Thanks for helping!
I'm not sure I understand this problem:
II need to find all seventh roots of unity and sketch them on this axes.
http://www.google.com/imgres?imgurl=http%3A%2F%2Fuser-content.enotes.com%2F7c4540d5f581a45d753fc2cff4180d9b3cb69630_thumb.png&imgrefurl=http%3A%2F%2Fwww.enotes.com%2Fhomework-help%2Ffind-all-seventh-roots-unity-sketch-them-axes-436533&h=287&w=282&tbnid=aGzfxmKJbQW5JM%3A&zoom=1&docid=WUxe1gxLnyuV-M&ei=pg55U7atMtOSqAbAgYGIAQ&tbm=isch&client=safari&ved=0CFUQMygBMAE&iact=rc&uact=3&dur=849&page=1&start=0&ndsp=32

- anonymous

- schrodinger

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- anonymous

since \(1^7=1\) you know one answer, namely 1
`

- anonymous

divide the circle up in to seven equal parts, with 1 as one of them

- anonymous

Is there a formula to follow?

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## More answers

- anonymous

- anonymous

no, that is all

- anonymous

if you want the nth roots of 1, divide the circle in to n parts
finish

- anonymous

not sure I understand :(

- anonymous

it is the same thing we did before

- anonymous

but of 1?

- anonymous

one answer is 1 right?

- anonymous

if you want to be real silly you can write
\[1=\cos(0)+i\sin(0)\]

- anonymous

divide the angle by 7 and you still get
\[\cos(0)+i\sin(0)\]

- anonymous

then add \(2\pi\) , divide by 7 and get
\[\cos(\frac{2\pi}{7})+i\sin(\frac{2\pi}{7})\]

- anonymous

wouldn't it be like this? https://www.wolframalpha.com/input/?i=x^7-1%3D0

- anonymous

all you are doing is dividing the circle up in to seven equal parts

- anonymous

okay, so if I do the same as before I'll end up with different results to graph?

- anonymous

yeah look at the wolf picture
they have the unit circle divided in to seven equal pieces

- anonymous

So dividing by 7 and adding 2pi does the same?

- jim_thompson5910

You'll have 7 roots of unity of the form
\[\Large \cos(\frac{2\pi}{7}*k)+i\sin(\frac{2\pi}{7}*k)\]
where k is an integer and k runs from k = 0 to k = 6

- anonymous

I thought theta was 0

- jim_thompson5910

For n roots of unity, you'll have n roots of the form
\[\Large \cos(\frac{2\pi}{n}*k)+i\sin(\frac{2\pi}{n}*k)\]
k will be an integer from k = 0 to k = n

- anonymous

According to satellites thing

- anonymous

satellite's

- jim_thompson5910

that happens when k = 0
there are 6 other roots of unity though

- anonymous

gotcha

- anonymous

So I make it equal 1, 2, .. and solve?

- jim_thompson5910

something like this
|dw:1434596805847:dw|

- jim_thompson5910

plug in k = 0 through k = 6

- anonymous

but is there a way to measure the coordinates specifically?

- jim_thompson5910

yeah using
\[\Large \cos(\frac{2\pi}{7}*k)+i\sin(\frac{2\pi}{7}*k)\]

- jim_thompson5910

plug in k = 0 through k = 6

- jim_thompson5910

if k = 0, the whole thing is 1+0i

- jim_thompson5910

|dw:1434597029810:dw|

- anonymous

Is (1,2pi/7) one?

- jim_thompson5910

next plug in k = 1
\[\Large \cos(\frac{2\pi}{7}*k)+i\sin(\frac{2\pi}{7}*k)\]
\[\Large \cos(\frac{2\pi}{7}*1)+i\sin(\frac{2\pi}{7}*1)\]
\[\Large \cos(\frac{2\pi}{7})+i\sin(\frac{2\pi}{7})\]
that approximates to 0.6234898+0.78183148i

- jim_thompson5910

|dw:1434597231701:dw|

- anonymous

Is there a way to graph that accurately on the graph above?

- jim_thompson5910

if you want, you can raise 0.6234898+0.78183148i to the 7th power
you should get 1 as a result

- jim_thompson5910

with software, yes

- jim_thompson5910

you can use geogebra as that's what I would use

- anonymous

Geogebra doesn't seem to have a function for this

- jim_thompson5910

it supports trig functions like sin, cos, etc

- jim_thompson5910

this is what I get with geogebra

##### 1 Attachment

- anonymous

Oh, I'll use this then. So for work, I can just show cis(2pi/7*k)? for k=0-6?

- anonymous

Thanks for the help!

- jim_thompson5910

cis(2pi/7*k) doesn't work
so type in cos(2pi/7*k)+i*sin(2pi/7*k)
and replace k with 0 through 6

- anonymous

For work in my worksheet I meant, sorry. Thanks for the help!

- jim_thompson5910

oh gotcha

- anonymous

would it work for work?

- jim_thompson5910

yeah your teacher should accept cis as shorthand for cos + i*sin

- anonymous

awesome :D Thanks for the help!

- jim_thompson5910

np

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