What is the equivalent of e in the rotational context, $e^i$ or $e^{i \tau }$?

Question

I am wondering: if, instead of making a value grow in size through multiplication by $e$ ("scaling"), I make it change in direction through multiplication by $e^i$ "(rotation") but to the very same extent, what other exponent -if any- should accompany i in the second expression:

(a) none, so the rotational analogue of scaling $e$ would be simply $e^i$, i.e. rotation by 1 radian? or (b) $\tau = 2\pi$, so the rotational analogue of scaling $e$ would be $e^{i \tau}$, i.e. rotation by a full cycle, a 360 degree turn?

It seems as if it were implied that the answer were (a) but lately I am convincing myself that it should be (b). The reason is that e with the scaling meaning entails no change of direction but the most optimized (continuous) change of modulus at 100% rate within a given time period. Likewise, e with the rotational meaning should imply no change of modulus but the most optimized change of direction, which would confront us with a full turn. In other words, in both cases we should be reaching the limit, making the most of a growth by 100%, either in terms of modulus or direction...

Is this a matter that is sometimes discussed, does it have an established answer?

EDIT: I know that (a) is rotation by 1 radian, whereas (b) is rotation by a full cycle = 2pi. The question is which one of the two is the rotational analogue of scaling by e. Imagine that you scale a value into e by employing for this purpose a given amount of energy. Now you take the same amount of energy but employ it in rotating the same value, how much would it turn, 1 radian or a full rotation?

Answer 1

The number $e^{i\theta}$ is a rotation by $\theta$ radians in the complex plane. Thus $e^{i}$ we be a rotation by $1$ radian, $e^{2\pi i}$ by $2 \pi$ radians, and so on.

To see why this must be the case, note that the complex number $a+ ib$ can be represented by the matrix $\begin{pmatrix} a &b \\ -b & a\\\end{pmatrix}$. By Euler's identity we have $e^{i\theta} = \cos\theta + i\sin\theta$, which has the matrix representation $\begin{pmatrix} \cos\theta &\sin\theta \\ -\sin\theta & \cos\theta\\\end{pmatrix}$. This is exactly the form of a rotation matrix in 2D.

Answer 2

After some reflection, I have hit on the answer. Contrary to what I was lately thinking, the equivalent of $e$ is $e^{i}$. The reason is clear. Rotation (in this case by 1 radian) is the overall result, but if you apply step by step each instruction, you find that:

• implicitly, $e$ is a factor multiplying $x = 1$;
• as number e instructs, you are "doubling continuously" (up to $e = 2.718...$), but that means creating an added part to the base, though the base remains as it is;
• what you add on top of that is the arc, whose length is $1$ radian $=$ the length of the radius and since you started with $x = 1$, you have doubled the length, but shouldn't you go farther until $2.718...$?
• no, because this added part has been created, as the exponent i instructs, by constraining your compound and continuous growth towards the imaginary axis, which means that you have been employing what we could call the "euler energy" (the extra instruction that e contains beyond pure "doubling") in that effort of changing direction, so it was not capable of increasing the size of the line any more beyond a total of $2$.

Overall, we regard the whole operation as simply rotating by $1$ radian, but this detailed approach is what answers my question: in what sense are $e$ and $e^{i}$ equal and in what sense do they differ?

Well, not very technical -I admit- but it is at least the answer I was looking for...