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

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.
