[6gvONxR4sf7o]

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!).