anonymous
  • 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
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
since \(1^7=1\) you know one answer, namely 1 `
anonymous
  • anonymous
divide the circle up in to seven equal parts, with 1 as one of them
anonymous
  • anonymous
Is there a formula to follow?

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anonymous
  • anonymous
anonymous
  • anonymous
no, that is all
anonymous
  • anonymous
if you want the nth roots of 1, divide the circle in to n parts finish
anonymous
  • anonymous
not sure I understand :(
anonymous
  • anonymous
it is the same thing we did before
anonymous
  • anonymous
but of 1?
anonymous
  • anonymous
one answer is 1 right?
anonymous
  • anonymous
if you want to be real silly you can write \[1=\cos(0)+i\sin(0)\]
anonymous
  • anonymous
divide the angle by 7 and you still get \[\cos(0)+i\sin(0)\]
anonymous
  • anonymous
then add \(2\pi\) , divide by 7 and get \[\cos(\frac{2\pi}{7})+i\sin(\frac{2\pi}{7})\]
anonymous
  • anonymous
wouldn't it be like this? https://www.wolframalpha.com/input/?i=x^7-1%3D0
anonymous
  • anonymous
all you are doing is dividing the circle up in to seven equal parts
anonymous
  • anonymous
okay, so if I do the same as before I'll end up with different results to graph?
anonymous
  • anonymous
yeah look at the wolf picture they have the unit circle divided in to seven equal pieces
anonymous
  • anonymous
So dividing by 7 and adding 2pi does the same?
jim_thompson5910
  • 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
  • anonymous
I thought theta was 0
jim_thompson5910
  • 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
  • anonymous
According to satellites thing
anonymous
  • anonymous
satellite's
jim_thompson5910
  • jim_thompson5910
that happens when k = 0 there are 6 other roots of unity though
anonymous
  • anonymous
gotcha
anonymous
  • anonymous
So I make it equal 1, 2, .. and solve?
jim_thompson5910
  • jim_thompson5910
something like this |dw:1434596805847:dw|
jim_thompson5910
  • jim_thompson5910
plug in k = 0 through k = 6
anonymous
  • anonymous
but is there a way to measure the coordinates specifically?
jim_thompson5910
  • jim_thompson5910
yeah using \[\Large \cos(\frac{2\pi}{7}*k)+i\sin(\frac{2\pi}{7}*k)\]
jim_thompson5910
  • jim_thompson5910
plug in k = 0 through k = 6
jim_thompson5910
  • jim_thompson5910
if k = 0, the whole thing is 1+0i
jim_thompson5910
  • jim_thompson5910
|dw:1434597029810:dw|
anonymous
  • anonymous
Is (1,2pi/7) one?
jim_thompson5910
  • 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
  • jim_thompson5910
|dw:1434597231701:dw|
anonymous
  • anonymous
Is there a way to graph that accurately on the graph above?
jim_thompson5910
  • 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
  • jim_thompson5910
with software, yes
jim_thompson5910
  • jim_thompson5910
you can use geogebra as that's what I would use
anonymous
  • anonymous
Geogebra doesn't seem to have a function for this
jim_thompson5910
  • jim_thompson5910
it supports trig functions like sin, cos, etc
jim_thompson5910
  • jim_thompson5910
this is what I get with geogebra
anonymous
  • anonymous
Oh, I'll use this then. So for work, I can just show cis(2pi/7*k)? for k=0-6?
anonymous
  • anonymous
Thanks for the help!
jim_thompson5910
  • 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
  • anonymous
For work in my worksheet I meant, sorry. Thanks for the help!
jim_thompson5910
  • jim_thompson5910
oh gotcha
anonymous
  • anonymous
would it work for work?
jim_thompson5910
  • jim_thompson5910
yeah your teacher should accept cis as shorthand for cos + i*sin
anonymous
  • anonymous
awesome :D Thanks for the help!
jim_thompson5910
  • jim_thompson5910
np

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