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+1 vote
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in Mathematics by (30 points)
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if n is an odd integer i=√-1 then (1+i)6n + (1-i)6n is equal to

1 Answer

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Best answer

(1+i)6n+(1−i)6n 

=((1+i)2)3n+((1−i)2)3n

=(2i)3n+(−2i)3n

=(2i)3n(1+(−1)3n)

Since, n is odd, 3n is also odd. We know that the odd powers of -1 are equal to -1

=(2i)3n(1+(−1))

=0

Thus,

(1+i)6n+(1−i)6n=0,

where n is an odd integer.

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