Question 128835
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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{{{z^6=1}}}

Get 0 on the right:

{{{z^6 - 1 = 0}}}

Write as the difference of two squares:

{{{(z^3)^2-1^2=0}}}

Factor as the difference of squares:

{{{(z^3-1)(z^3+1)=0}}}

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

{{{(z^3-1^3)(z^3+1^3)=0}}}

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

{{{(z-1)(z^2+z+1)(z+1)(z^2-z+1)=0}}}

Set each factor = 0

{{{z-1=0}}} yields root {{{z=1}}}

{{{z+1=0}}} yields root {{{z=-1}}}

{{{z^2+z+1=0}}}, by the quadratic formula

yields roots {{{z=(-1 + i*sqrt(3))/2}}} and {{{z=(-1 - i*sqrt(3))/2}}}

{{{z^2-z+1=0}}}, by the quadratic formula

yields roots {{{z=(1 + i*sqrt(3))/2}}} and {{{z=(1 - i*sqrt(3))/2}}}.

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So the 6 roots are:

{{{1}}}, {{{-1}}}, {{{(-1 + i*sqrt(3))/2}}}, {{{(-1 - i*sqrt(3))/2}}}, {{{(1 + i*sqrt(3))/2}}}, {{{(1 - i*sqrt(3))/2}}}

Add them up:

{{{(1) + (-1)+(-1 + i*sqrt(3))/2+(-1 - i*sqrt(3))/2 + (1 + i*sqrt(3))/2+(1 - i*sqrt(3))/2}}} =

{{{1-1+(-1 + i*sqrt(3)-1 - i*sqrt(3)+ 1 + i*sqrt(3)+1 - i*sqrt(3))/2}}} = {{{0}}}

Edwin</pre>