Find the value of ∑r=1412−ar where ak(k=0,1,2,3,4,5) are fifth roots of unity.

Dayton Burnett

Dayton Burnett

Answered

2022-01-21

Find the value of r=1412ar where ak(k=0,1,2,3,4,5) are fifth roots of unity.

Answer & Explanation

cevarnamvu

cevarnamvu

Expert

2022-01-22Added 12 answers

Another proof:
P(x)=x51=profr=04(xαr)
Hence
P(x)P(x)=5x4x51=r=041xαr
Now put x=2 in this formula.
eris0cg

eris0cg

Expert

2022-01-23Added 13 answers

Since a0,a1,a2,,a4 are roots of the equation
x51=0
You can apply Transformation of Roots to find a equation whose roots are
12α0,12α1,12α4
Let P(y) represent the polynomial whose roots are 12ak
y=12ak=12xx=2y1y
Put in (1)
(2y1y)51=0(2y1)5y5=0
Use Binomial Theorem to find coefficient of y5 and y4.You will get sum of the roots using Vieta's Formulas.
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

If a5=1,12α=c0+c1α+...+c4α4 where 2c0c4=1 2c0c4=1 2c1c0=0 2c2c2=0 2c3c2=0 2c4c3=0 The solution of this is c0=16/31, c1=8/31, c=4/31, c3=2/31, c4=1/31 Then r=1412αr=1631r=141+831r=14ar+431r=14a2r+231r=14a3r+131r=14a4r =16314831431231131=4931

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