Question 232188
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i^6]


In general


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i^2\ =\ -1\ \ ] (definition of *[tex \LARGE i])


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


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i^n\ =\ i^{n\,\text{mod}\,4}]


Where *[tex \LARGE n\,\text{mod}\,m] is the remainder when integer *[tex \LARGE n] is divided by integer *[tex \LARGE m]


6 divided by 4 leaves a remainder of 2, so *[tex \LARGE 6\,\text{mod}\,4\ =\ 2],

 
so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ i^6\ =\ i^{6\,\text{mod}\,4}\ =\ i^2\ =\ -1]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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