# Finding a conformal mapping with certain points explicitly mapped

I have experience finding mappings from a domain in $\mathbb{C}$ to another domain in $\mathbb{C}$ by composing conformal maps that I do know. However, I do not have any experience with the following type of problem and am curious how to enforce the conditions that certain points map to certain points.

Find a conformal map that sends the upper half plane $\{\Im z > 0\}$ to the unit disk $\{|z| < 1\}$ such that it sends $1 + i$ to $0$ and $−1 +i$ to $1/\sqrt{2}$.

My first idea was to use the conformal mapping $$G(z) = \cfrac{z-i}{z+i}$$ which maps the upper half plane to the unit disk. Multiply it by a scaling factor $\alpha$ with $|\alpha| = 1$ will preserve this mapping, but I'm not sure how to proceed or even if this is the correct idea in the first place.

• Personally, I like working in the UHP vs. the unit disk when doing this. I would (1) find which points in the UHP are getting sent to $0$ and $1/\sqrt{2}$, and then (2) define a map $H$ from UHP to UHP that sent $1+i$ and $-1+i$ to those points. Then $G\circ H$ is the map you want. – Steve D Jun 19 '17 at 17:44
• @SteveD Yes that sounds like a solid plan. If you expand this into an answer (describing how to find such an $H$), I would accept it. – NoseKnowsAll Jun 19 '17 at 18:46

You need a special bilinear transformation of the type $w(z)=k.\frac{z-a}{z-\overline{a}}$ where $a$ is a zero of $w(z)$. In your case $a=1+i$
so $w(z)=k.\frac{z-(1+i)}{z-(1-i)}$. To determine $k$, use $w(-1+i)=\frac{1}{\sqrt 2}$