Let $\bar{\partial}$ be the Cauchy–Riemann operator that is for $z=x+i y$, $$\bar{\partial} = \frac{1}{2i}\left( i\partial_x - \partial_y\right).$$ Now assume that $z=u+ve^{ia}$, where $u$ and $v$ are real variables, $a\in (0,1)$ a real constant, and we want to calculate $\bar{\partial} f(z)$ in terms of $\partial_u$ and $\partial_v$.
My question is that the following claim correct? $$\bar{\partial} f(z)=\frac{1}{2 i \sin a} \left( e^{i a} \partial_u f-\partial_v f\right).$$ In my attempt, I used the chain rule. I found the same first term on r.h.s, but I got a different term for the second.
Edit: Here is my calculation:
Write $z=u+ve^{ia}=(u+v\cos a)+i (v\sin a)=u'+iv'$. Then $$\bar{\partial} f(z)=\frac{1}{2 i} \left(i \partial_{u'} f-\partial_{v'} f\right).$$ By the chain rule, we have $\partial_{u'} f=\partial_{u} f + \frac{1}{\cos a} \partial_{v} f$ and $\partial_{v'} f=-\cot a \;\partial_{u} f + \frac{1}{\sin a} \partial_{v} f$. When replacing we get $$\bar{\partial} f(z)=\frac{1}{2 i \sin a} \left( e^{i a} \partial_u f- (i \tan a -1)\partial_v f\right).$$
Your idea seems to be allright, but I think you made some mistake using the chain rule (I'm still not sure what the error was).
Remember the chain rule says
$$\begin{aligned} \partial_x&=\frac{\partial u}{\partial x}\partial_u+\frac{\partial v}{\partial x}\partial_v \\ \partial_y&=\frac{\partial u}{\partial y}\partial_u+\frac{\partial v}{\partial y}\partial_v. \end{aligned}$$ (There's no need to introduce $u'$ and $v'$, since these are the usual $x$ and $y$.)
The partial derivatives that appear in these equations tell you that you need $u$ and $v$ in terms of $x$ and $y$. Solving from your equation $$x+iy=(u+v\cos a)+i(v\sin a),$$ you get $$\begin{aligned} u&=x-y\cot a \\ v&=y/\sin a. \end{aligned}$$
Can you finish?