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anonymous

  • one year ago

Thanks for helping! I need to simplify this equation to a+bi form. (1/2(cos(72 degrees)+isin(72 degrees))^5 To solve I should do 1/2^5 and multiply 72 degrees by 5 right? Thanks for helping!

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  1. anonymous
    • one year ago
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    @jim_thompson5910 @ganeshie8 @Nnesha @wio @abb0t @zepdrix @Whitemonsterbunny17 @mathmate @jagr2713 @iki @Mehek14 @ikram002p

  2. anonymous
    • one year ago
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    @ikram002p

  3. anonymous
    • one year ago
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    multiply \(72\) by \(5\)

  4. anonymous
    • one year ago
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    so if I do that and simplify it is right?

  5. anonymous
    • one year ago
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    you also have to take \(\left(\frac{1}{2}\right)^5\)

  6. anonymous
    • one year ago
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    if by "simplify" you mean "evaluate the functions" then yes

  7. anonymous
    • one year ago
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    yup, that is what I meant

  8. anonymous
    • one year ago
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    @satellite73 Can you help me find the complex fifth roots of 5-5sqrt(3)*i?

  9. anonymous
    • one year ago
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    yeah first write \[5-5\sqrt{3}i\] in trig form do you now how to do that?

  10. anonymous
    • one year ago
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    r=sqrt((5)^2+(-5sqrt(3)^2)) for r

  11. anonymous
    • one year ago
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    and -5sqrt(3)/5=tan(theta) for theta

  12. anonymous
    • one year ago
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    and then I make a rcis(theta)?

  13. anonymous
    • one year ago
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    @satellite73

  14. anonymous
    • one year ago
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    yeah \(r=\sqrt{a^2+b^2}\)

  15. anonymous
    • one year ago
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    from there what do I do?

  16. anonymous
    • one year ago
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    did you find \(\theta\)?

  17. anonymous
    • one year ago
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    theta=-pi/3 Sorry for being a little late. My browser froze :(

  18. anonymous
    • one year ago
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    r=10

  19. anonymous
    • one year ago
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    yeah looks like you got it

  20. anonymous
    • one year ago
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    From there how do I fin the complex fifth roots?

  21. anonymous
    • one year ago
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    find*

  22. anonymous
    • one year ago
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    divide the angle by 5

  23. anonymous
    • one year ago
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    -pi/15

  24. anonymous
    • one year ago
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    yeah

  25. anonymous
    • one year ago
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    do I get the 5th root of r?

  26. anonymous
    • one year ago
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    just say it \[\sqrt[5]{10}\]

  27. anonymous
    • one year ago
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    you can't really evaluate any of this, just write it in trig form

  28. anonymous
    • one year ago
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    Is there more to it?

  29. anonymous
    • one year ago
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    It asks me for the complex fifth roots

  30. anonymous
    • one year ago
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    there are five of them

  31. anonymous
    • one year ago
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    go around the circle again, then repeat

  32. anonymous
    • one year ago
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    so I add 2pik to the angle?

  33. anonymous
    • one year ago
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    right, the original angle then divide by 5 again

  34. anonymous
    • one year ago
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    or else you can divide the circle in to 5 equal parts, with \[-\frac{\pi}{15}\]as one of them

  35. anonymous
    • one year ago
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    each time I add +2pi, should I divide r by 5, or is that a one time thing?

  36. anonymous
    • one year ago
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    add \(2\pi\) to \(-\frac{\pi}{3}\) then divide that one by \(5\)

  37. anonymous
    • one year ago
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    then repeat

  38. anonymous
    • one year ago
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    oh, does r change at all?

  39. anonymous
    • one year ago
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    no it is going to be \(\sqrt[5]{10}\) each time

  40. anonymous
    • one year ago
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    okay

  41. anonymous
    • one year ago
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    oh wait, i think your \(r\) is wrong, check it again

  42. anonymous
    • one year ago
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    So would the first one be \[\sqrt[5]{10}(cis(\pi/3))?\]

  43. anonymous
    • one year ago
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    oh, okay

  44. anonymous
    • one year ago
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    Are you sure, I got 10 again

  45. anonymous
    • one year ago
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    \[\sqrt{5^2+(5\sqrt{3})^2}\]

  46. anonymous
    • one year ago
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    \[\sqrt{25+75}=10\]

  47. anonymous
    • one year ago
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    you are right sorry

  48. anonymous
    • one year ago
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    it's okay, you're the one helping me :) Would the first answer be \[\sqrt[5]{10}(cis(\pi/3))\]

  49. anonymous
    • one year ago
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    ?

