# Map Quadrant Conformally onto the Unit Disc and find $|g'(1+i)|$.

If $w = g(z)$ maps the quadrant $\{z = x + iy \; : \; x,y >0\}$ conformally onto $|w|<1$ with $g(1) = 1$, $g(i) = -1$, and $g(0) = -i$, find $|g'(1+i)|$.

My initial reaction was to directly compute the transform and then take the derivative, i.e. set $g(z) = \frac{az + b}{cz + d}$ and use the given data to solve for the coefficients. I was expecting something similar to $\frac{z^2 - i}{z^2 + i}$ but I ended up getting $g(z) = (1+i)z - i$. This satisfies the given data but doesn't seem to map the quadrant to the unit disc...

I'm not sure if I am making some kind of algebra mistake or my approach is wrong.

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Why do you set $g(z) = \frac{az + b}{cz + d}$? You may try $g(z) = \frac{az^2 + b}{cz^2 + d}$ instead. – 23rd May 13 '13 at 4:52
Thanks. This helped. – Bohring May 13 '13 at 6:00
You are welcome. Would you like to post an answer by yourself? – 23rd May 13 '13 at 6:18
@Landscape how did you conclude that form? could you tell me a bit more detail? – La Belle Noiseuse May 13 '13 at 9:28
@Tsotsi: Firstly, the map $z\mapsto z^2$ maps the first quadrant comformally to the upper half plane; secondly, a linear fractional transformation maps lines to circles/lines. – 23rd May 13 '13 at 10:11

The first step is to map the quadrant to the upper half-plane via the map $T_1(z) = z^2.$ This fixes $0$ and $1$, while mapping $i$ to $-1$.
Now we want to find a map of the form $T_2(z)=\frac{az+b}{cz+d}$ such that $T_2(0) = -i$, $T_2(-1) = -1$ and $T_2(1) = 1$. Given the conditions we find $T_2(z) = \frac{z - i}{-iz + 1}$.
Thus, composing the above two functions we see that $g(z) = \frac{z^2 - i}{-iz^2 + 1}$. Then $g'(z) = \frac{4z}{(-iz^2 + 1)^2}$ so that $|g'(1+i)| = \frac{4}{9}\sqrt{2}.$