Conformal map from upper half plane to the slit unit disk

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.

• Such a slit points to the possible use of the (principal branch of the complex) logarithm function. Commented Feb 11, 2019 at 0:09
• But for the logarithm to be conformal, the domain must be restricted to have a slit, defeating the purpose, am I wrong? Commented Feb 11, 2019 at 0:45
• Good methodology is given here for similar issues : math.stackexchange.com/questions/604877/… and math.stackexchange.com/q/264865 Commented Feb 11, 2019 at 8:01

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?
• Surely you mean the right semicircle. Then the linear fractional transformation should be taken as $(\pm i + z)/(\pm i - z)$. Commented Feb 13, 2019 at 10:40