Math, asked by purnima7080, 9 months ago

Find the smallest positive integer value of n. Such that [1+iota÷1-iota]^n =1

Answers

Answered by Anonymous

0

Answer:

Explanation :

n=2 , because (1+i)2n=(1−i)2n

(1+i1−i)2n=1

(i)2n=1 which is possible if n=2 [i4=1]

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