z 8 6 + z 1 7 5 + z 2 8 9 + z 9 1 8 + z 2 0 1 7
If z is a non-real root of z 7 = − 1 , then find the value of above expression.
For more , try this set .
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Since z is a non-real root of 7 − 1 it implies z 7 = − 1
Now z 8 6 = ( z 7 ) 1 2 × z 2 = ( − 1 ) 1 2 × z 2 = z 2
z 1 7 5 = ( z 7 ) 2 5 = ( − 1 ) 2 5 = − 1
z 2 8 9 = ( z 7 ) 4 1 × z 2 = ( − 1 ) 4 1 × z 2 = − z 2
z 9 1 8 = ( z 7 ) 1 3 1 × z = ( − 1 ) 1 3 1 × z = − z
z 2 0 1 7 = ( z 7 ) 2 8 8 × z = ( − 1 ) 2 8 8 × z = z
Adding all of these will get z 2 + ( − 1 ) + ( − z 2 ) + ( − z ) + z = − 1
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χ = z 8 6 + z 1 7 5 + z 2 8 9 + z 9 1 8 + z 2 0 1 7 = z 1 4 × 6 + 2 + z 1 4 × 1 2 + 7 + z 1 4 × 2 0 + 9 + z 1 4 × 6 5 + 8 + z 1 4 × 1 4 4 + 1 = z 2 + z 7 + z 9 + z 8 + z 1 = z 2 + z 7 + z 7 + 2 + z 7 + 1 + z 1 = z 2 + − 1 − z 2 − z 1 + z 1 = − 1 Note that z = 7 − 1 ⟹ z 7 = − 1 ⟹ z 1 4 = 1 Note that z 7 = − 1