ok ok ok >.<
Lemme just uhhh go through it real quick.
And you can stop me if a spot is confusing.

Answer 21

okay thank you so much omg

Answer 22

\[\Large\rm \left[8(\cos(216^o)+\mathcal i \sin(216^o)\right]^{1/3}\]Step 1: Rewrite our angles to allow for rotations.\[\large\rm \left[8(\cos(216^o+360^ok)+\mathcal i \sin(216^o+360^ok)\right]^{1/3},\quad k=0,1,2\]

Answer 23

ok got it

Answer 24

Distribute the 1/3 power to the 8 and to the other part.
Step 2: Apply De'Moivre's Theorem, bringing the 1/3 into the angle.\[\large\rm 8^{1/3}\left(\cos\left(\frac{216^o}{3}+\frac{360^ok}{3}\right)+\mathcal i \sin\left(\frac{216^o}{3}+\frac{360^ok}{3}\right)\right),\quad k=0,1,2\]

Answer 25

ohhh

Answer 26

Simplify the 8^1/3.
Step 3: Plug in the different k values to get the roots.\[\large\rm k=0:\quad 2\left(\cos\left(\frac{216^o}{3}+0\right)+\mathcal i \sin\left(\frac{216^o}{3}+0\right)\right)\]So your first root is nice and simple:\[\Large\rm 2\left(\cos72^o+\mathcal i \sin72^o\right)\]That is your first root, that is where you stop with it.
See how it doesn't simplify down to a #?

Answer 27

okay yeah I understand!

Answer 28

okay so what about the second root now?

Answer 29

\[\large\rm k=1:\quad 2\left(\cos\left(\frac{216^o}{3}+\frac{360^o(1)}{3}\right)+\mathcal i \sin\left(\frac{216^o}{3}+\frac{360^o(1)}{3}\right)\right)\]

Answer 30

2(cos(72+120) + isin(72+120)
so it's 2(cos192) + isin(192) ?

Answer 31

Ok great!
You need an extra set of braces though to show that the 2 is multiplying both terms.
2((cos192) + isin(192))

Answer 32

Oh oh nevermind, you had one in front of the cosine, you just didn't close it. my bad.

Answer 33

oh oops lmao thanks. what about the third?

Answer 34

\[\large\rm k=2:\quad 2\left(\cos\left(\frac{216^o}{3}+\frac{360^o(2)}{3}\right)+\mathcal i \sin\left(\frac{216^o}{3}+\frac{360^o(2)}{3}\right)\right)\]Just plug in 2's for your k's ya dingus! >.<

Answer 35

OH RIGHT.

Answer 36

thank u dr. steve brule!

Answer 37

Working in degrees is actually really nice because you don't have to worry about getting a common denominator.

When you're in radians you'll often have angles like: \(\Large\rm \dfrac{2\pi}{6}+\dfrac{2k \pi}{3}\)
So you're left with an extra step of combining the fractions.

Fer yer health.

Answer 38

so is it 2((cos)(312) + i(sin)(312) ??

Answer 39

Again your brackets are a total mess LOL
But yes you've got the correct angle :)

Answer 40

lmao ur right okay I'll clean them up ahahaha. thank you so much. you're very smart and kind <3

Answer 41

See how the brackets should match up? :o