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If $f(z) = z + 1$ and $g(z) = -\frac{1}{z}$ show that $$ g f g^{-1}(z) = \frac{z}{1-z}. $$

I don't know how to solve this question please help.

Sammy Black
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John Doe
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    By gfg-1, do you mean the composition $g\circ f\circ g^{-1}$? If so, can you compute $g^{-1}$? –  Nov 26 '13 at 04:59
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    There are no tricks here other than symbol manipulation. You have $z = - \frac{1}{g(z)}$. Let $g^{-1}(w) = z$. – copper.hat Nov 26 '13 at 07:44

1 Answers1

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Let's get this out of the unanswered queue, although comments already pointed out the relevant aspects

First you need to find $g^{-1}$. Assume $g(z)=w$, then $g^{-1}(w)=z$. If you have $w=-\frac1z$, you can solve that for $z$ (i.e. transform it such that it reads $z=\ldots$) to obtain a formula for $g^{-1}$.

Then you have all three formulas, so you can simply nest them and compute their result:

$$g\Bigl(f\bigl(g^{-1}(z)\bigr)\Bigr)=\ldots$$

Simply evaluate the innermost function, plug the result into the next, and so on. Simplify the final result until you obtain the desired form.

MvG
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