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I am trying to find a conformal map from $\mathbb{H}=\{z:Im(z)>0\}$ onto $\mathbb{D}$ take away $(-1,0]$, where $\mathbb{D}=\{z:|z|<1\}$ (the slit unit disk).

I have found that $f(z)=\frac{z-i}{z+i}$ takes $\mathbb{H}$ onto $\mathbb{D}$. Since it is a linear fractional transform, it is conformal.

However, I have no idea how to approach the slit.

Edit: One can map $\mathbb{H}$ onto $\mathbb{C}$ take away the positive imaginary axis by composing the conformal mappings $f(z)=e^{\pi i}z$ (maps to the bottom half plane) and $g(z)=z^{2}$ (doubling all angles), and then $h(z)=e^{\pi/2i}z$.

After this we can apply the fractional linear (Moebius) transformation $F(z)=\frac{z-i}{z+i}$. Since $F(0)=-1$ and $F(i)=0$, we can see that the slit of the positive imaginary axis maps to the negative real axis, exactly as we desire. However, $F$ maps the real axis to the boundary of the unit circle. This unfortunately means that $\phi(z)=\frac{e^{\pi/2i}(e^{\pi i}z)^2-i}{e^{\pi/2i}(e^{\pi i}z)^2+i}$ maps $\mathbb{H}$ onto $\{w:|w|\leq1\}-(-1,0]$, which is not quite $\mathbb{D}-(-1,0]=\{w:|w|<1\}-(-1,0]$. Please help.

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    $\begingroup$ Such a slit points to the possible use of the (principal branch of the complex) logarithm function. $\endgroup$ – Jean Marie Feb 11 at 0:09
  • $\begingroup$ But for the logarithm to be conformal, the domain must be restricted to have a slit, defeating the purpose, am I wrong? $\endgroup$ – Tejas Rao Feb 11 at 0:45
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    $\begingroup$ Good methodology is given here for similar issues : math.stackexchange.com/questions/604877/… and math.stackexchange.com/q/264865 $\endgroup$ – Jean Marie Feb 11 at 8:01
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It is easier if we transform the slit circle into $\Bbb H$. The principal branch of $\sqrt z$ transforms the slit circle into the right semicircle. Multiplication by $i$ takes it to the upper semicircle. Now, the LTF $(1+z)/(1-z)$ takes the semicircle to the first quadrant. ¿Can you finish from here?

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  • $\begingroup$ Surely you mean the right semicircle. Then the linear fractional transformation should be taken as $(\pm i + z)/(\pm i - z)$. $\endgroup$ – Maxim Feb 13 at 10:40
  • $\begingroup$ @Maxim Yes, that is what I meant. Thankyou for commenting. $\endgroup$ – Julián Aguirre Feb 13 at 12:34

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