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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 $$\bar{\partial} f(z)=\frac{1}{2 i \sin a} \left( e^{i a} \partial_u f- (i \tan a -1)\partial_v f\right).$$

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    $\begingroup$ Yes, the claim is correct. I think you should write your calculation so we can point out what went wrong. $\endgroup$ – Jackozee Hakkiuz Feb 9 at 17:57
  • $\begingroup$ Done! Thank you. $\endgroup$ – Migalobe Feb 10 at 19:52
  • $\begingroup$ I removed the "Cauchy-Riemann equations" tag and replaced it with "coordinate systems" tag, since I think it is more appropiate for the central point of the question. $\endgroup$ – Jackozee Hakkiuz Feb 10 at 22:44
  • $\begingroup$ Suggested title: "$\bar\partial$ operator of a function under change of coordinates" or something like that. $\endgroup$ – Jackozee Hakkiuz Feb 10 at 22:51
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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?

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    $\begingroup$ My mistake was in $\partial_x v$, which is $0$, when I calculate it, I replaced $y$ from the equation of $u$ to get dependence with $x$. Thanks again. $\endgroup$ – Migalobe Feb 11 at 8:55

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