That version of euler's formula might make a nice case for half turns. Then it's just
(-1)^x = ncos(x) + i nsin(x)
It's obvious how to handle it for integers (an even number of half turns is 1, an odd number is -1), and the extension to real numbers aids the intuition.
Or, depending on your focus, quarter turns are very clean too:
i^x = ncos(x) + i nsin(x)
Either way, turns > radians (it's what I think in when doing most fourier kinds of work anyways!).
(-1)^x = ncos(x) + i nsin(x)
It's obvious how to handle it for integers (an even number of half turns is 1, an odd number is -1), and the extension to real numbers aids the intuition.
Or, depending on your focus, quarter turns are very clean too:
i^x = ncos(x) + i nsin(x)
Either way, turns > radians (it's what I think in when doing most fourier kinds of work anyways!).