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Alexis1994_13

u=[root2(1-i)/2]30 Find the number u.

  • 2 years ago
  • 2 years ago

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  1. satellite73
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    \[\sqrt{\frac{1-i}{2}}\]?

    • 2 years ago
  2. satellite73
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    where does the "30" go?

    • 2 years ago
  3. Alexis1994_13
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    wait,I will use equation

    • 2 years ago
  4. myininaya
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    u=\[\sqrt{\frac{(1-i)}{2}}*30\]

    • 2 years ago
  5. Alexis1994_13
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    \[[(\sqrt{2}(1-i)/2 ]^{30}\]

    • 2 years ago
  6. satellite73
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    aaah much easier!

    • 2 years ago
  7. satellite73
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    \[(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)^{30}\]

    • 2 years ago
  8. myininaya
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    \[2^\frac{30}{2}*\frac{(1-i)^{30}}{2^{30}}=2^{15}*\frac{(1-i)^{30}}{2^{30}}=\frac{(1-i)^{30}}{2^{15}}\]

    • 2 years ago
  9. satellite73
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    makes the idea even easier than what was written

    • 2 years ago
  10. satellite73
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    oh dear no my myininaya

    • 2 years ago
  11. satellite73
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    i bet it is what i wrote, because that is an easy one. rewrite in trig (polar, whatever) form. then multiply the angle by 30. the modulus (absolute value) is 1 so nothing to worry about since you are on the unit circle.

    • 2 years ago
  12. myininaya
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    leave me alone i don't know math

    • 2 years ago
  13. myininaya
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    its all been guesses

    • 2 years ago
  14. Alexis1994_13
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    wrong answer,I need to found 1-i. :S

    • 2 years ago
  15. satellite73
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    ok do you know how to write this in polar form?

    • 2 years ago
  16. satellite73
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    because once you do it is really easy

    • 2 years ago
  17. satellite73
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    the number is \[\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\]

    • 2 years ago
  18. satellite73
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    which sits right on the unit circle

    • 2 years ago
  19. satellite73
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    do you prefer degrees or radians for these?

    • 2 years ago
  20. Alexis1994_13
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    The answer is 1-i. But I don't know how to find it

    • 2 years ago
  21. satellite73
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    actually that is not the answer, but we will find it. it is not hard

    • 2 years ago
  22. satellite73
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    but before we begin, do you know how to write this in trigonometric form?

    • 2 years ago
  23. Alexis1994_13
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    Do you mean A(root2/2,-root2/2) ?

    • 2 years ago
  24. satellite73
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    because the idea is just to multiply the angle by 30 and see what you get. the absolute value of this number is 1, and 1 to the power of 30 is 1

    • 2 years ago
  25. satellite73
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    no i mean as \[r(\cos(\theta)+i\sin(\theta))\]

    • 2 years ago
  26. satellite73
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    in this case \[r=1\] because we are right on the unit circle

    • 2 years ago
  27. satellite73
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    \[|\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i|=1\]

    • 2 years ago
  28. Alexis1994_13
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    I haven't learned anything like this yet.

    • 2 years ago
  29. satellite73
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    so all we need it \[\theta\] which should be easy because we see these numbers all the time

    • 2 years ago
  30. satellite73
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    well if you have not, you have no way to raise this to the power of 30!

    • 2 years ago
  31. satellite73
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    i mean you are not expected to multiply this number by itself 30 times !

    • 2 years ago
  32. Alexis1994_13
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    Wait,see what is extcally writes :

    • 2 years ago
  33. satellite73
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    ok. but if you can write in trig from raising to the power of 30 is easy. maybe this is something else

    • 2 years ago
  34. Alexis1994_13
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    u=(\[(\sqrt{2/2}\times w)^{30}\]

    • 2 years ago
  35. Alexis1994_13
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    and we know that w=1-i

    • 2 years ago
  36. satellite73
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    yeah well that is what i wrote

    • 2 years ago
  37. Alexis1994_13
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    The answer is i **

    • 2 years ago
  38. satellite73
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    \[(\frac{\sqrt{2}}{2}(1-i))^{30}\] like this yes?

