## Loser66 one year ago Find cube root of i Please, help

1. zzr0ck3r

I am here to learn, I sux at this stuff. Something about a circle :)

2. anonymous

|dw:1441159218311:dw|

3. anonymous

take one third of that angle to find the first one the other two are evenly spaced around the circle

4. anonymous

|dw:1441159323165:dw|

5. anonymous

$\frac{\sqrt3}{2}+\frac{1}{2}i$ is that one two more to go

6. ganeshie8

geometry is awesome! you want to find a number which when cubed produces $$i$$ : $\large {(e^{i\theta})^3 = e^{i\frac{\pi}{2}}\\\implies 3\theta = \frac{\pi}{2}+2k\pi}$ you can solve $$\theta$$

7. ganeshie8

notice the trick with exponents when we rise a complex number to a power, the argument(angle) simply gets multiplied by the power

8. Loser66

Thanks for the guidance. Question: is there any link among the roots?

9. freckles

assume $\sqrt[3]{i}=x+yi \\ \text{ cube both sides } i=(x+yi)^3 \\ i=x^3+(yi)^3+3x^2(yi)+3x(yi)^2 \\ i=x^3-3xy^2+3x^2yi-y^3i \\ \text{ so we have } x^3-3xy^2=0 \text{ and } 3x^2y-y^3=1 \\ \text{ assume } x \neq 0 \text{ so } x^2=3y^2 \text{ so the second equation can be written as } \\ 3(3y^2)y-y^3=1 \\ 9y^3-y^3=1 \\ 8y^3=1 \\ y=\frac{1}{2} \\ x^2=3(\frac{1}{2})^2 \\ x^2=\frac{3}{4} \\$ $x=\pm \frac{\sqrt{3}}{2}$

10. freckles

$\sqrt[3]{i}=\pm \frac{\sqrt{3}}{2} + \frac{1}{2} i$

11. ganeshie8

Easy to see that these nth roots add up to $$0$$ is that the link you talking about ?

12. ganeshie8

|dw:1441160623986:dw|

13. ganeshie8

as you might know we can treat these literally same as vectors look at the geometry, they have symmetry, they must add up to 0

14. freckles

oh and I guess x=0 can be a solution to that one equation which makes $-y^3=1 \\ \text{ so } y^3=-1 \\ \text{ so } y=-1 \text{ and so the last solution would be } x+yi=0-i =-i$

15. Loser66

My computer is .... crazy!! ok, I got sum of them =-i not 0, how?

16. ganeshie8

i wouldn't know what you did to get -i haha must be some arithmetic error somewhere.. just double check :) but isn't the geometry so convincing ?

17. Loser66

I use algebraic way

18. freckles

$(\frac{\sqrt{3}}{2}+\frac{1}{2}i)+(\frac{-\sqrt{3}}{2}+\frac{1}{2}i)+(0-i) \\ (\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}+0)+(\frac{1}{2}+\frac{1}{2}-1)i$

19. ganeshie8

^

20. Loser66

Like what you said above the root is $$r = e^{i(\pi/6 + 2\pi n/3)}$$ , where n = 0,1,2

21. Loser66

replace n = 0, I get $$r_1 = e^{i\pi/6}=\sqrt{3}/2 + i/2$$

22. Loser66

n =1 , $$r_2 = -\sqrt{3}/2 -i/2$$ n =2 $$r_3 = -i$$ sum of them =-i

23. Loser66

I realize that @freckles has the same answer with mine.

24. freckles

25. freckles

$-\frac{\sqrt{3}}{2}+\frac{i}{2} \text{ \right ? }$

26. ganeshie8

yeah the second root is in second quadrant so its imaginary component must be positive

27. Loser66

$$r_2 = e^{i(\pi/6 + 2pi/3}= e^{i\pi/6}*e^{i2\pi/3} = (\sqrt {3}/2+i/2)*(cos 2pi/3 + isin2pi/3)$$ $$=(\sqrt3/2+i/2)(-1/2+i\sqrt3/2)$$ $$=-\sqrt3/4+i 3/4 -i/4 -\sqrt 3/4$$ $$= -\sqrt3/2 +i/2$$

28. Loser66

oh, yes, I got it =0. Man!! it drove me crazy!! Thanks you guys!!:)

29. Loser66

Question: no matter what the result is, to test whether I can get it right or not, the sum of roots =0, right? just make sure. :)

30. ganeshie8

wouldn't it be simpler to just add the exponents $$r_2 = e^{i(\pi/6 + 2pi/3}= e^{i5\pi/6} = \cos(5\pi/6)+i\sin(5\pi/6)$$

31. Loser66

Yes, it is!! but my prof asked us to try many ways but geometry. hehehe..

