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nubeer

  • 2 years ago

find all roots of z^5+2z^3+6z^2+12 =0 using demoivre theroem anyone plz?

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  1. lgbasallote
    • 2 years ago
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    the demoivre theorem i am familiar with is the one in Calculus. Is that the one you're pertaining to?

  2. vishal_kothari
    • 2 years ago
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    http://maths.ucd.ie/courses/math1200/algebra/algebranotes8-3.pdf

  3. nubeer
    • 2 years ago
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    yes its in caculus.

  4. mathmate
    • 2 years ago
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    You'll see why when you factor the left-hand side of the equation to (z^2+2)*(z^3+6) = 0

  5. nubeer
    • 2 years ago
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    ahh yes i think this is the factor this helps alot.. can u tell me how u got that... dont tell me its again hit and trial

  6. Mr.Math
    • 2 years ago
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    \[(z^5+2z^3)+(6z^2+12)=z^3(z^2+2)+6(z^2+2)=(z^3+6)(z^2+2).\]

  7. nubeer
    • 2 years ago
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    well thanks .. this way is also a good one.. can u tell me any website whichhave questions like these with solutions.

  8. Mr.Math
    • 2 years ago
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    This was just a simple factorization, You can search factorization techniques. You still need to solve \((z^3+6)(z^2+2)=0\), which implies \(z^2+2=0\) or \(z^3+6=0\).

  9. Mr.Math
    • 2 years ago
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    \(z^2+2=0\) is very easy to solve and you it gives the roots \(z=\pm \sqrt{2}i\).

  10. Mr.Math
    • 2 years ago
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    You can use the demoivre theorem to solve \(z^3+6\), (i.e find the third roots of \(-6\)).

  11. nubeer
    • 2 years ago
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    so this last term will give us three roots?

  12. Mr.Math
    • 2 years ago
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    Yep, The notes vishal kothari gave are very good. You should probably read them.

  13. nubeer
    • 2 years ago
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    ya i read them all.. they are helpful.. but still was kinda confused of how getting these roots.. but thank you.. do u know any website for more questions about this? with solutions?

  14. Mr.Math
    • 2 years ago
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    I don't, but you can search the internet for solutions of polynomials over complex numbers. I would look for you, but I have to leave now.

  15. arijit.mech
    • 2 years ago
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    put z=e^ix =cosx+isinx then proceed

  16. nubeer
    • 2 years ago
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    @Mr.Math . thank you very much have a nice day :)

  17. Mr.Math
    • 2 years ago
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    You're welcome!

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