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Loser66

  • one year ago

Find cube root of i Please, help

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  1. zzr0ck3r
    • one year ago
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    I am here to learn, I sux at this stuff. Something about a circle :)

  2. anonymous
    • one year ago
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    |dw:1441159218311:dw|

  3. anonymous
    • one year ago
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    take one third of that angle to find the first one the other two are evenly spaced around the circle

  4. anonymous
    • one year ago
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    |dw:1441159323165:dw|

  5. anonymous
    • one year ago
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    \[\frac{\sqrt3}{2}+\frac{1}{2}i\] is that one two more to go

  6. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Thanks for the guidance. Question: is there any link among the roots?

  9. freckles
    • one year ago
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    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
    • one year ago
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    \[\sqrt[3]{i}=\pm \frac{\sqrt{3}}{2} + \frac{1}{2} i\]

  11. ganeshie8
    • one year ago
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    Easy to see that these nth roots add up to \(0\) is that the link you talking about ?

  12. ganeshie8
    • one year ago
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    |dw:1441160623986:dw|

  13. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    My computer is .... crazy!! ok, I got sum of them =-i not 0, how?

  16. ganeshie8
    • one year ago
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    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
    • one year ago
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    I use algebraic way

  18. freckles
    • one year ago
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    \[(\frac{\sqrt{3}}{2}+\frac{1}{2}i)+(\frac{-\sqrt{3}}{2}+\frac{1}{2}i)+(0-i) \\ (\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}+0)+(\frac{1}{2}+\frac{1}{2}-1)i \]

  19. ganeshie8
    • one year ago
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    ^

  20. Loser66
    • one year ago
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    Like what you said above the root is \(r = e^{i(\pi/6 + 2\pi n/3)}\) , where n = 0,1,2

  21. Loser66
    • one year ago
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    replace n = 0, I get \(r_1 = e^{i\pi/6}=\sqrt{3}/2 + i/2\)

  22. Loser66
    • one year ago
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    n =1 , \(r_2 = -\sqrt{3}/2 -i/2\) n =2 \(r_3 = -i\) sum of them =-i

  23. Loser66
    • one year ago
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    I realize that @freckles has the same answer with mine.

  24. freckles
    • one year ago
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    your r_2 should be...

  25. freckles
    • one year ago
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    \[-\frac{\sqrt{3}}{2}+\frac{i}{2} \text{ \right ? }\]

  26. ganeshie8
    • one year ago
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    yeah the second root is in second quadrant so its imaginary component must be positive

  27. Loser66
    • one year ago
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    \(r_2 = e^{i(\pi/6 + 2pi/3}= e^{i\pi/6}*e^{i2\pi/3} = (\sqrt {3}/2+i/2)*(cos 2pi/3 + isin2pi/3)\) \(=(\sqrt3/2+i/2)(-1/2+i\sqrt3/2)\) \(=-\sqrt3/4+i 3/4 -i/4 -\sqrt 3/4\) \(= -\sqrt3/2 +i/2\)

  28. Loser66
    • one year ago
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    oh, yes, I got it =0. Man!! it drove me crazy!! Thanks you guys!!:)

  29. Loser66
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Yes, it is!! but my prof asked us to try many ways but geometry. hehehe..

  32. ganeshie8
    • one year ago
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    geometry is only to convince you you cannot draw a picture and say "it is a proof"

  33. Loser66
    • one year ago
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    To this stuff, the first step is geometry to define the arg z , right?

  34. Loser66
    • one year ago
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    like: find the square root of 7-6i

  35. Loser66
    • one year ago
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    We need geometry to define where it is and how to get the angle.

  36. ganeshie8
    • one year ago
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    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
    • one year ago
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    geometry helps in visualizing stuff it was not part of definition so strictly speaking you can totally avoid geometry

  38. freckles
    • one year ago
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    like you can say \[\sqrt{7-6i}=x+yi \\ \text{ sqaure both sides and then compare real and imagainry parts }\]

  39. freckles
    • one year ago
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    that is what I did to the previous one

  40. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    I got you. I have another problem. Let me post a new one.

  43. Loser66
    • one year ago
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    Ha!! the net doesn't allow me to close this post. hehehe

  44. freckles
    • one year ago
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    the net or openstudy? :p

  45. ganeshie8
    • one year ago
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    i can close it for you, one sec..

  46. Loser66
    • one year ago
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    Let z = cis (2pi/n) for n > = 2 Show that 1 + z + ...+ z^(n-1) =0

  47. freckles
    • one year ago
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    n is an integer?

  48. ganeshie8
    • one year ago
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    There are like dozens of proofs, il just give you my favorite and simple one

  49. Loser66
    • one year ago
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    yes

  50. ganeshie8
    • one year ago
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    consider below polynomial of degree \(n\) : \[f(x)=x^n-1\] remember the vieta's formulas for sum of roots ?

  51. Loser66
    • one year ago
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    No, but go ahead. I will try to follow

  52. ganeshie8
    • one year ago
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    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
    • one year ago
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    \[\sum\limits_{n=0}^{n-1}cis(2\pi/n) \] each term in above series is a distinct root of \(f(x) = x^n-1\)

  54. Loser66
    • one year ago
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    Don't get why and how

  55. ganeshie8
    • one year ago
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    given a polynomial you need to know how to find the sum of roots

  56. ganeshie8
    • one year ago
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    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
    • one year ago
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    thats precalculus stuff we can directly use it, don't really need to prove it as part of our current proof

  58. Loser66
    • one year ago
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    I just remember it is = -b/a where b is the second coefficient of x and a is the first one.

  59. ganeshie8
    • one year ago
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    Exactly!

  60. Loser66
    • one year ago
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    0

  61. ganeshie8
    • one year ago
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    what can you say about the sum of roots of the polynomial \[f(x) = x^n+0x^{n-1}+\cdots +0x-1\]

  62. ganeshie8
    • one year ago
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    right, and you know that the \(n\) terms in the given sum are the roots of above polynomial that ends the proof

  63. Loser66
    • one year ago
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    I lost!! how can you link 1 + z +.... + z^ (n-1) =0 with z^n -1

  64. ganeshie8
    • one year ago
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    teamviewer /skype ?

  65. Loser66
    • one year ago
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    yes

  66. Loser66
    • one year ago
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    skype

  67. ganeshie8
    • one year ago
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    pm me ur id

  68. ganeshie8
    • one year ago
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    meantime il login to skype, onesec..

  69. Loser66
    • one year ago
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    oh, I don't have skype for this computer. let me download it. hehhehehe

  70. ganeshie8
    • one year ago
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    wait

  71. ganeshie8
    • one year ago
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    just download teamviewer if u can it will be more fast

  72. ganeshie8
    • one year ago
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    you're on window or mac

  73. Loser66
    • one year ago
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    window

  74. ganeshie8
    • one year ago
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    here it is http://download.teamviewer.com/download/TeamViewer_Setup_en.exe

  75. Loser66
    • one year ago
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    what is your meeting id?

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