Finding the Roots of Z^6 + 64 = 0

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Homework Help Overview

The discussion revolves around finding the roots of the equation Z^6 + 64 = 0, which is a problem in complex numbers and polynomial equations. Participants note that there should be six solutions to this equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of de Moivre's theorem and suggest factoring the equation into simpler components. There are attempts to clarify the goal of finding all six values of z, with hints about further factoring and using trigonometric forms.

Discussion Status

The discussion is active, with participants providing various approaches and hints for solving the problem. There is an acknowledgment of the need to clarify the goal, and some guidance has been offered regarding methods to find the roots.

Contextual Notes

Some participants express confusion about the initial problem statement and the methods to be used, indicating a need for clearer definitions or assumptions regarding the approach to the roots.

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
 
Last edited:
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Are you familiar with de Moivre's theorem? The problem is pretty easy then. Alternatively, you can factor the equation [tex]Z^6 + 64=0[/tex] into
[tex](Z^3 - 8i)(Z^3+8i)[/tex] from which you can factor more.
 
Last edited by a moderator:
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 [itex]z^6 = r^6(cos 6\theta + isin 6\theta)[/itex] 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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