Finding the Roots of Z^6 + 64 = 0

AI Thread Summary
The equation Z^6 + 64 = 0 has six solutions that can be found using de Moivre's theorem. The equation can be factored into (Z^3 - 8i)(Z^3 + 8i), allowing for further factorization to find the roots. To solve for the six values of z, express z^6 in polar form as r^6(cos 6θ + isin 6θ). The discussion emphasizes the importance of factoring and using trigonometric identities to derive the solutions. Ultimately, the goal is to identify all six roots of the equation effectively.
MercuryRising
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Z^6 + 64 = 0
there are suppose to be 6 solutions :confused:
need help on how to find the roots
 
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Are you familiar with de Moivre's theorem? The problem is pretty easy then. Alternatively, you can factor the equation Z^6 + 64=0 into
(Z^3 - 8i)(Z^3+8i) from which you can factor more.
 
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opps, sorry, i must 've made myself unclear, the goal is to find the 6 values of z, but thanks for heling on the roots! :smile:
 
Write z^6 = r^6(cos 6\theta + isin 6\theta) and solve.
 
MercuryRising said:
opps, sorry, i must 've made myself unclear, the goal is to find the 6 values of z, but thanks for heling on the roots! :smile:

Vsage's point was that, from there, you can factor further, thus getting the six linear roots (hint: think about sum and difference of cubes).
 
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