  50. anonymous
    • one year ago
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    @satellite73

  51. anonymous
    • one year ago
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    it's fine he is still viewing the chat so just give him a minute or to

  52. anonymous
    • one year ago
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    oh sorry was away for a minute

  53. anonymous
    • one year ago
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    kool

  54. anonymous
    • one year ago
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    it's okay

  55. anonymous
    • one year ago
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    ok you found \(r=10,\theta =-\frac{\pi}{3}\)

  56. anonymous
    • one year ago
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    Is the thing I wrote earlier correct?

  57. anonymous
    • one year ago
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    so \[5-5\sqrt{3}i=10\left(\cos(-\frac{\pi}{3})+i\sin(-\frac{\pi}{3})\right)\]

  58. anonymous
    • one year ago
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    you want the five fifth roots

  59. anonymous
    • one year ago
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    \[\sqrt[5]{10}(cis(\pi/3))?\]

  60. anonymous
    • one year ago
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    one is \[\sqrt[5]{10}\left(\cos(-\frac{\pi}{15})+i\sin(-\frac{\pi}{15})\right)\]

  61. anonymous
    • one year ago
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    then add \(2\pi\) to \(-\frac{\pi}{3}\) to get \(\frac{5\pi}{3}\)

  62. anonymous
    • one year ago
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    oh yeah , what you said

  63. anonymous
    • one year ago
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    divide by 5 and the next one is \[\sqrt[5]{10}\left(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})\right)\]

  64. anonymous
    • one year ago
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    without adding 2pi?

  65. anonymous
    • one year ago
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    not sure what you mean

  66. anonymous
    • one year ago
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    you divided without adding 2pik where k=1

  67. anonymous
    • one year ago
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    to theta

  68. anonymous
    • one year ago
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    we now have two fifth roots (3 more to go) one is \[\sqrt[5]{10}\left(\cos(-\frac{\pi}{15})+i\sin(-\frac{\pi}{15})\right)\] and the other is \[\sqrt[5]{10}\left(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})\right)\]

  69. anonymous
    • one year ago
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    So I can +4pi and divide by 5 for another one?

  70. anonymous
    • one year ago
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    oh no we did add \(2\pi\) to the original angle of \(-\frac{\pi}{3}\) to get \(\frac{5\pi}{3}\) then we divided that one by 5 to get \(\frac{\pi}{3}\)

  71. anonymous
    • one year ago
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    yeah

  72. anonymous
    • one year ago
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    or add \(2\pi\) to \(\frac{5\pi}{3}\) same thing

  73. anonymous
    • one year ago
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    when you add you get \(\frac{11\pi}{3}\) divide that by 5 and get \(\frac{11\pi}{15}\) for the third root

  74. anonymous
    • one year ago
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    oh kk

  75. anonymous
    • one year ago
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    and 2pi+11pi/15

  76. anonymous
    • one year ago
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    lather, rinse, repeat

  77. anonymous
    • one year ago
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    oh no careful!

  78. anonymous
    • one year ago
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    don't add \(2\pi\) to \(\frac{11\pi}{15}\) add it to \(\frac{11\pi}{3}\)

  79. anonymous
    • one year ago
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    oh right, my bad

  80. anonymous
    • one year ago
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    hold on a second, it is not a mystery why you are doing this both sine and cosine are periodic with period \(2\pi\) that means the angle is NOT unique

  81. anonymous
    • one year ago
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    so all the original numbers are the same \[5-5\sqrt{3}i=10(\cos(-\frac{\pi}{3})+i\sin(-\frac{\pi}{3}))=10(\cos(\frac{5\pi}{3})+i\sin(\frac{5\pi}{3}))=...\] then are all equal

  82. anonymous
    • one year ago
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    you are just expressing the original number in trig form with different values of \(\theta\) then you divide each of them by 5 to get the different fifth roots

  83. anonymous
    • one year ago
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    yup, so I just continue doing what we were just doing?

  84. anonymous
    • one year ago
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    yes

  85. anonymous
    • one year ago
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    So the theta of the roots would be -pi/15, pi/3, 11pi/15, 17pi/15, and 23pi/15?

  86. anonymous
    • one year ago
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    @satellite73

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