In the text "Quantum Optics" by Gerry Knight. Regarding coherent states, he states:
"The completeness relation for the coherent states is given as an integral over the complex $\alpha$-plane according to $$\int | \alpha \rangle \langle \alpha | \frac{d^2 \alpha}{\pi} = 1$$ where $d^2 \alpha = d \text{Re}(\alpha)d \text{Im}(\alpha)$. The proof of this goes as follows: writing $$\int | \alpha \rangle \langle \alpha | d^2 \alpha = \int e^{- |\alpha|^2} \sum_n \sum_m \frac{\alpha^n \alpha^{*m}}{\sqrt{n! m!}}| n \rangle \langle m | d^2 \alpha$$ we tranform to polar coordinates setting $\alpha = r e^{i \theta}, d^2 \alpha = rdr d \theta$ so that $$\int | \alpha \rangle \langle \alpha | d^2 \alpha = \sum_n \sum_m \frac{\alpha^n \alpha^{*m}}{\sqrt{n! m!}} \int_{0}^{\infty}dr e^{-r^2}r^{n+m+1}\int_{0}^{2 \pi}d \theta e^{i (n-m)\theta} $$"
Question: How does he go from the integral with respect to $d^2 \alpha = d \text{Re}(\alpha)d \text{Im}(\alpha)$, to the polar coordinate transformation $d^2 \alpha = rdr d \theta$, how is this a valid transformation?
Considering that $\alpha=x+iy$, $d^2\alpha$ is defined as the surface element $d^2\alpha=dxdy$ in $\mathbb R^2$ as $x=\Re(\alpha)$ and $y=\Im(\alpha)$. Then, it's applied a coordinate transformation to polar coordinates
$x=r\cos\theta$
$y=r\sin\theta$
The imaginary unit is understood simply as a constant with not affecting the transformation. The surface element in polar coordinates is $rdrd\theta$
The last integral must be,
$$\int | \alpha \rangle \langle \alpha | d^2 \alpha = \sum_n \sum_m \frac{\vert n\rangle\langle m\vert}{\sqrt{n! m!}} \int_{0}^{\infty}dr e^{-r^2}r^{n+m+1}\int_{0}^{2 \pi}d \theta e^{i (n-m)\theta}$$