Given z = cos ( 2 n + 1 2 π ) + i sin ( 2 n + 1 2 π ) ,where n is a positive integer,find the equation whose roots are α = z + z 3 + z 5 + . . . + z 2 n − 1 and β = z 2 + z 4 + z 6 + . . . + z 2 n .
The equation will be of form: x 2 + x + k 1 s e c 2 ( 2 n + 1 π ) What is k?
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Here's the hint :-P!Well much more of an answer i'd say!