If `i=sqrt(-1)`, then `{i^(n)+i^(-n), n in Z}` is equal to

Options

A. `{0,2}`
B. `{0,-2}`
C. `{0,-2,2}`
D. `{0,-2i}`

Answer

C

Explanation

We have the folloiwng case:
Case 1: When n=4m, `m in Z`
In this case, we have
`i^(n)=i^(4)=(i^(4))^(m)=1` and `i^(-n)=1/i^(n)=1/1=1`
`therefore i^(n)+i^(-n)=2`
Case 2: When n=4m+1, `m in Z`
In this case, we have
`i^(n)=i^(4m+1)=(i^(4))^(m)i^(1)=i` and `i^(-n)=1/i^(n)=1/i=-i`
`therefore i^(n)+i^(-n)=i=i=0`
Case III: When `n=4m+2, m in Z`
In this case, we have
`i^(n)=i^(4m+2)=(i^(4))^(m)i^(2)=i^(2)=-1`
`therefore i^(n)+i^(-n)=(-1)+(-1)=-2`.
Case IV: When `n=4m+3, m in Z`
In this case , we have
`i^(n)=i^(4m+3)=(i^(4))^(m)i^(3)=i^(3)-i` and `i^(-n)=1/i^(n)=1/i^(3)=i`
`therefore i^(n)+i^(-n)=(-i)+(i)=0`.
Hence, `S={-2,0,2}`.