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Question Number 113185 by 175mohamed last updated on 11/Sep/20

Answered by mindispower last updated on 12/Sep/20

z^n −1=Π_(k=0) ^(n−1) (z−a_k ) z=−1⇒(−1)^n −1=Π_(k=0) ^(n−1) (−1−a^k )=(−1)^n Π_(k=0) ^(n−1) (1+a^k ) ⇒(1+a)(1+a^2 )....(1+a^(n−1) )=1−(−1)^n easy too see that for n=2k=0 since (−1) is root of unity ⇒(1+(−1))=0

zn−1=∏n−1k=0(z−ak)z=−1⇒(−1)n−1=∏n−1k=0(−1−ak)=(−1)n∏n−1k=0(1+ak)⇒(1+a)(1+a2)....(1+an−1)=1−(−1)neasytooseethatforn=2k=0since(−1)isrootofunity⇒(1+(−1))=0

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