Hi guys, find all complex numbers satisfying `z^6-1=0.`
I have seen someone do the working out as follows:
`z^6-1=(z^3-1)(z^3+1) = (z-1)(z^2+z+1) (z+1)(z^2-z+1).`
However, I am not sure what the next step is.
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Because of the factors `z-1` and `z+1,` we know that `+-1` are solutions.
Now we can use the quadratic formula for the remaining factors. For
`z^2+z+1,` we get
and for `z^2-z+1` we get
`z=(1+-i sqrt(3))/2` .
The six solutions are `+-1,(-1+-i sqrt(3))/2,(1+-i sqrt(3))/2.`
Note that if `z` is a solution, then so is ``the conjugate of `z.` This is true for any solution of `z^n=1` for any `n.` Also, if `z` is a solution, so is `-z.` This is the case for the solutions of `z^n=1` only when `n` is even (which is why it works for this example).
sorry, the working out that is missing:
z^6 - 1 = 0
z^3-1 = (z-1)(z^2+z+1)
z^3+1 = (z+1)(z^2-z+1)
z^6 - 1 = (z-1)(z+1)(z^2+z+1)(z^2-z+1)
however, I am not sure what the next step is.
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