# Change of variables in complex integration

Assume $z=x+iy$ and $\bar{z}=x-iy$. Then one might want to change the integration variables (for some arbitrary integral) over $dz d\bar{z}$ to something proportional to $dxdy$. Then it seems that $$dz d\bar{z} = 2i dxdy$$ but I do not see how this is obtained. What I know is that $dz = dx+idy$ and $d\bar{z}=dx-idy$ thus $$dz d\bar{z} =(dx+idy)(dx-idy)=dx^2+dy^2$$ so, my question is, which of the two is the correct one? It seems to be like a very trivial question which I am unable to get the answer.

Best way to see it is to consider it as a differential form. Notice that you have a minus sign off. Your $dx \, dy$ is really $dx \wedge dy$. Note that $dx \wedge dy = - dy \wedge dx$, so $dx \wedge dx = dy \wedge dy = 0$, (and same for $dz$ and $d\bar{z}$). So: $$dz \wedge d\bar{z} = (dx + i \,dy) \wedge (dx - i\, dy) = dx \wedge dx - i\, dx \wedge dy + i\, dy \wedge dx +i(-i) dy \wedge dy$$ $$= 0 - i \, dx \wedge dy - i\, dx \wedge dy +i(-i) 0 = -2i \, dx \wedge dy$$ $dx \wedge dy$ is for the standard orientation of $\mathbb R^2$. So when you use the two dimensional integration, you integrate $dx \wedge dy$ as $dx \, dy$ or perhaps $dA$, that is the standard area metric on the plane.
Once you start working in more dimensions, it is a lot more convenient to consider differential forms. Perhaps more importantly, you cannot in general just do arithmetic on the differentials like that as you must consider orientation, and that's where the $\wedge$ product comes in. Sometimes one gets lulled into false sense of being able to do this arithmetic beacause it works when working with a single $dx$ and multiplying by functions, etc...
• You are going the wrong direction. Think just the standard $x = u(x)$, then $du = u'(x) dx$. I will try to edit it for you. Commented Dec 13, 2016 at 15:57