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anonymous

  • one year ago

Express the complex number in trigonometric form. -3 + 3 square root of three i

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  1. jim_thompson5910
    • one year ago
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    I'm assuming the expression given is \(\Large -3+3\sqrt{3}*i\) If so, then it's the same as last time but now \[\Large a = -3\] \[\Large b = 3\sqrt{3}\]

  2. anonymous
    • one year ago
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    yeah, that is the correct expression

  3. Plasmataco
    • one year ago
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    quick question... what is the square root of i?

  4. Plasmataco
    • one year ago
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    I know... im a little slow with the noggin but still...

  5. jim_thompson5910
    • one year ago
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    the i isn't part of the square root

  6. Plasmataco
    • one year ago
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    i know, just curios.

  7. anonymous
    • one year ago
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    -1

  8. Plasmataco
    • one year ago
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    ...

  9. Plasmataco
    • one year ago
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    i thought that was the square... exponent 2.

  10. Plasmataco
    • one year ago
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    -1*-1=1 not i.

  11. jim_thompson5910
    • one year ago
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    \[\Large i = \sqrt{-1}\] \[\Large i^2 = (\sqrt{-1})^2\] \[\Large i^2 = -1\]

  12. anonymous
    • one year ago
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    \[\iota ^{2} = -1\]

  13. Plasmataco
    • one year ago
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    am i just the stupid idiot here or u genuinely dont know or connection problemos?

  14. Plasmataco
    • one year ago
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    square root. sry like radical.

  15. jim_thompson5910
    • one year ago
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    sorry @Plasmataco I'm not following

  16. Plasmataco
    • one year ago
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    i know but like -1 under a radical of 4

  17. jim_thompson5910
    • one year ago
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    were you able to find r and theta, @lxoser ?

  18. Plasmataco
    • one year ago
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    ...

  19. jim_thompson5910
    • one year ago
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    oh you mean \[\Large \sqrt{i} = \sqrt[4]{-1}\] @Plasmataco ??

  20. Plasmataco
    • one year ago
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    \[\sqrt[4]{-1}\]

  21. Plasmataco
    • one year ago
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    yeah

  22. anonymous
    • one year ago
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    im gonna need help finding r and theta @jim_thompson5910

  23. Plasmataco
    • one year ago
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    sry for interrupting...

  24. anonymous
    • one year ago
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    its okay

  25. jim_thompson5910
    • one year ago
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    \[\Large r = \sqrt{a^2 + b^2}\] \[\Large r = \sqrt{3^2 + (3\sqrt{3})^2}\] ... ... ... \[\Large r = ??\]

  26. anonymous
    • one year ago
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    r = 30?

  27. Plasmataco
    • one year ago
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    dont think so.

  28. Plasmataco
    • one year ago
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    simplify.

  29. Plasmataco
    • one year ago
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    it should be somthing a lot lower

  30. jim_thompson5910
    • one year ago
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    Hint: \[\Large (3\sqrt{3})^2=(3\sqrt{3})*(3\sqrt{3})\] \[\Large (3\sqrt{3})^2=(3*3)*(\sqrt{3}*\sqrt{3})\] \[\Large (3\sqrt{3})^2=(3*3)*\sqrt{3*3}\] \[\Large (3\sqrt{3})^2=9\sqrt{9}\] \[\Large (3\sqrt{3})^2=9*3\] \[\Large (3\sqrt{3})^2=27\]

  31. anonymous
    • one year ago
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    r = 6

  32. jim_thompson5910
    • one year ago
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    yes

  33. Plasmataco
    • one year ago
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    yup.

  34. Plasmataco
    • one year ago
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    horay! now my question.

  35. anonymous
    • one year ago
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    for theta, i got -60?

  36. jim_thompson5910
    • one year ago
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    incorrect

  37. anonymous
    • one year ago
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    is the result suppose to be in radians or degrees?

  38. jim_thompson5910
    • one year ago
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    it depends on what the instructions say

  39. jim_thompson5910
    • one year ago
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    does it say which mode they want?

  40. anonymous
    • one year ago
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    no, but these are the answer choices

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  41. jim_thompson5910
    • one year ago
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    ok so they want radian form

  42. anonymous
    • one year ago
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    so would theta be - pi/3

  43. jim_thompson5910
    • one year ago
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    yes, now because our point is in Q2 (see graph) we add pi radians to our angle to land in the right quadrant |dw:1442707471746:dw|

  44. jim_thompson5910
    • one year ago
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    - pi/3 is in Q4 |dw:1442707591362:dw|

  45. jim_thompson5910
    • one year ago
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    so we have to add 180 degrees or pi radians to get it into the right quadrant |dw:1442707637525:dw|

  46. anonymous
    • one year ago
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    2pi/3?

  47. anonymous
    • one year ago
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    @jim_thompson5910

  48. jim_thompson5910
    • one year ago
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    correct

  49. anonymous
    • one year ago
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    so would the answer be 6(cos 2pi/3 + i sin 2pi/3) ?

  50. jim_thompson5910
    • one year ago
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    yes

  51. anonymous
    • one year ago
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    thank you so much!! :")

  52. jim_thompson5910
    • one year ago
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    no problem

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