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kalli

  • one year ago

Use W and Z to solve Z= (-5sqrt3)/2+(5/2)i W= 1+(sqrt3) i A. Convert Z and W to polar form

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  1. freckles
    • one year ago
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    \[r^2=x^2 +y^2 \implies r=\sqrt{x^2+y^2} \\ \tan(\theta)=\frac{y}{x} \\ \text{ do } \theta=\arctan(\frac{y}{x}) \text{ in first and forth quadrants } \\ \text{ do } \theta=\arctan(\frac{y}{x}) +\pi\text{ in second and third quadrants }\]

  2. freckles
    • one year ago
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    though you can add 2pi *n to those results depending on what range you want theta in

  3. kalli
    • one year ago
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    @freckles I think i figured it out by myself .... for Z at least Converted to polar form it would be z=5(cos(5pi/6)+isin(5pi/6) right??

  4. freckles
    • one year ago
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    \[Z=\frac{-5 \sqrt{3}}{2}+\frac{5}{2}i \\ r=\sqrt{(\frac{-5 \sqrt{3}}{2})^2+(\frac{5}{2})^2}=\sqrt{\frac{25(3)}{4}+\frac{25}{4}}=\sqrt{\frac{25}{4}}\sqrt{3+1}=\frac{5}{2}\sqrt{4}=5\] your r looks amazing

  5. freckles
    • one year ago
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    one sec checking your theta

  6. kalli
    • one year ago
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    ok :)

  7. freckles
    • one year ago
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    \[\theta=\arctan(\frac{\frac{5}{2}}{\frac{-5 \sqrt{3}}{2}}) +\pi \\ \theta=\arctan(\frac{5}{2} \cdot \frac{2 }{-5 \sqrt{3}})+ \pi \\ \theta=\arctan(\frac{-1}{\sqrt{3}})+\pi \\ \theta=\frac{-\pi}{6}+\pi\] and your theta looks good as well

  8. kalli
    • one year ago
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    Ok, can you help me on W? I got pi/3 but i feel like that may not be right

  9. freckles
    • one year ago
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    so since W is in the first quadrant all that you need to do to calculate theta is do arctan(y/x) y is sqrt(3) and x=1

  10. freckles
    • one year ago
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    so your theta appears to be right

  11. freckles
    • one year ago
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    can you find r

  12. kalli
    • one year ago
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    Yay!! Can you help with a second question using these equations? It asks me to convert ZW using De Moivre Theorem but I dont know what de moivre theorem is

  13. freckles
    • one year ago
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    \[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2)) \\ AB=r_1 r_2 (\cos(\theta_1+\theta_2)+i \sin(\theta_1+\theta_2))\]

  14. freckles
    • one year ago
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    so you multiply your r's and add your thetas

  15. freckles
    • one year ago
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    when multiplying

  16. freckles
    • one year ago
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    if you wanted to do \[\frac{A}{B} \\ \text{ well that is } \frac{r_1}{r_2}(\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2))\]

  17. kalli
    • one year ago
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    is that the form im supposed to use?

  18. freckles
    • one year ago
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    Well you asked to find AB not A/B I was just sharing that last bit just in case you might find it useful later in the class

  19. freckles
    • one year ago
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    And I say you want to use the AB one because you said you want to find ZW

  20. freckles
    • one year ago
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    do you understand this: \[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2)) \\ AB=r_1 r_2 (\cos(\theta_1+\theta_2)+i \sin(\theta_1+\theta_2))\] ?

  21. kalli
    • one year ago
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    Yeah kind of, I just don't know where to plug in what

  22. freckles
    • one year ago
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    your r's for Z and W were 5 and 2 respectively your theta's for Z and W were 5pi/6 and pi/3 respectively your r_1=5 and r_2=2 and your theta_1=5pi/6 and your therta_2=pi/3

  23. freckles
    • one year ago
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    \[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2)) \\ AB=r_1 r_2 (\cos(\theta_1+\theta_2)+i \sin(\theta_1+\theta_2))\] \[ZW=5\cdot2(\cos(\frac{5\pi}{6}+\frac{\pi}{3})+i \sin(\frac{5\pi}{6}+\frac{\pi}{3}))\]

  24. freckles
    • one year ago
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    multiply the 5 and 2 there and add the 5pi/6 and the pi/3 there

  25. kalli
    • one year ago
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    25 and 3.14??

  26. freckles
    • one year ago
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    5(2) is the same as saying 5+5 or 2+2+2+2+2 both of these should add up to be 10

  27. freckles
    • one year ago
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    ok hmm...if you aren't sure how to add 5pi/6 and pi/3 maybe just look at 5/6 and 1/3 and that then attach a pi

  28. kalli
    • one year ago
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    oooooh lol oops

  29. freckles
    • one year ago
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    \[\frac{5\pi}{6}+\frac{\pi}{3}=\pi(\frac{5}{6}+\frac{1}{3})=\pi(\frac{5}{6}+\frac{1}{3} \cdot \frac{2}{2}) \\ =\pi(\frac{5}{6}+\frac{2}{6})=?\]

  30. kalli
    • one year ago
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    pi(5/6+2/6) = 7pi/6??

