# $\bar{\partial}$ operator of a function

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

• Yes, the claim is correct. I think you should write your calculation so we can point out what went wrong. – Jackozee Hakkiuz Feb 9 at 17:57
• Done! Thank you. – Migalobe Feb 10 at 19:52
• 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. – Jackozee Hakkiuz Feb 10 at 22:44
• Suggested title: "$\bar\partial$ operator of a function under change of coordinates" or something like that. – Jackozee Hakkiuz Feb 10 at 22:51

\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}
• 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. – Migalobe Feb 11 at 8:55