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Mathematics 22 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Is there a formula to follow?

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

no, that is all

OpenStudy (anonymous):

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

OpenStudy (anonymous):

not sure I understand :(

OpenStudy (anonymous):

it is the same thing we did before

OpenStudy (anonymous):

but of 1?

OpenStudy (anonymous):

one answer is 1 right?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

jimthompson5910 (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

OpenStudy (anonymous):

I thought theta was 0

jimthompson5910 (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

OpenStudy (anonymous):

According to satellites thing

OpenStudy (anonymous):

satellite's

jimthompson5910 (jim_thompson5910):

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

OpenStudy (anonymous):

gotcha

OpenStudy (anonymous):

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

jimthompson5910 (jim_thompson5910):

something like this |dw:1434596805847:dw|

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