How to simplify exponential imaginary numbers?

i^247








I remember that there was a shortcut where you would have to take the number (in this case 247) and divide it by something, and the answer can be found judging by its remainder.How to simplify exponential imaginary numbers?
Remember the following:


i^1 = i


i^2 = -1


i^3 = (i^2)(i) = -i


i^4 = (i^2)(i^2) = 1





Divide the exponent by 4, then use the following table to find the answer:


remainder = 1 ==%26gt; answer is i


remainder = 2 ==%26gt; answer is -1


remainder = 3 ==%26gt; answer is -i


remainder = 0 ==%26gt; answer is 1





The reason is this: suppose the number is i^x,


i^x = i^(4n + k)


where x and n are positive integers and k is an integer fom 0 to 3.





i^x = i^(4n)*i^k


i^x = ((i^4)^n)*i^k


i^x = (1^n)*i^k


i^x = i^k





Thus 247/4 = 61 remainder 3, and


i^247 = i^3 = -i





because


i^247 = i^(4*61 + 3)


i^247 = ((i^4)^61)*i^3


i^247 = (1^61)*i^3


i^247 = i^3


i^247 = -i