## anonymous one year ago 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

1. anonymous

since $$1^7=1$$ you know one answer, namely 1 `

2. anonymous

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

3. anonymous

Is there a formula to follow?

4. anonymous

@satellite73

5. anonymous

no, that is all

6. anonymous

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

7. anonymous

not sure I understand :(

8. anonymous

it is the same thing we did before

9. anonymous

but of 1?

10. anonymous

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

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

18. anonymous

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

19. 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

20. anonymous

I thought theta was 0

21. 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

22. anonymous

According to satellites thing

23. anonymous

satellite's

24. jim_thompson5910

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

25. anonymous

gotcha

26. anonymous

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

27. jim_thompson5910

something like this |dw:1434596805847:dw|

28. jim_thompson5910

plug in k = 0 through k = 6

29. anonymous

but is there a way to measure the coordinates specifically?

30. jim_thompson5910

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

31. jim_thompson5910

plug in k = 0 through k = 6

32. jim_thompson5910

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

33. jim_thompson5910

|dw:1434597029810:dw|

34. anonymous

Is (1,2pi/7) one?

35. 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

36. jim_thompson5910

|dw:1434597231701:dw|

37. anonymous

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

38. jim_thompson5910

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

39. jim_thompson5910

with software, yes

40. jim_thompson5910

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

41. anonymous

Geogebra doesn't seem to have a function for this

42. jim_thompson5910

it supports trig functions like sin, cos, etc

43. jim_thompson5910

this is what I get with geogebra

44. anonymous

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

45. anonymous

Thanks for the help!

46. 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

47. anonymous

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

48. jim_thompson5910

oh gotcha

49. anonymous

would it work for work?

50. jim_thompson5910

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

51. anonymous

awesome :D Thanks for the help!

52. jim_thompson5910

np