Question

*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°).

Answer 1

My work so far:

Answer 2

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

Answer 3

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

Answer 4

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 \[3(\cos \frac{ \theta }{ 3 }+i \sin \frac{ \theta }{ 3 })\]

Answer 5

Im stuck on finding the second and third roots..

Answer 6

HI!!

Answer 7

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

Answer 8

Correct @misty1212

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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:

\[\sqrt[3]{27}(\cos \frac{ \theta+2 \pi(1)}{ 3 }+ i \sin \frac{ \theta+2 \pi(1) }{ 3})\]

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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?

Answer 22

Oh ok

Answer 23

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

Answer 24

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

Answer 25

330/3 = 110

Answer 26

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

Answer 27

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

Answer 28

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 \(360\)

Answer 29

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?

Answer 30

yeah just keep going around

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

ready to find the next root? there are three total

Answer 37

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

Answer 38

exactly!

Answer 39

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

Answer 40

hmmm no

Answer 41

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

Answer 42

Yes

Answer 43

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

Answer 44

Yes

Answer 45

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

Answer 46

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

Answer 47

350

Answer 48

you keep going around the circle

Answer 49

right

Answer 50

not sure where the 196.6 came from

Answer 51

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!

Answer 52

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

Answer 53

Crystal clear, thank you!

Answer 54

you're welcome dear