To find n such that,
`((2i)/(1+i))^n` is positive
Rationalizing
`((2i)/(1+i) * (1-i)/(1-i))^n`
`rArr (((2i)(1-i))/(1-i^2))^n`
`rArr (((2i)(1-i))/(1-(-1)))^n`
`rArr (((2i)(1-i))/(2))^n`
`rArr(i-i^2)^n`
`rArr(i+1)^n`
If n = 2
`(i-i^2)^n = 1+ i^2 + 2i = 2i`
So n=2 is not the required result as it gives complex value.
If n = 3
`(i-i^2)^n = (1+i)^2(1+i) = 2i(1+i) = 2i - 2`
So n = 3 is not the required result as it gives complex value.
If n = 4
`(i-i^2)^n = ((1+i)^2)^2 = (2i)^2 = 4i^2 = -4`
So n = 4 is not the required result as it gives real but negative value.
Now this value can square at n = 8 only, when the result will become positive.
If n = 8
`(i-i^2)^n = ((i-i^2)^4)^2 = (-4)^2 = 16`
Hence, n=8 is the required result.
