I am struggling a bit when it comes to conformal mappings. I have the general idea: preserves angle, analytic, non-zero derivative, etc . . . but the specifics of actually coming up with mapping are eluding me.


Let $D_1=\{z \in \mathbb{C}: |z-1|>1 \text{ and } |z+1|>1\}$ and $D_2=\{z \in \mathbb{C}: |z|<1\}$

Find a conformal bijection from $D_1$ onto $D_2$.


ng So the image set is the unit disc, I am having some trouble visualizing but I believe we are looking at the set $D_1=\{z \in \mathbb{C}: -2\Re(z) < |z|^2 < 2\Re(z)\}$

I know the map $\frac{z-1}{z+1}$ maps the unit disc onto the half-plane but I don't have a general strategy for coming up with conformal mappings.

  • 1
    $\begingroup$ Regarding the shape of $D_1$: you are looking at $\mathbb{C}$ with the two closed unit disks around $(-1,0)$ and $(1,0)$, respectively, removed. $\endgroup$ – Thomas Mar 20 '16 at 17:16
  • $\begingroup$ Try to include images in your question $\endgroup$ – user297008 Mar 20 '16 at 17:59
  • $\begingroup$ Note that \begin{align}|z+1|&>1 \iff (z+1)(\bar{z}+1)>1\iff |z|^2>-2\Re z,\\ |z-1|&>1\iff (z-1)(\bar{z}-1)>1\iff |z|^2>2\Re z. \end{align} Why did you recognize $D_1$ as $\{z \in \mathbb{C}: -2\Re(z) < |z|^2 < 2\Re(z)\}$ ? $\endgroup$ – ts375_zk26 Mar 21 '16 at 23:43

We consider several transformations $\varphi , \phi, \psi$ and $g$. So \begin{align} \zeta&=\varphi (z)=\frac{1}{z},\quad \varphi : D_1\to G_1=\{\zeta:-\frac{1}{2}<\operatorname{Re}\,\zeta<\frac{1}{2}\},\\ \xi&=\phi(\zeta)=i\pi \zeta, \quad \phi : G_1\to G_2=\{\xi : -\frac{\pi}{2}<\operatorname{Im}\,\xi<\frac{\pi}{2}\},\\ \eta&=\psi (\xi)=e^\xi,\quad \psi : G_2\to G_3=\{\eta : \operatorname{Re}\, \eta>0\}\\ w&=g(\eta)=\frac{\eta-1}{\eta+1}, \quad g : G_3\to D_2. \end{align} See the diadram below. Then we get $$ w=g\circ \psi\circ\phi\circ\varphi (z)=\frac{\exp\left(\displaystyle\frac{i\pi}{z}\right)-1}{\exp\left(\displaystyle\frac{i\pi}{z}\right)+1}$$ as a conformal bijection from $D_1$ onto $D_2$. enter image description here

Addendum: The function $\zeta=\varphi (z)=\frac{1}{z}$ maps $\{z : |z-1|>1\}$ onto $\{\zeta : \operatorname{Re}\, \zeta<\frac{1}{2}\}$ and maps $\{z : |z+1|>1\}$ onto $\{\zeta : \operatorname{Re}\, \zeta>-\frac{1}{2}\}$. Therefore $\varphi $ maps $D_1$ onto $\{\zeta : -\frac{1}{2}<\operatorname{Re}\, \zeta<\frac{1}{2}\}$.

  • $\begingroup$ Could you provide latex code for the drawing please? $\endgroup$ – S. Maths Jan 22 at 1:31

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