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

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lgbasalloteBest ResponseYou've already chosen the best response.0
the demoivre theorem i am familiar with is the one in Calculus. Is that the one you're pertaining to?
 2 years ago

vishal_kothariBest ResponseYou've already chosen the best response.10
http://maths.ucd.ie/courses/math1200/algebra/algebranotes83.pdf
 2 years ago

mathmateBest ResponseYou've already chosen the best response.0
You'll see why when you factor the lefthand side of the equation to (z^2+2)*(z^3+6) = 0
 2 years ago

nubeerBest ResponseYou've already chosen the best response.0
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
 2 years ago

Mr.MathBest ResponseYou've already chosen the best response.1
\[(z^5+2z^3)+(6z^2+12)=z^3(z^2+2)+6(z^2+2)=(z^3+6)(z^2+2).\]
 2 years ago

nubeerBest ResponseYou've already chosen the best response.0
well thanks .. this way is also a good one.. can u tell me any website whichhave questions like these with solutions.
 2 years ago

Mr.MathBest ResponseYou've already chosen the best response.1
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 \(z^3+6=0\).
 2 years ago

Mr.MathBest ResponseYou've already chosen the best response.1
\(z^2+2=0\) is very easy to solve and you it gives the roots \(z=\pm \sqrt{2}i\).
 2 years ago

Mr.MathBest ResponseYou've already chosen the best response.1
You can use the demoivre theorem to solve \(z^3+6\), (i.e find the third roots of \(6\)).
 2 years ago

nubeerBest ResponseYou've already chosen the best response.0
so this last term will give us three roots?
 2 years ago

Mr.MathBest ResponseYou've already chosen the best response.1
Yep, The notes vishal kothari gave are very good. You should probably read them.
 2 years ago

nubeerBest ResponseYou've already chosen the best response.0
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?
 2 years ago

Mr.MathBest ResponseYou've already chosen the best response.1
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.
 2 years ago

arijit.mechBest ResponseYou've already chosen the best response.0
put z=e^ix =cosx+isinx then proceed
 2 years ago

nubeerBest ResponseYou've already chosen the best response.0
@Mr.Math . thank you very much have a nice day :)
 2 years ago
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