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marigirl

  • one year ago

write z=1 - √3 i in polar form I got my r = 2 and i got arg (theta) value as -pi/3

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  1. marigirl
    • one year ago
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    the answer reads: (g) The modulus of 1 - √3 i is √12 + (√3)2 = 2 and its argument θ satisfies tan θ = - √3/1 = - √3, whence θ is equal to either 2π/3 or 5π/3. Since y=-√3 is negative, we have θ = 5π/3. The polar form of -3 + 3i is thus 2 (cos 5π/3 + i sin 5π/3). ....

  2. marigirl
    • one year ago
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    |dw:1441322582245:dw| Thats what i thought

  3. marigirl
    • one year ago
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    this example too seems to be adding when i thought arg (theta) is positive \[-\pi < \theta \le \pi \]

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  4. marigirl
    • one year ago
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    unless im just confused.

  5. marigirl
    • one year ago
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    @jdoe0001

  6. anonymous
    • one year ago
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    im still confused

  7. anonymous
    • one year ago
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    that has to be the ugliest circle i have ever seen

  8. anonymous
    • one year ago
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    correct answer though

  9. marigirl
    • one year ago
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    Umm no that's not a circle

  10. marigirl
    • one year ago
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    I was trying to show that when the Arg theta value is between 0 and 180deg then it's positive

  11. freckles
    • one year ago
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    |dw:1441326450077:dw| your point is in the 4th quadrant

  12. freckles
    • one year ago
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    |dw:1441326527937:dw|

  13. marigirl
    • one year ago
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    Yes so should it be a negative and I got negative pi/3

  14. freckles
    • one year ago
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    |dw:1441326541656:dw|

  15. freckles
    • one year ago
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    right -pi/3 is right

  16. freckles
    • one year ago
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    but

  17. marigirl
    • one year ago
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    Why haa the model answera added 3pi/2 to pi/3

  18. freckles
    • one year ago
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    it looks like they wanted you to write theta in between 0 and 2pi based on their answer

  19. freckles
    • one year ago
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    so you just do -pi/3+2pi

  20. freckles
    • one year ago
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    this will put you between 0 and 2pi

  21. freckles
    • one year ago
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    i see you wanted to add pi for some reason but that would put you in the second quadrant

  22. freckles
    • one year ago
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    |dw:1441326754891:dw| doing theta+pi will put you in second quadrant as you can see

  23. freckles
    • one year ago
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    and we definitely want to be in 4th

  24. freckles
    • one year ago
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    theta+2pi would get us back there to fourth quadrant

  25. marigirl
    • one year ago
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    If the question doesn't specify will -pi/3 also be a sufficient answer

  26. freckles
    • one year ago
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    yes if they don't specify what they want theta between yes -pi/3 will do and so would -pi/3+2pi*n where n is an integer as long as you keep your r positive that is

  27. marigirl
    • one year ago
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    Also can u look at the image I have attached above...I guess they have demonstrated both methods

  28. freckles
    • one year ago
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    \[(r,\theta)=(r,\theta+2 \pi n) \\ \text{ or } \\ (r,\theta)=(-r,\theta+\pi+2 \pi n)\]

  29. freckles
    • one year ago
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    when it comes to polar coordinates there is not a unique way to express one point

  30. freckles
    • one year ago
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    not a unique way meaning there are infinite amount of ways to express one point

  31. freckles
    • one year ago
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    so you could haven chosen (-2,2pi/3)

  32. anonymous
    • one year ago
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    on account of sine and cosine are periodic functions with period \(2\pi\)

  33. freckles
    • one year ago
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    or (2,5pi/3) or (2,-pi/3)

  34. freckles
    • one year ago
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    and so on

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