  31. freckles
    • one year ago
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    yeah

  32. freckles
    • one year ago
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    \[ZW=2 \cdot 5 (\cos(\frac{5\pi}{6}+\frac{\pi}{3})+i \sin(\frac{5\pi}{6}+\frac{\pi}{3})) \\ ZW=10(\cos(\frac{7 \pi}{6})+i \sin(\frac{ 7\pi}{6}))\]

  33. kalli
    • one year ago
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    Thank you so much! Do you think you could help with one last part to this?? Its similar to what we just did so mybe ill have a better understanding

  34. freckles
    • one year ago
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    ok

  35. kalli
    • one year ago
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    C. Calculate z/w using de moivre theorem

  36. freckles
    • one year ago
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    ok well actually I mentioned that quotient above with A and B

  37. freckles
    • one year ago
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    let me copy and paste one sec

  38. freckles
    • one year ago
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    \[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2))\] \[\frac{A}{B} \\ \text{ well that is } \frac{r_1}{r_2}(\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2))\]

  39. freckles
    • one year ago
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    Identify your r1 and r2 and theta1 and theta2 and plug in the formula for A/B

  40. freckles
    • one year ago
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    where A=Z and B=W in this case

  41. kalli
    • one year ago
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    are the r's still 5 and 2?

  42. freckles
    • one year ago
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    yep because we are still using Z And W

  43. kalli
    • one year ago
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    and my thetas are 5pi/6 and pi/3 yes?

  44. freckles
    • one year ago
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    yep

  45. kalli
    • one year ago
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    (cos(5pi/6 -5pi/6) + isin(pi/3-pi/3) ??

  46. freckles
    • one year ago
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    for the inside you are suppose to do Z's theta - W's theta and you still need Z's r / W's r on the outside of the ( )

  47. freckles
    • one year ago
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    \[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2))\] \[\frac{A}{B} \\ \text{ well that is } \frac{r_1}{r_2}(\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2))\] do you how for A/B on the outside you have A's r divided by B's r and the angles inside the cos and sin thing is A's theta- B's theta?

  48. kalli
    • one year ago
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    no im so confused right now lol

  49. freckles
    • one year ago
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    do you understand in that formula r1 is A's radius?

  50. freckles
    • one year ago
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    and that r2 is B's?

  51. freckles
    • one year ago
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    and that theta1 is A's angle thing and theta2 is B's angle thing?

  52. freckles
    • one year ago
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    you are doing Z/W so your r1 and theta1 will come from the information you got from Z and r2 and theta2 will come from the information you got from W

  53. freckles
    • one year ago
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    earlier you told me r1=5 and r2=2 and theta1=5pi/6 and theta2=pi/3 just plug in the formula

  54. freckles
    • one year ago
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    \[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2))\] \[\frac{A}{B} \\ \text{ well that is } \frac{r_1}{r_2}(\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2))\] \[Z=5(\cos(\frac{5 \pi}{6})+ i \sin(\frac{5 \pi}{6})) \\ W=2 (\cos(\frac{\pi}{3})+i \sin(\frac{\pi}{3})) \\ r_1=5,r_2=2 ,\theta_1=\frac{5\pi}{6}, \theta_2=\frac{\pi}{3} \\ \text{ plug into the formula } \\ \frac{Z}{W}=\frac{r_1}{r_2}(\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2))\]

  55. freckles
    • one year ago
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    replace the r1 with 5 replace the r2 with 2 theta1 with 5pi/6 theta 2 with pi/3

  56. kalli
    • one year ago
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    5(1)/2(2) = (cos(5pi/6-pi/3) +isin(5pi/6-pi/3) ??

  57. freckles
    • one year ago
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    ok but there shouldn't be an equal sign in between r1/r2 and (cos(thet...blah blah) also where does the extra 1 and 2 come from?

  58. kalli
    • one year ago
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    oh wait i misunderstood its suppose to be 5/2

  59. freckles
    • one year ago
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    yep

  60. freckles
    • one year ago
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    and also remember to omit the equal sign

  61. freckles
    • one year ago
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    between the r1/r2 and the (cos...blah blah)

  62. kalli
    • one year ago
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    5/2(cos(5pi/6-pi/3) +isin(5pi/6-pi/3) ??

  63. freckles
    • one year ago
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    yep

  64. kalli
    • one year ago
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    is tht my answer

  65. freckles
    • one year ago
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    well just like we did 5pi/6+pi/3 earlier you could do 5pi/6-pi/3

  66. kalli
    • one year ago
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    pi/2

  67. freckles
    • one year ago
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    \[\frac{5\pi}{6}-\frac{\pi}{3}=\pi(\frac{5}{6}-\frac{1}{3})=\pi(\frac{5}{6}-\frac{2}{6})=\pi(\frac{3}{6})=\pi \frac{1}{2}= \frac{\pi}{2}\]

  68. freckles
    • one year ago
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    oh yes

  69. freckles
    • one year ago
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    5/2 can be written as 2.5 but I think it looks just as pretty as 5/2

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