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

Question

lgbasallote · Answer

the demoivre theorem i  am familiar with is the one in Calculus. Is that the one you're pertaining to?

vishal_kothari · Answer

http://maths.ucd.ie/courses/math1200/algebra/algebranotes8-3.pdf

nubeer · Answer

yes its in caculus.

mathmate · Answer

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

nubeer · Answer

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

mr.math · Answer

$$(z^5+2z^3)+(6z^2+12)=z^3(z^2+2)+6(z^2+2)=(z^3+6)(z^2+2).$$

nubeer · Answer

well thanks .. this way is also a good one.. can u tell me any website whichhave questions like these with solutions.

mr.math · Answer

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 $z^3+6=0$.

mr.math · Answer

$z^2+2=0$ is very easy to solve and you it gives the roots $z=\pm \sqrt{2}i$.

mr.math · Answer

You can use the demoivre theorem to solve $z^3+6$, (i.e find the third roots of $-6$).

nubeer · Answer

so this last term will give us three roots?

mr.math · Answer

Yep, The notes vishal kothari gave are very good. You should probably read them.

nubeer · Answer

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?

mr.math · Answer

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.

anonymous · Answer

put z=e^ix =cosx+isinx then proceed

nubeer · Answer

@Mr.Math . thank you very much have a nice day :)

mr.math · Answer

You're welcome!