I know in general if $f: \mathbb{R}^n \rightarrow \mathbb{R}$, the differential at $x$ is a mapping from the tangent space $T_x\mathbb{R}^n$ to $T_{f(x)}\mathbb{R}$ which can be rewritten as $$df|_x = \sum_i \frac{\partial f}{\partial r_i}\bigg|_x {dr_i}|_x$$ where $\phi=(r_1,...,r_n)$ could be seen as a global chart on $\mathbb{R}^n$.
Now let $f(r) = e^{iar}$, suppose I want to calculate the differential of the map. I think naively writing $$df|_x = \frac{\partial e^{iar}}{\partial r}\bigg|_x dr|_x = iae^{iax}dr|_x$$ is not correct since here the codomain of $f$ is $\mathbb{C}$ which is a 2-dimensional manifold. Now should I rewrite $$f(r) = u(r) + iv(r) = \cos(ar) + i\sin(ar),$$ then calculate $du|_x, dv|_x$ separately? This will give $$df|_x= du|_x + idv|_x = -a\sin(ax)dr|_x + ia\cos(ar) dr|_x = -a\sin(ax) + ia\cos(ax) dr|_x. $$ In general, is this the right way of doing it?
Let me translate your problem into a completely real one: Your map $f$ is just given by $f : x \mapsto (\cos ax, \sin ax)$ in cartesian coordinates. So the derivative is $df = (a\cos(ax)dx, a\sin(ax))dx)$ in these coordinates. You could also use other coordinates (for instance polar coordinates) to calculate the derivative, but this does not change the 1-form itself.