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in Complex Numbers by (29.7k points)
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The least positive integer n such that \(\Big(\frac{2i}{1+i}\Big)^n\) is a positive integer, is 

A. 16 

B. 8 

C. 4

D. 2

1 Answer

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Best answer

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

\(\Big(\frac{2i}{1+i}\Big)^n\) n= (1+i)

Let check the value of (1 + i)n for different value of n 

at n =1 , 1+ i (no) 

at n =2 , (1 + i)2 = 1 + i2 + 2i = 2i (no) 

at n =3 , (1 + i)2(1 + i) = (1 + i)(2i) = 2i – 2 (no) 

at n =4 , (1 + i)2(1 + i)2 = (2i)2 = -4 (no) 

at n =5 , (1 + i)4(1 + i) = -4(1 + i) (no) 

at n =6 , (1 + i)4(1 + i)2 = -4(2i) (no) 

at n =7 , (1 + i)6(1 + i) = -8i(1 + i) = -8i + 8 (no) 

at n =8 , (1 + i)4(1 + i)4 = (-4)(-4) = 8 (yes) 

So, we can say that n = 8 is the least positive integer for which \(\Big(\frac{2i}{1+i}\Big)^n\) is positive integer

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