Change of variable in integral for $f(z,\bar{z})$

Question

Consider the integral of a function of two variables :

$$I=\int dx dy f(x,y)$$

Now we do the change of variables : $z=\frac{x+iy}{\sqrt{2}}$, $\bar{z}=\frac{x-iy}{\sqrt{2}}$.

In my course it is written that we have then :

$$I=-i\int dz d\bar{z} f(x(z,\bar{z}),y(z,\bar{z}))$$

How to we know that the change of variables is $ dxdy \rightarrow -i dz d\bar{z}$.

I studied holomorphic functions theory but a holomorphic function is a function of $z$ only. Thus I don't know how to do a change of variables if it depends on $z$ and $\bar{z}$. I tried to do an "analogy" as what we do in $\mathbb{R}^n$ using a Jacobian but in the jacobian we have a modulus so we shouldn't have this $-i$ appearing.

How to give a sense to it, and the most important : how to compute change of variables of such integrals ?

Answer 1

Well I think an explanation is the Jacobian, (assume of course that $z=x+iy$)

\begin{align*} dz \, d\bar{z} &= \begin{vmatrix} \partial_x z & \partial_y z \\ \partial_x \bar{z} & \partial_y \bar{z} \end{vmatrix} \, dx \, dy \\ &= \begin{vmatrix} 1 & i \\ 1 & -i \end{vmatrix} dx \, dy \\ &= -2i \, dx \, dy \end{align*}

Edit Notwithstanding the above I offer a bit more information into this.

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.