The least positive integer n such that \(\Big(\frac{2i}{1+i}\Big)^n\)�

1.	The least positive integer n such that \(\Big(\frac{2i}{1+i}\Big)^n\) is a positive integer, is A. 16 B. 8 C. 4D. 2
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

The least positive integer n such that \(\Big(\frac{2i}{1+i}\Big)^n\) is a positive integer, is A. 16 B. 8 C. 4D. 2

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