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anonymous

  • one year ago

Which expression is a cube root of -1 + i sqrt(3)? A. cubert(2) (cos(120degrees) + i sin(120 degrees)) B. cubert(2) (cos(40 degrees) + i sin(40 degrees)) C. cubert(2) (cos(280 degrees) + i sin(280 degrees)) D. cubert(2) (cos(320 degrees) + i sin(320 degrees)) ***My Answer: D*** Please Help!!!

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  1. anonymous
    • one year ago
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    \[-1+\sqrt{3}=r \left( \cos \theta+\iota \sin \theta \right)\] \[r \cos \theta =-1,r \sin \theta =\sqrt{3}\] square and add and find r divide find theta

  2. anonymous
    • one year ago
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    So I would do r = -1-cos(theta), r = 3sin^2(theta)?

  3. anonymous
    • one year ago
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    \[r^2\cos ^2\theta =1,r^2\sin ^2\theta=3\] \[r^2\left( \cos ^2\theta+\sin ^2\theta \right)=1+3=4\] r=2

  4. anonymous
    • one year ago
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    Oh ok

  5. anonymous
    • one year ago
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    r is always positive ,so \[\cos~ \theta~ is~ negative~and~\sin \theta~is~positive.\] so angle lies in second quadrant. now divide to find theta

  6. anonymous
    • one year ago
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    Would \[\theta = 120\] be the angle?

  7. anonymous
    • one year ago
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    no guess ,show me your calculations.

  8. anonymous
    • one year ago
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    Ok \[r ^{2}(\cos ^{2}\theta + \sin ^{2}\theta)\]\[2 (\cos ^{2}\theta + \sin ^{2}\theta)\] \[-1 + i \sqrt{3}\] \[\theta = \frac{ \sqrt{3} }{ -1}\]\[\theta = -\frac{ \pi }{ 3}\]

  9. anonymous
    • one year ago
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    That is all I know of. Not exactly sure if it is correct. Sorry

  10. anonymous
    • one year ago
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    \[\frac{ r \sin \theta }{ r \cos \theta }=\frac{ \sqrt{3} }{ -1 }=-\sqrt{3}=-\tan 60=\tan \left( 180-60\right) =\tan 120\]

  11. anonymous
    • one year ago
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    \[\tan \theta=\tan 120,\theta=120\]

  12. anonymous
    • one year ago
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    Oh wow I was far off but at the same time I kinda sorta knew what I was doing. Haha. I answer would now be A I'm assuming. Thank you for taking the time to explain this to me. I desperately needed the help.

  13. anonymous
    • one year ago
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    \[-1+i \sqrt{3}=2\left( \cos 120+i \sin 120 \right)\]

  14. anonymous
    • one year ago
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    yw

  15. anonymous
    • one year ago
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    Oh wait the answer would be B

  16. anonymous
    • one year ago
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    oh sorry you wanted cube root ,i have not noticed

  17. anonymous
    • one year ago
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    It's ok :) I already hit the submit button haha

  18. anonymous
    • one year ago
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    Wouldn't you just divide 120/3 to get 40?

  19. anonymous
    • one year ago
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    \[1+i \sqrt{3}=2\left( \cos( 120+360n)+i \sin \left( (120+360n \right) \right)\] \[or Z ^{\frac{ 1 }{ 3 }}=2^1/3e ^{\frac{ 360n+120 }{ 3 }}\] put n=0,1,2 ,you get all the three cube roots

  20. anonymous
    • one year ago
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    \[\left( 1+i \sqrt{3} \right)^{\frac{ 1 }{ 3 }}=2^{\frac{ 1 }{ 3 }}\left\{ \cos \left( \frac{ 360n+120 }{ 3 }+i \sin \left( \frac{ 360n+120 }{ 3 } \right)\right)\right\}\] put n=0,1,2 you get all the values.

  21. anonymous
    • one year ago
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    Thank you so much! I very much appreciate it!

  22. anonymous
    • one year ago
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    yw

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