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Algebra
IF
@1 @2 @3 , ............................. @8 are the 8th roots of unity then which of the following is(are) correct,
A) @1 + @2 + @3 +......................@ 8 = 1
B) (@1)^3 + (@2)^3 +...................................(@8)^3 = 24
C) (@1)^50 + (@2)^50 +.............................................. (@8)^50 = 0
D) (@1)^48 + (@2)^48 +.............................................(@8)^48 = 0
@1-----------------------> means that alpha 1 (in the subscript and not coefficient of alpha)
PLEASE ATTEMPT AND ANSWER. RATES ARE ASSURED FOR CORRECT ANSWER+PROOF.
(i do not know its answer.) .
Comments (2)
simple .. only (C) is correct.
we have @1 +@2 + .. +@8 =0 ... ( (A) can't be correct )
(@1)^3 + (@2)^3 + .... +(@8)^3 = 3(@1 . @2 .... @8) =3 ( (B) can't be correct )
also see that (@1)^8 = (@2)^8 = .... (@8)^8=1
so (@1)^48 + (@2)^48 + ... (@8)^48 = 8 ( (D) can't be correct )
Now observe that each of (@i)^2 is a 4th root of unity .
So (@1)^50 + (@2)^50 +.... (@8)^50
= (@1)^2 + (@2)^2 + ... (@8)^50
= 2* (sum of the 4th roots of unity)
=0
So (C) is the ONLY correct answer.
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8 roots of unity can be found easily..
they all lie on x^2 + y^2 = 1..
on complex no. graph..
they r:
i -i 1 -1 e^45(deg) e^(-45)deg e^(135)deg and e ^(-135)deg..
u can check them directly..!!
they also satisfy x^8 = 1...........general eq..
thrfr the sum of roots is 0 as welll as the product is -1...
this is clear from the eq...!!
rest substitute and get..!!