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 a Möbius transform is a map that goes from the extended complex plane to the extended complex plane, given by some $\omega = \frac{az + b}{cz + d}$, where $ad - bc \neq 0$. In my notes, underneath the definition, I have:

Ex: $ \frac{1z + 2}{2z + 4} = \frac{1}{2}$.

How does this work? Isn't ad - bc = 0? Or have I made some kind of mistake in writing down the notes properly? Where does the fraction come from?

share|improve this question
    
The example illustrates why we want $ad - bc\neq 0$ since we would like these to not be constant functions. –  Tobias Kildetoft Dec 13 '12 at 18:08
    
I think the "ex" is probably the reason that we require $ad-bc\neq 0$ - you get problematic cases if $ad-bc=0$, and this is an example of why it is problematic. (Another example, $c=d=0$ is obviously problematic.) –  Thomas Andrews Dec 13 '12 at 18:08
    
Where does the 1/2 come into it then? –  Kaish Dec 13 '12 at 18:10
1  
$2z+4=2(1z+2)$, so $\frac{1z+2}{2z+4}=\frac{1}{2}$ –  Thomas Andrews Dec 13 '12 at 18:10
add comment

1 Answer

up vote 2 down vote accepted

As per your definition, the map is not a Möbius transform. As for how it's $1/2$? $2z+4=2(1z+2)$.

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.