i^61 ???

i^61

\(i^{61} = i^{60}\times i\\ = (i^{4})^{15} \times i\\ = [(i^2)^2]^{15} \times i\\ = [(-1)^2]^{15} \times i\\ = [1]^{15} \times i\\ = 1\times i\\ =i\)

Thanks, Melody for that answer....

Here's another way to evalute this for    i    raised to any  "n"   positve integer

n mod 4  =  1    →   i

n mod 4  =  2   →   -1

n mod 4  = 3   →   - i

n mod 4  = 4   →    1

So

i^61  =      61 mod 4   = 1    →   i            just as Melody found   !!!