#i = sqrt(-1)#
#i^2 = (color(white)(.)sqrt(-1)color(white)(.))^2 = -1#
#color(red)("Explanation expanded from this point on")#
But if we had #i^2xxi^2# that would be #(-1)xx(-1)=+1#
So #color(blue)(i^(2+2)=i^4=+1)#
#color(green)("So the question is: how many "i^4" are there in "i^87#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Consider the powers for a moment")#
#color(brown)(color(white)(.......)87/4=21 + 3/4 -> (i^4)^21xxi^3)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#"So the "color(blue)((i^4)^21 =+1)#
#color(purple)("Putting it all together")#
#i^87 = (i^4)^21xxi^3 = (+1)^21xxi^2xxi#
giving:# (+1)xx(-1)xxi#
So the final solution is #-i#
#color(blue)("'~~~~~~~~~~~~~~~~~Foot Note ~~~~~~~~~~~~~~~~~~~~~~~~~~")#
Raising #i# to an even power does not always mean that you have +1.
However, raising #i^2# to an even power does. #color(green)("If n is some even number then")# #color(red)((i^2)^n =(-1)^n = +1)#
