I am investigating q-differentiability of complex functions and I am trying to find a q-analog of the Cauchy Riemann conditions. Here is what I am trying to emulate for regular differentiability: $$\frac{d}{dx}f(z=x+iy)=\frac{df}{dz}\frac{dz}{dx}=\frac{df}{dz}$$ $$\frac{d}{dy}f(z=x+iy)=\frac{df}{dz}\frac{dz}{dy}=i\frac{df}{dz}$$ Which gives us the Cauchy Riemann conditions when we set $f(x+iy)=u(x,y)+iv(x,y)$. To this end I want to establish a chain rule for translations. In "Quantum Calculus" by Kac and Cheung they remark that a simple and general chain rule for q-derivatives is unknown (or even impossible), but they don't exclude the possibility of specific cases existing.
So just messing around, I wrote down $$D_q\big[f(x+a)\big]=\frac{f(q(x+a))-f(x+a)}{(q-1)x}=\frac{f(q(x+a))-f(x+a)}{(qx+a)-(x+a)}\frac{(qx+a)-(x+a)}{(q-1)x}$$ which looks like $\bigg[\big(D_qf\big)\circ(x+a)\bigg]D_q(x+a)$.
Assuming I have not made a simple error, is it fair to then claim the following? $$(1)\quad\frac{d_q}{dx_q}(f(z=x+iy))=\frac{d_qf}{dz_q},$$ $$(2)\quad\frac{d_q}{dy_q}(f(z=x+iy))=i\frac{d_qf}{dz_q}.$$ These are just idle thoughts but any help is appreciated.