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
since \(1^7=1\) you know one answer, namely 1 `
divide the circle up in to seven equal parts, with 1 as one of them
Is there a formula to follow?
@satellite73
no, that is all
if you want the nth roots of 1, divide the circle in to n parts finish
not sure I understand :(
it is the same thing we did before
but of 1?
one answer is 1 right?
if you want to be real silly you can write \[1=\cos(0)+i\sin(0)\]
divide the angle by 7 and you still get \[\cos(0)+i\sin(0)\]
then add \(2\pi\) , divide by 7 and get \[\cos(\frac{2\pi}{7})+i\sin(\frac{2\pi}{7})\]
all you are doing is dividing the circle up in to seven equal parts
okay, so if I do the same as before I'll end up with different results to graph?
yeah look at the wolf picture they have the unit circle divided in to seven equal pieces
So dividing by 7 and adding 2pi does the same?
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
I thought theta was 0
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
According to satellites thing
satellite's
that happens when k = 0 there are 6 other roots of unity though
gotcha
So I make it equal 1, 2, .. and solve?
something like this |dw:1434596805847:dw|
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