*WILL FAN AND MEDAL* Can someone please take a look at my work and help me work out the rest of this problem?
Find the cube roots of 27(cos 330° + i sin 330°).

- anonymous

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- anonymous

My work so far:

- anonymous

\[z = r(\cos \theta + i \sin \theta)\]

- anonymous

\[\sqrt[n]{r}(\cos \frac{ \theta+2\pi k }{ n }+i \sin \frac{ \theta+2\pi k }{ n })\]

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## More answers

- anonymous

I substituted the values into the formula above. K is any real number so I chose to use k= 0, 1, 2. For my first root working with k=0, I got \[3(\cos \frac{ \theta }{ 3 }+i \sin \frac{ \theta }{ 3 })\]

- anonymous

Im stuck on finding the second and third roots..

- misty1212

HI!!

- misty1212

\[\frac{330}{3}=111\] right?

- anonymous

Correct @misty1212

- misty1212

since you seem to be working in degrees, which is weird, but whatever

- misty1212

then to get the next one, go around the circle again

- misty1212

\[330+360=690\] and
\[\frac{690}{3}=230\]

- misty1212

so next angle is \(230\)
then go around yet again

- misty1212

i can see why it is confusing, because you are using that \(\frac{\theta+2k\pi}{n}\) thing, but you using degrees

- misty1212

if you were using radians, then that is what you would use
but since you are using degrees, keep adding 360 then divide
not \(2\pi\)

- anonymous

Im using DeMoivre's Theorem, but I was just confused because I'm not sure if I should substitute the angle 330 degrees in place of theta in the expression I have so far. In finding my second root, I used my second value for k, which is 1. Im just stuck on whether or not im supposed to substitute in 330 degrees in place of theta. This is the work I have built around my second root:
\[\sqrt[3]{27}(\cos \frac{ \theta+2 \pi(1)}{ 3 }+ i \sin \frac{ \theta+2 \pi(1) }{ 3})\]

- anonymous

I cant give my answers in degrees on this problem though.

- anonymous

Because the standard form of the expression i have is \[z=r(\cos \theta + i \sin \theta)\]

- misty1212

if you are working in degrees, then you have to add 360 not \(2\pi\)

- misty1212

and the problem was given in degrees, so probably you are supposed to answer in degree
otherwise it is just too weird

- misty1212

if you want to work in radians, then you would need to convert \(330\) in to radians as step one

- anonymous

For my first root I got \[3(\cos \frac{ \theta }{ 3 } +i \sin \frac{ \theta }{ 3 })\]
So if i am to give my answer in degree form, how do I get that from this answer?

- anonymous

Oh ok

- misty1212

\(\theta\) is the angle
in this case \(\theta=330\)

- anonymous

Alright, so i substitute 330 in place of theta and simplify?

- anonymous

330/3 = 110

- misty1212

looks like you may be confusing yourself with the formula
in simple english to take the cubed root, divide the angle by 3

- misty1212

so first answer is
\[3\left(\cos(110^\circ)+i\sin(110^\circ)\right)\]

- misty1212

to get the second answer, as i wrote above, add \(360\) to \(330\)
then divide by 3
IF you were working in radians (which you are not) then you would add \(2\pi\) but since you are working in degrees, add \(360\)

- anonymous

I see what you mean here by how confusing it is. So I could solve this just by using a unit circle and finding coterminal angles and dividing them by 3 to get cube roots?

- misty1212

yeah just keep going around

- misty1212

don't forget for the trig form of a complex number the angle is not unique, since sine and cosine are periodic

- anonymous

Okay, so my first root would be 330/3= 110?

- misty1212

no, dear, that is your first ANGLE
the root is \[3\left(\cos(110^\circ)+i\sin(110^\circ)\right)\]

- anonymous

OH! Okay, so I do that process but just substitute in my answer for theta?

- misty1212

you have no idea what that number is, since you know nether \(\cos(110^\circ)\) or \(\sin(110^\circ)\)
just leave it in that form

- misty1212

ready to find the next root? there are three total

- anonymous

Right, thats what I mean. So my second root would look like this:
\[3(\cos(230)+i \sin (230))\]

- misty1212

exactly!

- anonymous

Oh, and then my last root would look like this:
3(cos(196.6)+isin(196.6))

- misty1212

hmmm no

- misty1212

lets back up a second
first angle was \(330\div 3=110\)

- anonymous

Yes

- misty1212

second one was \(\frac{330+360}{3}=230\)

- anonymous

Yes

- misty1212

third one add 360 again
\[\frac{330+360+360}{3}\]

- misty1212

or more to your formula
\[\frac{330+2\times 360}{3}\]

- anonymous

350

- misty1212

you keep going around the circle

- misty1212

right

- misty1212

not sure where the 196.6 came from

- anonymous

Oh my gosh, thank you so much for explaining that. I really needed it. You are by far the most persistent person yet. Thank you so much!

- misty1212

\[\huge \color\magenta\heartsuit\]
btw i hope it is now clear who easy this is

- anonymous

Crystal clear, thank you!

- misty1212

you're welcome dear

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