Conceptually, why does $e^{i\phi}$ describe a rotation?

Question

I've never understood conceptually why $e^{i\phi}$ describes a rotation by parameter $\phi$. I've seen this several times, mostly for more abstract objects like $e^{itA}$ where $A$ is an operator, I was always told it creates a rotation but I've never understood why.

If someone could explain this to me for the basic case $e^{i\phi}$ where $\phi\in\mathbb{R}$, then that would be really great, I can just generalize that concept to operators if I understood it.

As a side note, for the operator case, considering Stone's Theorem I can understand how unitary operators describe rotations (because the norm doesn't change, so it must just rotate the vector), but however I still don't understand conceptually how $e^{itA}$ would describe a rotation.

Answer 1

If $\phi\in\Bbb R$ then the map$$\begin{array}{rccc}r\colon&\Bbb C&\longrightarrow&\Bbb C\\&z&\mapsto&e^{i\phi}z\end{array}$$is a rotation of angle $\phi$ around the origin. That's so because $e^{i\phi}=\cos(\phi)+\sin(\phi)i$ and because, if $x,y\in\Bbb R$,\begin{align}r(x+yi)&=(\cos(\phi)+\sin(\phi)i)(x+yi)\\&=\cos(\phi)x-\sin(\phi)y+\bigl(\cos(\phi)y+\sin(\phi)x\bigr)i\end{align}and the map$$\begin{array}{ccc}\Bbb R^2&\longrightarrow&\Bbb R^2\\(x,y)&\mapsto&\bigl(\cos(\phi)x-\sin(\phi)y,\cos(\phi)y+\sin(\phi)x\bigr)\end{array}$$is a rotation of angle $\phi$ around the origin.

Answer 2

This is obvious, using Euler's formula: $e^{i \phi} = sin \phi + i sin \phi$, since the right side describes a point on the unit circle in the complex plane.

However, I think what you are asking about is perhaps about the intuition behind Euler's formula. The standard proof is based on the observation, that the Taylor expansions of both sides turn out to be the same, as answered here: Intuition behind euler's formula, but I don't think it is intuitive at all. So, this is only a sort of semi-answer; perhaps others more knowledgeable can give a better answer?

Euler's formula was based on an earlier result by Roger Cotes, who according to Wikipedia gave a geometric argument, but I haven't been able to find it.

Answer 3

For me, the most convincing argument is of differential nature.

Let us begin, for a 2D rotation, by the addition property:

$$R(\phi)R(d\phi)=R(\phi+d\phi)\tag{1}$$

which can be transformed into:

$$\dfrac{R(\phi+d\phi)-R(\phi)}{d\phi}=R(\phi)\dfrac{R(d\phi)-Id}{d\phi}\tag{2}$$

Accept for the moment that (2) expresses the fact that:

$$R'(\phi)=R(\phi).J \ \ \text{where} \ \ J:=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\tag{3}$$

(where the "prime" symbol means "derivative wrt $\phi$)

In (3), we recognize the differential equation for the classical (1D...) exponential. From there, it is rather normal to still denote the solution by an exponential. But there is more to it. Indeed, if we consider :

$$S(\phi):= \begin{pmatrix}\cos \phi&-\sin \phi\\ \sin \phi&\cos \phi\end{pmatrix}\tag{4}$$

and replace $R(\phi)$ by $S(\phi)$ in (3), we get an identity : $S(\phi)$ is a solution of differential equation (3) (and all other solutions being multiple of this one, it is the only solution such that $S(0)=Id$).

It remains, coming back to (2), to understand what the hell is this $J$ coming from...

Plainly, comparing the form of $J$ with (4), we see that it expresses a $+\pi/2$ rotation (which is, btw, the role devoted to multiplication by number $i$, hehe). Now the final step, why is

$$\dfrac{R(d\phi)-Id}{d\phi} \ \text{identifiable with} \ J \ ?$$

Just because:

$$\dfrac{R(d\phi)-Id}{d\phi}=\dfrac{1}{d\phi}\left(\begin{pmatrix}\cos d\phi&-\sin d\phi\\ \sin d\phi&\cos d\phi\end{pmatrix}-\begin{pmatrix}1&0\\0&1\end{pmatrix}\right)$$

has a first order Taylor expansion (recall we need first order only for the derivative, for example no $(d\phi)^2$ terms):

$$\dfrac{R(d\phi)-Id}{d\phi}=\dfrac{1}{d\phi}\begin{pmatrix}1+... -1 &d\phi+...\\ -d\phi+...&1...-1\end{pmatrix}$$

where we recognize indeed $J$ (alias, under its mask, complex number $i$)!