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If $[ $$\frac{1+i}{1-i} ] ^ x$ =$1$ then

(A) x = 2n + 1, where n is any positive integer

(B) x = 2n, where n is any positive integer

(C) x = 4n + 1, where n is any positive integer

(D) x = 4n, where n is any positive integer
in Linear Algebra by Boss
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1 Answer

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$\frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{-2i}{2} = -i$

Now we get  [-i]= 1, now for x=4n where n is positive integer we get [-i]x=1. So answer is option D.

by Loyal
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