But this does not depend in the unit.

> But this does not depend in the unit.

It does!

e^z, defined as the series \sum_{n = 0}^\infty z^n/n!, can only be a function of a dimensionless number z.

sin(z) and cos(z), defined as power series, technically also work this way. And that's OK, because angles are dimensionless: a radian is just C/(2πr), where C is the circumference of a circle of radius r. But it is sometimes convenient to pick your favorite number of radians, like π/180 of them, and call that a degree, and then to say that sin(x degrees) is the same as sin(xπ/180 radians).

With this convention, where the left-hand side of e^(ix) = sin(x) + icos(x) is a function of a dimensionless variable, and the right-hand side can be viewed as a function of a dimensioned argument only in the sense written above, it really is the case that the equation written is true, but the equation e^(ix) = sin(x degrees) + icos(x degrees) is false.

(On the other hand, you could make the case that e^(ix) is really a function of an angle, where its value is the complex number that lies on the unit circle at that angle. Then you do recover a "dimensioned" version of e^(ix) = sin(x) + i*cos(x) that's valid even if you measure angles in degrees.)

If we use a degrees version of sin and cos (call them sind and cosd), then we cannot have e on the left side without a conversion factor.

      (iπx/180)
     e            = cosd x + i sind x

            ix
      π/180 
  -> e            = cosd x + i sind x


                   π/180 
  -> let   f =   e

       ix
     f            = cosd x + i sind x

Probem is, f doesn't have nice properties like:

  d    x             x
  -  f        /=   f
  dx

There is something uniquely special about the unit circle, and about using the unscaled distance around the unit circle as the measure of the angle.

Radians have the property that if we step x by some tiny amount δ, then the cos/sin coordinates will move by that same distance around the unit circle:

   |[cos(x+δ) + i sin(x+δ)] - [cos(x) + i sin(x)]| = δ

This is also related to how we can estimate sin(x) = x for small values next to zero, if using radians.

In radians, the derivative sin'(x) is cos(x), and cos'(x) is -sin(x). Derviation just shifts the waveform left by ninety degrees. In units other than radians, we get wacky constant terms that change at each step.

That's related to how e^x is its own derivative.