Why does e to pi i equal -1? (2015) [video]

In that sense I think I have the same problem with this proof that I do with the standard one, where you add the Maclaurin series of cos(θ) and i * sin(θ) and match term-by-term with the series for e^(i * θ). The problem is, at the point you can actually show equality, the things on one side aren't obviously a rotation and the things on the other side aren't obviously an exponential.

I'm not just hear to yell at clouds. I was given a proof that I truly love by a professor I adore, which I think really does give insight into what all these operators are doing. The best video I can find with it is here:

https://www.youtube.com/watch?v=-dhHrg-KbJ0 (Skip to 7:30 if you're already comfortable with the limit definition of e^x)

The basic summary is:

1) e^iθ is equal to (1 + iθ/n) ^ n for large n

2) That base, (1 + iθ/n), plotted as a complex number, has length approaching 1, angle approaching θ/n

3) The base squared, (1 + iθ/n)^2, by de moivre's theorem, forms another point as if the transformation from (0, 1) were repeated twice — that is, the length stays one, and another tiny angle is added for a total of 2θ/n

4) The full result is therefore n transformations, taking the path along the unit circle, traveling θ and arriving at cos(θ) + i * sin(θ)

It is not immediately obvious to me how skewing could define such an operation in the general case; or how flipping could define such an operation in most cases.

In any case, the system of adders and multipliers in 2-dimension he describes, even if not the only reasonable 2D generalization, is certainly a reasonable generalizaion, and one that has proved useful.

>It's actually at the heart of why rotation (and not some other geometric operation) is key to e^iπ.

In this video, e^x is defined as a mapping between multipliers and adders. Multipliers are defined to be a combination of rotation and scaling. Any relationship between exponential, calculus, infinite sums, etc is purely a result of these definitions (except, perhaps, the motivation for choicing i * pi to be the principle multiplier mapping that gets mapped to -1 under e.)

The primary motivation for the exponential function is to be the inverse of the logarithm function. And the motivation of logarithms is to convert multiplication problems to addition problems, so they could be solved with table lookups (later performed on a slide rule) instead of difficult arithmetic. That is, log ab = log a + log b. Which is to say, exp(c + d) = (exp c)(exp d). Just setting this constraint along with the derivative exp’ 0 = 1 is enough to characterize the exponential function.

Or you can get to this function in many other ways, e.g. by solving the differential equation d/dx exp x = exp x; by the series 1 + x + x^2/2 + x^3/6 + ...; or by defining the logarithm as the definite integral of 1/x starting at 1, and then taking the exponential function to be its inverse. I like defining the (complex) exponential as a conformal mapping between the cylinder and the plane minus a point.

No you don't. You might notice this by simply working with imaginary numbers. You might invent imaginary numbers for rotation [0].

Alternatively, consider the multiplication (x + yi)(a + bi), as the value (a + bi) performing a transformation on (x + yi). We want (x + yi)(a + bi) = xy - by + ayi + bxi. If we consider (x + yi) to be a 2 dimensional matrix (with basis 1 and i), we can write the above equation as a matrix multiplication:

    [ x   y ] [  a   b ] = [ ax-by ; ay + bx]
                  -b  a

    [  a   b  ] 
      -b  a

[0] That is, define a group (in the group-theory sense) of functions of rotations, denoted as xi for real numbers x, and a group of group of functions for sliding, denoted x for real numbers x. You might then notice that you can combine these groups in a field structure, that happens to have 1i * 1i = -1, and that the subfield of elements with no i component happens to be isomorphic the the reals.