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I have a question on a change of variable discussed in the article PI.

Let $A$ be a positive non-decreasing smooth function on $(1,\infty)$.

Let $w (\rho,z)$ denotes the solution to $\frac{\partial^2 w}{\partial \rho^2}+\frac{\partial^2 w}{\partial z^2}=0$ in $\{(\rho,z) \mid |z|<A(\rho),\ \rho>1 \} \subset \mathbb{R}^2$, with $w(\rho, \pm A(\rho))=\pm 1$ for $\rho>1$, and $w_{\rho}(1,z)=0$, for $|z|<A(1)$.

The authors of PI make the change of variable $(\rho,z) \mapsto (r,w)$, where $r=1$ on $\rho=1$ and $dr=w_z \,d\rho-w_{\rho}\,dz$.

I know that the map $D:=\{(\rho,z) \mid \rho>1,\ |z|<A(\rho)\} \ni (\rho,z) \mapsto (r,w) \in \mathbb{C}$ is a holomorphic map.

I would like to know that whether $w_\rho^2+w_{z}^2>0$ holds on $D$.

Thank you in advance.

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    $\begingroup$ You should probably replace $\Bbb{R}^2$ by $\Bbb{C}$ and start with $w(s) = \Re(W(s))$ with $W$ complex analytic ($s = \rho+iz$). Then $w_z = \Re(i W'(s)), w_\rho = \Re(W'(s))$ so $r =C+\Im(W(s))$ and $(\rho,z) \mapsto (r,w)$ is really $s \mapsto W(s)$ which is conformal away from the zeros of $W'(s)$ $\endgroup$ – reuns Apr 11 at 12:32
  • $\begingroup$ @reuns Thank you for your comment. I understood that $w$ is the real part of a holomorphic function $W$ and $r$ is essentially the imaginary part of $W$. But what else should I prove to show that $D \ni (\rho,z) \mapsto (r,w) \in \{(r,w) \mid r>1,\ |w|<1\}$ is a conformal map? $\endgroup$ – sharpe Apr 12 at 0:59
  • $\begingroup$ @reuns I understood it now. To show the map $(\rho,z)\mapsto (r,w)$ is conformal, it is enough to show that $W'(s) \neq 0$ on $D$. $\endgroup$ – sharpe Apr 12 at 1:11

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