e^(i*x) = cos(x) + i*sin(x)
into something you can kinda understand by staring at the complex plane: -1^(2x) = cost(x) + i*sint(x)
Credit to justinpombrio for this: https://news.ycombinator.com/item?id=32986869 e^(i*x) = cos(x) + i*sin(x)
into something you can kinda understand by staring at the complex plane: -1^(2x) = cost(x) + i*sint(x)
Credit to justinpombrio for this: https://news.ycombinator.com/item?id=32986869(cost x, sint x) is a point on the unit circle x turns counterclockwise from (1,0).
cost x + i sint x is a point in the complex plane x turns counterclockwise from 1.
Now look at integer powers of i, a point in the complex plane 1/4 turn from 1:
i^0 = 1 (0 turns from 1)
i^1 = i (1/4 turn from 1)
i^2 = -1 (2/4 turn from 1)
i^3 = -i (3/4 turn from 1)
and we define complex exponentiation such that, for all real x,
i^x = cost (x/4) + i sint (x/4) (x/4 turn from 1)
e^(i*pi)+1 = 0
..which is a result of the more general formula. e^(i*x) = cos(x) + i*sin(x)
Pi is hiding there in the sin and cos functions implicitly, because the unit radian is defined by 2*pi. In comparison, the version you mentioned that takes x in "turns". -1^(2x) = cost(x) + i*sint(x)
It got rid of pi and e, which already seems a win for simplicity. i is still there for the imaginary component, or y in the complex plane. So the need for pi was removed thanks to the "turn", defined by 1 as the whole circle or cycle.Multiplying -1 to itself every half turn makes it an alternating series of 1 and -1.. Weird, but it is visually clear to understand, without involving e. Though I still don't see where e went. Oh, this comment explains:
> If we rearrange the products in the exponent we get
2πix πi2x ( πi ) 2x
e -> e -> (e )
> Where e^(πi) is -1. That shows there is something to the turns units; we can express the analog of the Euler identity using exponentiation using a base and factor which are integers.Yeah I get it now, a "turn" acts like a dimensionless unit to the circle/cycle.