Question 487766
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i^0\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i^1\ =\ i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i^2\ =\ -1\ \ ]by definition of *[tex \Large i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i^3\ =\ i^2\,\cdot\,i^1\ =\ -i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i^4\ =\ i^2\,\cdot\,i^2\ =\ 1\ =\ i^0]


Therefore


If *[tex \Large x\ mod\ 4\ =\ 0] then *[tex \Large i^x\ =\ 1]


If *[tex \Large x\ mod\ 4\ =\ 1] then *[tex \Large i^x\ =\ i]


If *[tex \Large x\ mod\ 4\ =\ 2] then *[tex \Large i^x\ =\ -1]


If *[tex \Large x\ mod\ 4\ =\ 3] then *[tex \Large i^x\ =\ -i]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ mod\ n]


is the remainder when *[tex \Large x] is divided by *[tex \Large n] using integer division.


next


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^{-n}\ =\ \frac{1}{x^n}]


and finally, if you end up with an *[tex \Large i] in your denominator, multiply your fraction by *[tex \Large \frac{i}{i}] so as to rationalize your denominator.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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