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I need help solving this problem:

For $z$ a complex number, let $g(z) = \frac{1 + 2z }{1 + z}$. Find a function $h_1(z)$ such that $h_1(g(z)) = z$ and another (possibly the same) $h_2(z)$ such that $g(h_2(z)) = z$. Use Maple to simplify yout expressions.

Any ideas?

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    @DennisGulko fixed2012-10-09

4 Answers 4

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Because you got instruction to do (at least some of) it in Maple...

restart:  g := s -> (1+2*s)/(1+s);     # g is an operator (ie. function)                               1 + 2 s                         s -> -------                               1 + s   g(z);                        # g(z) is an expression                              1 + 2 z                             -------                              1 + z   isolate(g(z)=t, z);          # isolate g(z) for z                                  -1 + t                           z = - ------                                 -2 + t  rhs(%);                      # an expression                                -1 + t                             - ------                               -2 + t  h1:=unapply(%, t);           # h1 is also an operator                                 -1 + t                         t -> - ------                                -2 + t  h1(g(z));                    # composition of operators (functions)                                  1 + 2 z                            -1 + -------                                  1 + z                            - ------------                                 1 + 2 z                            -2 + -------                                  1 + z   simplify(%);                 # you've shown h1(g(z))=z                                 z  g(h1(z));                    # composition of operators                               2 (-1 + z)                          1 - ----------                                -2 + z                            --------------                                -1 + z                              1 - ------                                  -2 + z    simplify(%);                 # h1 serves as desired h2, as well                                 z 
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Denote $w=\frac{1+2z}{1+z}$. Now isolate $z$: $(1+z)w=1+2z\Rightarrow z(w-2)=1-w\Rightarrow z=\frac{1-w}{w-2}$ Now denote $h_1(w)=h_2(w)=\frac{1-w}{w-2}$. This is called finding the inverse.
Compute $g(h_2(z))$ and $h_1(g(z))$.

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    @JavenP: I'm not sure whether you know this or not (since you are new to Math.SE), but once you got an answer that suites you, you should click the check mark below the answer to accept it, so other people on the site will know that a satisfying answer was given to your question. I write this as a general recommendation and not as a request to accept my answer.2012-10-09
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Basically, invert the matrix: $\begin{pmatrix} 1 & 2 \\ 1 & 1\end{pmatrix}$.

(You can check that composition of these rational functions straightly corresponds to 2x2 matrix multiplication.)

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    @Jave$n$P To add a bit onto this answer, you might want to take a look at [this](http://en.wikipedia.org/wiki/M%C3%B6bius_transformation#Projective_matrix_representations).2012-10-09
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If $f(z)= \frac{az+b}{cz+d} , ad-bc \not= 0$, then $f^{-1}(z)=\frac{dz-b}{-cz+a}$.