proof of Box–Muller transform (polar form) There're proofs of Box–Muller transform available online but my book (pattern recognition and machine learning) seems to have put it in a different form.


I didn't follow the derivation of equation 11.12, can anyone please help? Thanks!
EDIT
As mentioned in Nadiels's answer, there's a mistake in formula 11.10 and 11.11, as logarithm has to take in a positive number (PRML errata).

So first thing there is an error in equations $(11.10)$ and $(11.11)$ and in fact you should have the transformations
$$
y_i = z_i \left( \frac{-2 \ln r^2 }{r^2 } \right)^{1/2}
$$
and in particular we have
\begin{align*}
\exp\left( -\frac{1}{2} \left(y_1^2 + y_2^2 \right) \right) &=\exp\left( \left( z_1^2 +z_2^2\right)\frac{\ln(r^2)}{r^2} \right) = r^2,
\end{align*}
which using the inverse function theorem tells us that if 
$$
\mathbf{J} =\begin{bmatrix} \frac{\partial y_1}{\partial z_1} & \frac{\partial y_1}{\partial z_2} \\
\frac{\partial y_2}{\partial z_1} & \frac{\partial y_2}{\partial z_2}\end{bmatrix},
$$
then to get the desired result we want to show that $\left| \operatorname{det}(\mathbf{J}) \right| = 2/r^2$. Or
$$
\left| \left(\frac{\partial y_1}{\partial z_1}\right)\left(\frac{\partial y_2}{\partial z_2}\right) - \left(\frac{\partial y_1}{\partial z_2}\right)^2 \right|= \frac{2}{r^2}.
$$
Let
$$
y_i = z_i h(r^2), \qquad \mbox{where } h(r^2) = \left(-\frac{2\ln r^2}{r^2} \right)^{1/2}
$$
then
\begin{align*}
\left(\frac{\partial y_1}{\partial z_1}\right)\left(\frac{\partial y_2}{\partial z_2}\right) - \left(\frac{\partial y_1}{\partial z_2}\right)^2 &= h(r^2)^2+2r^2h'(r^2)h(r^2) \\
&=h(r^2)^2 + \frac{2}{r^2}\left( \ln(r^2) - 1 \right)\\
&=\frac{-2\ln(r^2)}{r^2} + \frac{2 \ln r^2}{r^2} - \frac{2}{r^2} \\
&= -\frac{2}{r^2}.
\end{align*}
as desired.
