Change of variables from complex variables to real ones I´m trying to prove a result shown on many physics books:
$\int {dz^* \over (2\pi i)^{1/2}} {dz \over (2\pi i)^{1/2}} e^{-z^* b z} = {1\over b}$,
$z$ and $z^*$ complex.
It is usually obtained by making $z = x + i y$,  $z^* = x + i y$ and $dx dy = -{1\over 2} i dz^* dz$.
It is this last part I don't understand. The Jacobian determinant of the change from $dz^* dz$ to $dx dy$, can be $2 i$ or $-2 i$ depending on the ordering of $dz^* dz$ and $dx dy$, and if we take the absolute value the complex $i$ goes away. In other words, why is the following wrong:
$dx dy = {1\over 2} dz^* dz$ ?
I know this probably a very basic question of complex analysis, but I can't find a single reference that treats it in a simple way (without going into differential forms and such)

Well, if you write $z(x, y) = x+iy$, then by "cal 3 chain rule" we have that
\begin{align}
dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y}dy = dx+idy
\end{align}
and likewise
\begin{align}
dz^\ast = dx-idy.
\end{align}
Hence it follows
\begin{align}
-i\frac{1}{2}dzdz^\ast = \frac{1}{2i}[dx^2+dy^2] \neq dxdy .
\end{align}
Note, I have used the fact $[dx, dy] = 0$. So, regardless whether you use $\frac{1}{2}$ or $\frac{1}{2i}$, you will  not get what you want. What the expression should actually read is as follows
\begin{align}
\frac{1}{2i}dz\wedge dz^\ast =&\ \frac{1}{2i}(dx+idy)\wedge (dx-idy)\\
 =&\ \frac{1}{2i}[dx\wedge dx +idy\wedge dx-idx\wedge dy +dy \wedge dy]\\
 =&\ \frac{1}{2i}[idy\wedge dx-idx\wedge dy]\\
=&\ \frac{1}{2i}[-idx\wedge dy-idx\wedge dy]= -dx\wedge dy = -dxdy
\end{align}
which means
\begin{align}
\frac{1}{2i}dz^\ast \wedge dz = dx\wedge dy =dxdy.
\end{align}
Hence order matters and the $1/i$ matters. (Note: $\wedge$ is the wedge product). 
