How to Solve z^3 = -8 in Complex Numbers?

[sara_87]

find all complex numbers satisfying z^3=-8
i have no idea how to solve this can someone help please?
thank u

 

[trambolin]

Think -8 as -8 + 0i... Then convert this into polar coordinates.

 

[sara_87]

in polar coordinates it will be:
-8=8(cos0+isin0)
giving
-8=8
?

 

[Marco_84]

i think that if you write down z=rexp(ix) would be easier...
you find three roots on the cricle with radius r=(-8)^(1/3).

bye
marco

 

[HallsofIvy]

Are you saying you have been asked to find cuberoots and have never heard of D'Moivre's theorem? Why that's awful! Almost as bad posting a homework problem under the mathematics thread!

 

[Shooting Star]

He actually doesn't need D'Moivre's theorem, only the three cube roots of unity.

Hi sara_87,

Have you done the complex cube roots of one in class? You know, omega, omega^2 etc.? Show some work.

 

[rock.freak667]

Or just basic remainder and factor theorem...f(z)=z^3+8=0. But the other ways are faster.