# On a change of variable and conformal mappings

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 $$\{(\rho,z) \mid |z|1 \} \subset \mathbb{R}^2$$, with $$w(\rho, \pm A(\rho))=\pm 1$$ for $$\rho>1$$, and $$w_{\rho}(1,z)=0$$, for $$|z|.

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| is a holomorphic map.

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

• 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 $W'(s)$ – reuns Apr 11 at 12:32
• @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? – sharpe Apr 12 at 0:59
• @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 $D$. – sharpe Apr 12 at 1:11