How do i solve this complex number? i^-6 + i^-5?

Question

read as " i raised to negative 6 plus i raised to negative 5"

Answer 1

i^4 = 1, so powers of i are modulo 4.
1/i = 1i/i² = i/(-1) = -i
1/i³ = i/i^4 = i/1 = i

6/4 = 1 with remainder 2 so i^6 = i² = -1
1/-1 = -1

5/4 = 1 with remainder 1 so i^5 = i¹ = i
1/i = -i

Only the remainders matter.

So i^(-6) + i^(-5) = -1 - i

Answer 2

i^-6 + i^-5
= i^-6(1 + i)
= i^-2(1+i) because i^4 = 1
= -(1+i) because i^2=-1 ---> 1 = -i^-2
= -1 - i

Answer 3

i^-6

i^-4=1
i^-2=-1 so i^-6=1*-1=-1

i^-5=1*1/i=-i

i^-6+i^-5=-1-i