If $a\in \mathbb C^*$, i need a relationship between $\int_{\mathbb C} f(az)dz$ and $\int_{\mathbb C} f(z)dz$ If $a\in \mathbb R^*$,  by identification of $\mathbb C$ with $\mathbb R^2$ and identification of $f(z)$ with $f(x,y)$ forall $z=x+iy$, I can write that
$\displaystyle\int_{\mathbb C} f(az)dz=\iint_{\mathbb R^2} f(ax,ay) dx dy=\frac 1{a^2}\iint_{\mathbb R^2} f(x,y) dx dy= \frac 1{a^2}\int_{\mathbb C} f(z)dz$.
But if $a\in \mathbb C^*$, what is the relationship between $\displaystyle\int_{\mathbb C} f(az)dz$ and $\displaystyle\int_{\mathbb C} f(z)dz$?
 A: Identifying $\mathbb{C}$ with $\mathbb{R}^2$ in the usual way, an element $a = \sigma + it \in \mathbb{C}^\times$ acts on $\mathbb{R}^2$ by multiplication by $(\sigma + it)(x + iy) = \sigma x - ty + i(tx + \sigma y)$; or, in matrix form (mapping $x + iy\in \mathbb{C}$ to the vector $(x, y)\in \mathbb{R}^2$)
\begin{align*}
a.\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}\sigma & -t \\ t & \sigma\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}.
\end{align*}
Thus by (e.g.) the change of variables theorem,
\begin{align*}
\int_\mathbb{C} f(az)\; dz 
   &= \int_{\mathbb{R}^2} f(g(x, y))\; dx \; dy
   = \int_{g^{-1}(\mathbb{R}^2)} f(x, y) |\det Dg(x, y)|^{-1}\; dx \; dy,
\end{align*}
for
\begin{align*}
g(x, y) &= \begin{pmatrix}\sigma & -t \\ t & \sigma\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}
\end{align*}
and sufficiently nice $f$. Thus
\begin{align*}
\int_\mathbb{C} f(az)\; dz 
   &= \frac{1}{\sigma^2 + t^2} \int_{\mathbb{R}^2} f(x, y) \, dx\, dy
   = \frac{1}{|a|^2} \int_{\mathbb{C}} f(x, y).
\end{align*}
