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if n is an odd integer i=√-1 then (1+i)6n + (1-i)6n is equal to |
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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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