    • 2 years ago
  39. Alexis1994_13
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    yes

    • 2 years ago
  40. satellite73
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    well we can do it easily, but if you have not seen trig from i don't know how to explain it

    • 2 years ago
  41. Alexis1994_13
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    write it

    • 2 years ago
  42. satellite73
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    i mean you certainly are not going to multiply this number by itself 30 times

    • 2 years ago
  43. satellite73
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    ok this number is \[\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\] \[=\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4})\]

    • 2 years ago
  44. Alexis1994_13
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    I have an idea. We can do this: (\frac{\sqrt{2}}{2}(1-i))^{29}*\frac{\sqrt{2}}{2}

    • 2 years ago
  45. satellite73
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    so when you raise to the power of 30 you just multiply the angle by 30

    • 2 years ago
  46. Alexis1994_13
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    I have an idea. We can do this: (\frac{\sqrt{2}}{2}(1-i))^{29}*(\frac{\sqrt{2}}{2}(1-i)

    • 2 years ago
  47. Alexis1994_13
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    omg!

    • 2 years ago
  48. satellite73
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    \[(\frac{\sqrt{2}}{2}(1-i))^{29}*\frac{\sqrt{2}}{2}(1-i)\]

    • 2 years ago
  49. satellite73
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    ok that is what you wrote yes?

    • 2 years ago
  50. Alexis1994_13
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    YE!

    • 2 years ago
  51. satellite73
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    then what?

    • 2 years ago
  52. Alexis1994_13
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    *-(-\[\sqrt{2}/2+\sqrt{2}/2\]

    • 2 years ago
  53. satellite73
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    ok you have to write \[\frac{7\pi}{4}\times 30=\frac{105\pi}{2}\] and realize that this is the same place on the unit circle as \[\frac{\pi}{2}\] and so the answer is \[i\]

    • 2 years ago
  54. satellite73
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    without polar form or trig form i really don't know how you are expected to do this problem.

    • 2 years ago
  55. satellite73
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    sorry that was not much help.

    • 2 years ago
  56. Alexis1994_13
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    thanks

    • 2 years ago
  57. satellite73
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    yw

    • 2 years ago
  58. myininaya
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    omg you guys made this totally harder than is should be \[(1-i)*(1-i)=1-2i-i^2=1-2i-(-1)=1-2i+1=-2i\] \[(1-i)^2=-2i\] \[(1-i)^{2*15}=(-2i)^{15}\] \[(1-i)^{30}=(-2i)^{15}\] \[(1-i)^{30}=(-2)^{15}i^{15}=-2^{15}i^{4*3+2}=-2^{15}i^2=-2^{15}(-1)=2^{15}\] \[(1-i)^{30}=2^{15}\] \[(\frac{\sqrt{2}}{2}(1-i))^{30}=\frac{2^{15}}{2^{30}}(1-i)^{30}=\frac{1}{2^{15}}*2^{15}=1\]

    • 2 years ago
  59. myininaya
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    i made a typo in first line \[(1-i)(1-i)=1-2i+i^2=1-2i+(-1)=-2i\]

    • 2 years ago
  60. myininaya
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    :)

    • 2 years ago
  61. satellite73
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    actually i am not kicking myself yet

    • 2 years ago
  62. myininaya
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    lol

    • 2 years ago
  63. satellite73
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    since this is wrong

    • 2 years ago
  64. myininaya
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    what?

    • 2 years ago
  65. satellite73
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    oh wait. i was looking at the first line.

    • 2 years ago
  66. satellite73
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    but you are right it must be a "typo"

    • 2 years ago
  67. satellite73
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    you do get \[(1-i)(1-i)=-2i\]

    • 2 years ago
  68. myininaya
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    yep

    • 2 years ago
  69. Alexis1994_13
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    the answer is i.

    • 2 years ago
  70. satellite73
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    everything looks good but your answer!

    • 2 years ago
  71. satellite73
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    because i happen to know that it is i, not 1

    • 2 years ago
  72. myininaya
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    i will look at it again

    • 2 years ago
  73. myininaya
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    but the approach looks good

    • 2 years ago
  74. satellite73
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    taht is ok you just forgot the \[i^{15}\] in your answer

    • 2 years ago
  75. satellite73
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    no it is right, you just forgot one part i think at the end. much much easier in polar form really

    • 2 years ago
  76. myininaya
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    \[i^{4*3+3}\] omg i put a 2 where that 3 was

    • 2 years ago
  77. satellite73
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    to hell with IE9!

    • 2 years ago
  78. myininaya
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    yes we get i :)

    • 2 years ago
  79. myininaya
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    did you see my mistake above \[(-2)^{15}*i^{15}=-2^{15}*i^{4*3+3}=-2^{15}*i^3=-2^{15}*(-i)=2^{15}*i\] \[\frac{1}{2^{15}}*(1-i)^{30}=\frac{1}{2^{15}}*(2^{15}*i)=i\]

    • 2 years ago
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