Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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? –  Bunuelian Trick 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
show 1 more comment

1 Answer

up vote 2 down vote accepted

I figured I should answer this myself so this question can be closed.

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 $|g'(1+i)| = \frac{4}{9}\sqrt{2}.$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.