SOLUTION: Find the roots of z^6=1 for z complex. Show that the sum of the roots you found is zero.

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Question 128835: Find the roots of z^6=1 for z complex. Show that the sum of the roots you found is zero.
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Find the roots of z^6=1 for z complex. Show that the sum of the roots you found is zero. 

z%5E6=1

Get 0 on the right:

z%5E6+-+1+=+0

Write as the difference of two squares:

%28z%5E3%29%5E2-1%5E2=0

Factor as the difference of squares:

%28z%5E3-1%29%28z%5E3%2B1%29=0

Write the first factor as the difference of two cubes
and the second factor as the sum of two cubes:

%28z%5E3-1%5E3%29%28z%5E3%2B1%5E3%29=0

Factor each by the rule for factoring the difference
(and sum) of two cubes

%28z-1%29%28z%5E2%2Bz%2B1%29%28z%2B1%29%28z%5E2-z%2B1%29=0

Set each factor = 0

z-1=0 yields root z=1

z%2B1=0 yields root z=-1

z%5E2%2Bz%2B1=0, by the quadratic formula

yields roots z=%28-1+%2B+i%2Asqrt%283%29%29%2F2 and z=%28-1+-+i%2Asqrt%283%29%29%2F2

z%5E2-z%2B1=0, by the quadratic formula

yields roots z=%281+%2B+i%2Asqrt%283%29%29%2F2 and z=%281+-+i%2Asqrt%283%29%29%2F2.

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So the 6 roots are:

1, -1, %28-1+%2B+i%2Asqrt%283%29%29%2F2, %28-1+-+i%2Asqrt%283%29%29%2F2, %281+%2B+i%2Asqrt%283%29%29%2F2, %281+-+i%2Asqrt%283%29%29%2F2

Add them up:

 =

 = 0

Edwin