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anonymous

  • one year ago

Thanks for helping! I need to calculate wz using the De Moivre's theorem. z=sqrt(3)-i and w=-2-2i. I need to rewrite it in polar form. Thank You!

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  1. anonymous
    • one year ago
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    @Ashleyisakitty @campbell_st @zepdrix @TheSmartOne @Nnesha @Luigi0210 @Compassionate @wio @sammixboo

  2. zepdrix
    • one year ago
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    In general when we have:\[\Large\rm z_1=r_1(\cos \alpha+i \sin \alpha)\]\[\Large\rm z_2=r_2(\cos \beta+i \sin \beta)\]Then multiplying will give us:\[\Large\rm z_1\cdot z_2=r_1 \cdot r_2 \left(\cos(\alpha + \beta)+i \sin(\alpha + \beta)\right)\]

  3. zepdrix
    • one year ago
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    Multiply the radial lengths, add the angles. So we gotta do some work to get it into polar form first, ya? :d

  4. anonymous
    • one year ago
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    already done :D

  5. anonymous
    • one year ago
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    2*2sqrt(2)(cos(-pi/6+5pi/4)+isin(-pi/6+5pi/4))

  6. anonymous
    • one year ago
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    Then I just calculate that?

  7. zepdrix
    • one year ago
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    Mmm ok ya, looks good!

  8. anonymous
    • one year ago
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    so for theta I would put 13pi/12?

  9. anonymous
    • one year ago
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    and r is 4sqrt(2)

  10. zepdrix
    • one year ago
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    and you can use CIS if you wanna be lazy.\[\Large\rm z=2 cis\left(-\frac{\pi}{6}\right),\qquad w=2\sqrt2 cis\left(\frac{5\pi}{4}\right)\]I like to do that sometimes hehe

  11. anonymous
    • one year ago
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    Yup, but my teacher wants full form unfortunately :(

  12. zepdrix
    • one year ago
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    Yes that looks correct! :)

  13. anonymous
    • one year ago
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    awesome! Thanks for the help! Can I have help with one more question really quickly?

  14. zepdrix
    • one year ago
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    sure

  15. anonymous
    • one year ago
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    z^2 so (sqrt(3)-i)^2. I think it is 3+i, but I have a feeling I did it wrong.

  16. zepdrix
    • one year ago
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    \[\Large\rm z=2cis\left(-\frac{\pi}{6}\right)\]Then,\[\Large\rm z^2=2^2cis\left(-\frac{\pi}{6}\right)^2\]De'Moivre's let's us bring the exponent inside,\[\Large\rm z^2=4cis\left(-\frac{2\pi}{6}\right)\] So that angle is -pi/3 ya? So our imaginary part should be negative, we're still in quadrant 4.

  17. zepdrix
    • one year ago
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    Oh I guess NOT converting to polar is probably better :) lol is that what you did?

  18. anonymous
    • one year ago
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    yup

  19. anonymous
    • one year ago
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    oh crap, sorry, I forgot to say I needed to write in a+bi form

  20. anonymous
    • one year ago
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    my bad

  21. zepdrix
    • one year ago
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    \[\Large\rm z=(\sqrt3-i)\]\[\Large\rm z^2=(\sqrt3-i)(\sqrt3-i)\]\[\Large\rm z^2=3-\sqrt3 i-\sqrt3 i+i^2\]Something like that?

  22. anonymous
    • one year ago
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    Is that all I need to do? I thought there was some formula or something. I did (a)^+(b+^2

  23. anonymous
    • one year ago
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    (b)^2*

  24. zepdrix
    • one year ago
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    No, that's when you're multiplying complex `conjugates`.\[\Large\rm (a+bi)(a-bi)=a^2+b^2\]This is not the same as squaring something! :)

  25. zepdrix
    • one year ago
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    \[\Large\rm (a+bi)(a-bi)\ne(a+bi)^2\]

  26. anonymous
    • one year ago
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    Oh, so I would leave my results as i^2-sqrt(3)i-sqrt(3)i+3?

  27. zepdrix
    • one year ago
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    No no :) We can simplify further. I just wanted to show the FOIL process

  28. zepdrix
    • one year ago
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    So we have some like-terms in the middle which can be combined. And what do you get when you square i?

  29. anonymous
    • one year ago
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    i^2? or if you mean root, i

  30. anonymous
    • one year ago
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    -2sqrt(3)i

  31. zepdrix
    • one year ago
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    woops!

  32. anonymous
    • one year ago
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    -1?

  33. zepdrix
    • one year ago
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    \[\Large\rm \sqrt{-1}\cdot\sqrt{-1}=-1\]Yah that looks better :) I made a silly mistake lol

  34. zepdrix
    • one year ago
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    \[\Large\rm z^2=3-2\sqrt3 i-1\]

  35. zepdrix
    • one year ago
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    Simplify furtherrrrr :U

  36. anonymous
    • one year ago
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    -2sqrt(3)i+2

  37. zepdrix
    • one year ago
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    maybe we write the a first, a+(b)i = 2 + (-2sqrt3)i ok looks good! \c:/

  38. anonymous
    • one year ago
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    Awesome, I presume it's the exact same if I were to do w^4? Thanks for the help, I learned a lot :D

  39. zepdrix
    • one year ago
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    Expanding a 4th degree is really a pain in the butt, lot of brackets to multiply out. When you get above a 3rd power, you want to start doing it the other way: Change to polar, apply De'Moivre's Theorem.

  40. zepdrix
    • one year ago
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    \[\Large\rm w=2\sqrt{2}cis(5\pi/4)\]\[\Large\rm w^4=(2\sqrt{2})^4cis(5\pi/4)^4\]\[\Large\rm w^4=(2\sqrt{2})^4cis(4\cdot 5\pi/4)\]

  41. anonymous
    • one year ago
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    64(cis(5pi))?

  42. anonymous
    • one year ago
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    I apologize if I'm taking too much of your time, but may I ask a quick question? If I have to write the coordinates of (-3, pi/3) as a positive value (as in radius), would I write (3, pi/3)? Thanks!

  43. zepdrix
    • one year ago
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    ah sorry i ran away for a bit hehe

  44. anonymous
    • one year ago
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    That is fine :)

  45. zepdrix
    • one year ago
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    |dw:1434425783548:dw|

  46. zepdrix
    • one year ago
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    So if THAT direction is -3, which direction is positive 3? maybe you can label it on the graph.

  47. anonymous
    • one year ago
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    |dw:1434425967838:dw|

  48. zepdrix
    • one year ago
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    |dw:1434426009450:dw|Think about your regular x-y plane. if i go 3 in this direction,

  49. zepdrix
    • one year ago
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    |dw:1434426045344:dw|Then the opposite (negative) of that is to go 3 in the other direction, ya?

  50. anonymous
    • one year ago
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    yea

  51. zepdrix
    • one year ago
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    |dw:1434426087433:dw|So we want to go 3 in the opposite direction

  52. anonymous
    • one year ago
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    Oh, gotcha

  53. anonymous
    • one year ago
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    4pi/3?

  54. zepdrix
    • one year ago
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    |dw:1434426162247:dw|

  55. zepdrix
    • one year ago
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    Mmm ya that sounds right: pi/3 + pi = 4pi/3 or -2pi/3 would work also. the positive angle seems nicer though

  56. anonymous
    • one year ago
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    Awesome, once again thanks!

  57. zepdrix
    • one year ago
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    np

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