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My work so far:

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

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

Im stuck on finding the second and third roots..

HI!!

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

Correct @misty1212

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

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

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

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

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

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

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

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

Oh ok

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

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

330/3 = 110

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

yeah just keep going around

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

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

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

ready to find the next root? there are three total

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

exactly!

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

hmmm no

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

Yes

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

Yes

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

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

350

you keep going around the circle

right

not sure where the 196.6 came from

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

Crystal clear, thank you!

you're welcome dear