Show that if $z^n$ + $z^{n-1}$ + … + $z$ + 1 = 0 then $z$ $neq$ 1 and $z^{n+1}$ =1.
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Show that if $z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 0 then $z$ $neq$ 1 and $z^{n+1}$ =1. I thought I had managed to show that $z$ $neq$ 1 since if $z$ = 1 then $z^n$ + $z^{n-1}$ + ... + $z$ + 1 = 1 + 1 ... 1 + 1 = n $neq$ 0. But then when I tried to prove the second part of the question I got a contradiction as follows: Factorising $z^{n+1}$ - 1 = 0 gives ( $z$ - 1)( $z^n$ + $z^{n-1}$ + ... + $z$ + 1) = 0. But since $z$ $neq$ 1 how does this work?
complex-analysis proof-verification complex-numbers proof-explanation
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asked Nov 17 at 23:54
sam