0
$\begingroup$

I have really seen many examples about Möbius transformation but still stuck on special cases can you help me to outline them and solve this exercise

Find the Möbius transformation which carries the points $ {\mathrm{}}{-1}{\mathrm{,}}\mathrm{\infty}{\mathrm{,}}{i} \ $ into the points :

a) $ {i}{\mathrm{,}}{1}{\mathrm{,}}{1}\mathrm{{+}}{i} $.

b) $ \mathrm{\infty}{\mathrm{,}}{i}{\mathrm{,}}{1} $.

c) $ {0}{\mathrm{,}}\mathrm{\infty}{\mathrm{,}}{1} $.

  • 0
    Is it $1,\infty,i$ or $-1,\infty,i$ ? – 2017-01-25
  • 0
    Where are you stuck? Did you try to substitute the given values into the general form $T(z) = (az+b)/(cz+d)$, and solve for $a, b, c, d$? – 2017-01-25
  • 0
    In short, it is cases that deal with $\infty$ that bother you... – 2017-01-25
  • 0
    The values are $ {-}{1}{,}\infty{,}{i} $ – 2017-01-25
  • 0
    @JeanMarie yes I think that ...I saw some examples really couldn't understand the results . – 2017-01-25
  • 0
    @MartinR I could to solve the first one. Cause there is no infinity – 2017-01-25
  • 0
    Two rules: $z=\infty$ is sent to $a/b$ and $z=-d/c$ is sent to $\infty$ – 2017-01-25
  • 0
    In all cases (with hopefuly the cases where there is $\infty$ that I just described), writing $f(z_1)=Z_1,f(z_2)=Z_2,f(z_3)=Z_3$ gives you 3 equations in $a,b,c,d$ which is enough because in fact all solutions are proportional. – 2017-01-25
  • 0
    Because without loss of generality we can restrict to $ad-bc=1$ – 2017-01-25

1 Answers 1

0

A way that I find quite handy when dealing with $\infty$ in the complex field is to use the concept of homogeneous coordinates , which is tremendously powerful in affine geometry.

So put $$ z = \frac{{z'}} {{z_{\,0} }}\quad \left| {\;z_{\,0} \in \;\mathbb{R}\,\,} \right. $$ Now, when $z$ is a finite number, you normally take $z_{0}=1$ and $z'=z$, unless $z$ has rational components, in which case you may prefer and take $z_0$ to be the common denominator.
To accomodate for an infinite value instead, take $z_{0}=0 $ and for the numerator put any $z \ne 0$, or you can leave it undetermined.
$$ \infty = \frac{z}{{z_{\,0} \to 0}} $$ That premised, let's see a practical application to your case.

So we have $$ w = T(z) = \frac{{a\,z + b}} {{c\,z + d}}\quad \Rightarrow \quad - a\,z - b + c\,w\,z + d\,w = 0 $$ substitute for the homogeneous coordinates, and clear the denominators by multiplying by $z_{0} w_{0}$ (before taking the limit, they are not null), arriving to $$ \left( { - z' \cdot \,w_{\,0} } \right)\,a\, + \left( { - z_{\,0} \cdot \,w_{\,0} } \right)b\, + \left( {w' \cdot \,z'} \right)c\, + \left( {w' \cdot \,z_{\,0} } \right)d\, = 0 $$

Now you can pass and solve this equation for $(a,b,c,d)$, given the three original and relevant tranformed points.

For example in case a) you get $$ \begin{gathered} a)\;\; - 1,\;\infty ,\;\,i\quad \to \quad i,\;1,\,\;1 + i \hfill \\ \hfill \\ \left\{ \begin{gathered} \left( { - \left( { - 1} \right) \cdot \,1} \right)\,a\, + \left( { - 1 \cdot \,1} \right)b\, + \left( {i\, \cdot \left( { - 1} \right)} \right)c\, + \left( {i\, \cdot 1} \right)d\, = 0 \hfill \\ \left( { - z'\, \cdot 1} \right)\,a\, + \left( { - 0\, \cdot 1} \right)b\, + \left( {1 \cdot \,z'} \right)c\, + \left( {1 \cdot \,0} \right)d\, = 0 \hfill \\ \left( { - i \cdot \,1} \right)\,a\, + \left( { - 1 \cdot \,1} \right)b\, + \left( {\left( {1 + i} \right) \cdot \,i} \right)c\, + \left( {\left( {1 + i} \right) \cdot \,1} \right)d\, = 0 \hfill \\ \end{gathered} \right.\quad \Rightarrow \hfill \\ \Rightarrow \left\{ \begin{gathered} \,a\, - b\, - i\,c\, + i\,d\, = 0 \hfill \\ - z'\,\left( {a\,\, - c} \right)\,\, = 0 \hfill \\ - i\,a\, - b\, + i\,c\, - c + d\, + i\,d = 0 \hfill \\ \end{gathered} \right.\quad \hfill \\ \Rightarrow \left\{ \begin{gathered} a\,\, = 2 + i \hfill \\ b = 3 + 4\,i\, \hfill \\ c = 2 + i \hfill \\ d = 5 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} $$