Suppose $P_R(\alpha) f(x,y)=f(x\cos\alpha +y\sin\alpha, y\cos\alpha-x\sin\alpha)$
How to show that the operator $P_R(\alpha)$ has the form:
$$ P_R(\alpha)=1+\sum_{n=1}^\infty \frac{1}{n!}(-i\alpha)^n l_z^n $$
where $l_z=-i(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x})$.
I tried to expand the right hand side, but don't know how to collect the terms:
$$ f(x\cos\alpha +y\sin\alpha, y\cos\alpha-x\sin\alpha) =\sum_{m,n}\frac{1}{m!n!}f^{(m,n)}(0,0)(x\cos\alpha +y\sin\alpha)^m (y\cos\alpha-x\sin\alpha)^n $$
(up to factors of $i$): Start by showing that $l_z = {\partial\over{\partial \alpha}}$ and then show that $e^{\beta {\partial \over {\partial \alpha}} }$ is the shift operator, and satisfies $ e^{\beta {\partial \over {\partial \alpha}} } g(\alpha)=g(\alpha +\beta)$ for smooth enough $g's$. Once you've gotten that far you will notice that $P_R(\alpha)$ is essentially the shift operator.