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mathslover

  • 2 years ago

Convert into polar form : (-1/2) + i ( \(\frac{-\sqrt{3}}{2}\) )

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  1. mathslover
    • 2 years ago
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    (-1/2) + i ( \(-\frac{\sqrt{3}}{2}\) )

  2. mathslover
    • 2 years ago
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    I got : cos ( 4 pi/3 ) + sin ( 4 pi/3)

  3. mathslover
    • 2 years ago
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    But the book says : cos ( - 2pi/3 ) + i sin ( - 2pi/3). Which is also right...

  4. Azteck
    • 2 years ago
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    isin

  5. Azteck
    • 2 years ago
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    You're wrong.

  6. mathslover
    • 2 years ago
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    Sorry typing mistake I meant i sin 4pi/3

  7. mathslover
    • 2 years ago
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    ^ cos ( 4 pi/3 ) + i sin(4pi/3)

  8. Azteck
    • 2 years ago
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    Find the absolute value of this: \[\left| -\frac{ 1 }{ 2 } -\frac{ i \sqrt{3} }{ 2 }\right|\]

  9. mathslover
    • 2 years ago
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    So which should I prefer more ? -2pi/3 or 4 pi/3

  10. Azteck
    • 2 years ago
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    So \[=\sqrt{(-\frac{ 1 }{ 2 })^2+(-\frac{ \sqrt{3} }{ 2 })^2}\]

  11. Azteck
    • 2 years ago
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    So r=1

  12. Azteck
    • 2 years ago
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    Now draw it.

  13. Azteck
    • 2 years ago
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    Drawing it gives you a way better picture of what the angle is.

  14. mathslover
    • 2 years ago
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    \[\large{\sqrt{ (\frac{-1}{2} )^2 + ( \frac {- \sqrt{3}}{2})^2}}\] right so it is : \(\1\)

  15. Azteck
    • 2 years ago
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    |dw:1359153193735:dw|

  16. Azteck
    • 2 years ago
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    You took the other angle

  17. Azteck
    • 2 years ago
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    You go backwards meaning it will be negative.

  18. mathslover
    • 2 years ago
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    By the diagram I think that the angle theta will surely lie in the III quadrant i.e. 4 pi /3

  19. mathslover
    • 2 years ago
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    Ok got it thanks.

  20. Azteck
    • 2 years ago
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    If it's in the third of fourth quadrant go backwards.

  21. Azteck
    • 2 years ago
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    third or fourth*

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