32. ganeshie8

geometry is only to convince you you cannot draw a picture and say "it is a proof"

33. Loser66

To this stuff, the first step is geometry to define the arg z , right?

34. Loser66

like: find the square root of 7-6i

35. Loser66

We need geometry to define where it is and how to get the angle.

36. ganeshie8

not really complex numbers form a field, we define complex number as an ordered pair $$(a,b)$$ and give it structure by defining other necessary operations

37. ganeshie8

geometry helps in visualizing stuff it was not part of definition so strictly speaking you can totally avoid geometry

38. freckles

like you can say $\sqrt{7-6i}=x+yi \\ \text{ sqaure both sides and then compare real and imagainry parts }$

39. freckles

that is what I did to the previous one

40. ganeshie8

above equation,$$\sqrt{~z~}=w$$ is equivalent to $7 = x^2-y^2\\-6 = 2xy$ so any equation in complex variable can be split as a system of equations of real variables

41. ganeshie8

finding $$n$$th roots of a complex number is essentially same as solving a system of polynomial equations of degree $$n$$ unless you take advantage of the geometry, finding nth roots is very difficult by hand as solving $$n$$th degree polynomial equations is a major pain

42. Loser66

I got you. I have another problem. Let me post a new one.

43. Loser66

Ha!! the net doesn't allow me to close this post. hehehe

44. freckles

the net or openstudy? :p

45. ganeshie8

i can close it for you, one sec..

46. Loser66

Let z = cis (2pi/n) for n > = 2 Show that 1 + z + ...+ z^(n-1) =0

47. freckles

n is an integer?

48. ganeshie8

There are like dozens of proofs, il just give you my favorite and simple one

49. Loser66

yes

50. ganeshie8

consider below polynomial of degree $$n$$ : $f(x)=x^n-1$ remember the vieta's formulas for sum of roots ?

51. Loser66

No, but go ahead. I will try to follow

52. ganeshie8

we're done! by vieta's formulas, the sum of roots = 0 and its easy to see that the terms in given expression are precicely the roots of above polynomial

53. ganeshie8

$\sum\limits_{n=0}^{n-1}cis(2\pi/n)$ each term in above series is a distinct root of $$f(x) = x^n-1$$

54. Loser66

Don't get why and how

55. ganeshie8

given a polynomial you need to know how to find the sum of roots

56. ganeshie8

consider a generic polynomial $$f(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots + x+1$$ do you remember how to find the sum of its roots ?

57. ganeshie8

thats precalculus stuff we can directly use it, don't really need to prove it as part of our current proof

58. Loser66

I just remember it is = -b/a where b is the second coefficient of x and a is the first one.

59. ganeshie8

Exactly!

60. Loser66

0

61. ganeshie8

what can you say about the sum of roots of the polynomial $f(x) = x^n+0x^{n-1}+\cdots +0x-1$

62. ganeshie8

right, and you know that the $$n$$ terms in the given sum are the roots of above polynomial that ends the proof

63. Loser66

I lost!! how can you link 1 + z +.... + z^ (n-1) =0 with z^n -1

64. ganeshie8

teamviewer /skype ?

65. Loser66

yes

66. Loser66

skype

67. ganeshie8

pm me ur id

68. ganeshie8

meantime il login to skype, onesec..

69. Loser66

oh, I don't have skype for this computer. let me download it. hehhehehe

70. ganeshie8

wait

71. ganeshie8

72. ganeshie8

you're on window or mac

73. Loser66

window

74. ganeshie8
75. Loser66