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anonymous

  • one year ago

Find all three complex cube roots of (2i - 2)

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  1. anonymous
    • one year ago
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    |dw:1440071028488:dw| Am I at least on the right track?

  2. ganeshie8
    • one year ago
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    Yes \(\sqrt{8}\) will be the magnitude of roots, it will be same for all three roots

  3. anonymous
    • one year ago
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    So what is the next step? I'm lost at the moment :(

  4. ganeshie8
    • one year ago
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    what exactly are we trying to find here ?

  5. ganeshie8
    • one year ago
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    if you understand the question well, you can answer it easily

  6. anonymous
    • one year ago
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    "Find all three complex roots of (2i - 2)".

  7. ganeshie8
    • one year ago
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    Yes, what is a root ?

  8. ganeshie8
    • one year ago
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    the root is a number which when you multiply itself three times produces the number 2i-2

  9. anonymous
    • one year ago
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    I'm not sure how to find out those roots though. I did it previously in another question but I'm getting so confused here for some reason

  10. ganeshie8
    • one year ago
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    you must be knowing that angles add up when you multiply two complex numbers : \[e^{i\theta_1} *e^{i\theta_2} = e^{i(\theta_1+\theta_2)}\]

  11. ganeshie8
    • one year ago
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    we can use that trick to find the angles of roots

  12. anonymous
    • one year ago
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    I think I've seen that before :)

  13. ganeshie8
    • one year ago
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    whats the angle of 2i-2 ?

  14. anonymous
    • one year ago
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    I'm still trying to find it out. I'm at root8 at the moment

  15. ganeshie8
    • one year ago
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    in which quadrant does the number (-2, 2) lie in ?

  16. anonymous
    • one year ago
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    sine

  17. ganeshie8
    • one year ago
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    wat do you mean by sine

  18. anonymous
    • one year ago
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    I'll just show you what I've been trying to do:

  19. anonymous
    • one year ago
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    |dw:1440073112572:dw|

  20. ganeshie8
    • one year ago
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    again, in which quadrant does the number (-2, 2) lie in ?

  21. anonymous
    • one year ago
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    2nd

  22. ganeshie8
    • one year ago
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    good, can you show it in the graph |dw:1440073212144:dw|

  23. anonymous
    • one year ago
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    |dw:1440073236919:dw|

  24. ganeshie8
    • one year ago
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    right, what is its angle ? |dw:1440073291571:dw|

  25. anonymous
    • one year ago
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    135 degrees

  26. ganeshie8
    • one year ago
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    when you see the number, you can eyeball the angle, arctan formula and all is not needed

  27. ganeshie8
    • one year ago
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    yes 135 is correct

  28. ganeshie8
    • one year ago
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    so the magnitude of given complex number is \(\sqrt{8}\) and the angle is \(135\) we're ready to find the roots

  29. anonymous
    • one year ago
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    Alright cool :)

  30. ganeshie8
    • one year ago
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    lets find the angles first

  31. ganeshie8
    • one year ago
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    basically you want to find \(\theta\) such that \[\theta+\theta+\theta = 135 + 360k\]

  32. ganeshie8
    • one year ago
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    which is same as : \[3\theta = 135 + 360k\]

  33. anonymous
    • one year ago
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    So is theta 165?

  34. ganeshie8
    • one year ago
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    or \[\theta = \dfrac{135}{3} + \dfrac{360k}{3}\]

  35. ganeshie8
    • one year ago
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    or \[\theta = 45 + 120k\]

  36. ganeshie8
    • one year ago
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    letting \(k=0,\) you get \(\theta = 45+120*0 = \color{red}{45}\) letting \(k=1,\) you get \(\theta = 45+120*1 = \color{red}{165}\) letting \(k=2,\) you get \(\theta = 45+120*2 =\color{red}{ 285}\)

  37. ganeshie8
    • one year ago
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    the magnitude is same for all of the roots : \(\large (\sqrt{~8~})^{\frac{1}{3}}\)

  38. ganeshie8
    • one year ago
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    notice that \(\large (\sqrt{~8~})^{\frac{1}{3}}\) simplifies to \(\large \sqrt{~2~}\)

  39. anonymous
    • one year ago
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    Oh ok

  40. ganeshie8
    • one year ago
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    so the 3 roots are \[\sqrt{~2~}(\cos(45)+i\sin(45))\] \[\sqrt{~2~}(\cos(165)+i\sin(165))\] \[\sqrt{~2~}(\cos(285)+i\sin(285))\]

  41. ganeshie8
    • one year ago
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    simplify them if you want to by plugging in the trig values

  42. anonymous
    • one year ago
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    I'm amazed by how much easier it was to understand when I just saw it graphed.

  43. anonymous
    • one year ago
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    Thank you so much for helping me to understand this. I've been finding it difficult but I'll keep working on it :)

  44. ganeshie8
    • one year ago
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    complex numbers are really fun in polar form, roughly speaking : multiplying two complex numbers is same as adding the angles dividing two complex numbers is same as subtracting the angles

  45. anonymous
    • one year ago
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    Hopefully I'll find them fun one day!

  46. ganeshie8
    • one year ago
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    sure, you're on right track :) watch this video when you're free http://ocw.mit.edu/resources/res-18-008-calculus-revisited-complex-variables-differential-equations-and-linear-algebra-fall-2011/part-i/lecture-1-the-complex-numbers/

  47. anonymous
    • one year ago
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    Will do, thanks again for the help! :)

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