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  1. anonymous
    • one year ago
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    since \(1^7=1\) you know one answer, namely 1 `

  2. anonymous
    • one year ago
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    divide the circle up in to seven equal parts, with 1 as one of them

  3. anonymous
    • one year ago
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    Is there a formula to follow?

  4. anonymous
    • one year ago
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    @satellite73

  5. anonymous
    • one year ago
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    no, that is all

  6. anonymous
    • one year ago
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    if you want the nth roots of 1, divide the circle in to n parts finish

  7. anonymous
    • one year ago
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    not sure I understand :(

  8. anonymous
    • one year ago
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    it is the same thing we did before

  9. anonymous
    • one year ago
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    but of 1?

  10. anonymous
    • one year ago
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    one answer is 1 right?

  11. anonymous
    • one year ago
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    if you want to be real silly you can write \[1=\cos(0)+i\sin(0)\]

  12. anonymous
    • one year ago
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    divide the angle by 7 and you still get \[\cos(0)+i\sin(0)\]

  13. anonymous
    • one year ago
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    then add \(2\pi\) , divide by 7 and get \[\cos(\frac{2\pi}{7})+i\sin(\frac{2\pi}{7})\]

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

  15. anonymous
    • one year ago
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    all you are doing is dividing the circle up in to seven equal parts

  16. anonymous
    • one year ago
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    okay, so if I do the same as before I'll end up with different results to graph?

  17. anonymous
    • one year ago
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    yeah look at the wolf picture they have the unit circle divided in to seven equal pieces

  18. anonymous
    • one year ago
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    So dividing by 7 and adding 2pi does the same?

  19. jim_thompson5910
    • one year ago
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    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

  20. anonymous
    • one year ago
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    I thought theta was 0

  21. jim_thompson5910
    • one year ago
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    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

  22. anonymous
    • one year ago
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    According to satellites thing

  23. anonymous
    • one year ago
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    satellite's

  24. jim_thompson5910
    • one year ago
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    that happens when k = 0 there are 6 other roots of unity though

  25. anonymous
    • one year ago
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    gotcha

  26. anonymous
    • one year ago
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    So I make it equal 1, 2, .. and solve?

  27. jim_thompson5910
    • one year ago
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    something like this |dw:1434596805847:dw|

  28. jim_thompson5910
    • one year ago
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    plug in k = 0 through k = 6

  29. anonymous
    • one year ago
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    but is there a way to measure the coordinates specifically?

  30. jim_thompson5910
    • one year ago
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    yeah using \[\Large \cos(\frac{2\pi}{7}*k)+i\sin(\frac{2\pi}{7}*k)\]

  31. jim_thompson5910
    • one year ago
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    plug in k = 0 through k = 6

  32. jim_thompson5910
    • one year ago
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    if k = 0, the whole thing is 1+0i

  33. jim_thompson5910
    • one year ago
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    |dw:1434597029810:dw|

  34. anonymous
    • one year ago
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    Is (1,2pi/7) one?

  35. jim_thompson5910
    • one year ago
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    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

  36. jim_thompson5910
    • one year ago
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    |dw:1434597231701:dw|

  37. anonymous
    • one year ago
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    Is there a way to graph that accurately on the graph above?

  38. jim_thompson5910
    • one year ago
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    if you want, you can raise 0.6234898+0.78183148i to the 7th power you should get 1 as a result

  39. jim_thompson5910
    • one year ago
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    with software, yes

  40. jim_thompson5910
    • one year ago
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    you can use geogebra as that's what I would use

  41. anonymous
    • one year ago
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    Geogebra doesn't seem to have a function for this

  42. jim_thompson5910
    • one year ago
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    it supports trig functions like sin, cos, etc

  43. jim_thompson5910
    • one year ago
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    this is what I get with geogebra

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

  45. anonymous
    • one year ago
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    Thanks for the help!

  46. jim_thompson5910
    • one year ago
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    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

  47. anonymous
    • one year ago
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    For work in my worksheet I meant, sorry. Thanks for the help!

  48. jim_thompson5910
    • one year ago
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    oh gotcha

  49. anonymous
    • one year ago
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    would it work for work?

  50. jim_thompson5910
    • one year ago
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    yeah your teacher should accept cis as shorthand for cos + i*sin

  51. anonymous
    • one year ago
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    awesome :D Thanks for the help!

  52. jim_thompson5910
    • one year ago
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    np

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