isnt i^3 just i? of course, i fail at this, i did a little of it last year for my math project, but not the complete story
i^1 = i i^2 = -1 i^3 = -i i^4 = 1 so i^5 = i because you can take i^4 out making it 1 * i^1 therefore 1*i thus i^5 = i thats how I learned it. basically if the exponent is over 4 you divide it by 4 and whatever that is time the remainder. So i^39 = -i because you can take out an i^4 (4 goes into 36 leavinging 3 as the remainder) so i^4 = 1 1 * i^3 1*-i = -i I know, it didn't make sense to me either when I learned it, but thats how I got it right on the test. And I did that 4 years ago, how it remained fresh is beyond me..
i learned i via random books. anyways, its like a 2-d graph. horizontally, its the real numbers, vertically, its imaginary numbers, which is basically any numer with "i" at the end (like 8i, etc.) Ste is correct in his/her/its logic. (actually, in eigineering and physics, i is represented by j, idk why i guess they like curved-handle umbrellas better, like me) but, wat do i know, im a high school freshman T.T
Ok lesson time. When you take the square root of a number you find another number that when multiplied by itself gets you the original number. In Mathematics there no number that when multiplied by itself creats a negative number. This creates a problem when dealing with the square root of a negative number, so a Mathemitician (whos name I can't remember) created a new number callied i which is defined as being the square root of negative one. As for how i works mechanicly ste pretty much hit it no the nose.
