Question 762782
i^15


i^(12+3)


i^(12)*i^3


(i^(3*4))*i^3


(i^3)^4*i^3


(-i)^4*(-i)


(-1)^4*(i)^4*(-i)


(-1)^4*(i^2)^2*(-i)


(-1)^4*(-1)^2*(-i)


(-1)^4*1*(-i)


(-1)^4*(-i)


1*(-i)


-i


So i^15 = -i


Shortcut: divide 15 by 4 to get 3 remainder 3. The remainder of 3 means that i^(15) = i^3. Since i^3 = -i, this means i^15 = -i