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If $f(z)=\cfrac{z+1}{z-1}$ , then find $f^{1991}(2+i)$

Forgive me if the question is too short but really I don't know how to do this one.

That's what I have done so far:

$\left(f(2+i)\right)^{1991}=\left(\cfrac{3+i}{1+i}\right)^{1991}$

So now If I can find a polar form for $(3+i)$ and $(1+i)$ I can then apply the property that for any complex number I have $(s,\phi)^{1991}=(s^{1991},1991 \cdot\phi)$

The problem is that I can't find one,and by checking with wolfram alpha I've understood why.

Therefore there must be slick way I am not seeing.

Can you guys help ?

Mr. Y
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    You have a notation misconception: By $f^n(u)$ they mean $f(f^{n-1}(u))$, not $[f(u)]^n$. – Mark Fischler Dec 20 '15 at 15:23
  • Oh.Thanks for that....so my effort is totally flawed. – Mr. Y Dec 20 '15 at 15:24
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    If you don't see the 'trick' then there are other ways to get the result. The transformation is a so-called Möbius transformation. Such a transformation can be decomposed in terms of elementary transformations as: $\frac{z+1}{z-1} = [z + 1] \circ [2z] \circ [1/z] \circ [z - 1]$ which is a “translation by $1$” + “dilation by $2$” + “inversion” + “translation by $-1$”. In this picture (when you get some intuition) not too hard to see that applying it two times should give you back what you started with so $f\circ f = z$. – Winther Dec 20 '15 at 16:17

3 Answers3

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$$ f^2(z) = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = z $$ So $f^{2k}(z) = z$ and $f^{2k+1}(z) = \frac{z+1}{z-1}$

$$f^{1991}(2+i) = \frac{3+i}{1+i} = 2-i$$

Mark Fischler
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Hint:

Check that $f^2(z)=z$. What can you deduce from this relation?

Bernard
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  • I see what you mean,but now I am curious :why it does work for every even power? – Mr. Y Dec 20 '15 at 15:30
  • Because any power of identity is identity. – Bernard Dec 20 '15 at 15:31
  • Because $f^{2n}(z) = [f^2]^n(z)$ – Ben Grossmann Dec 20 '15 at 15:32
  • @Omnomnomnom Where can I find a usefull resource about that ?I don't really get much from a comment (I need to see all the theory behind ...) – Mr. Y Dec 20 '15 at 15:39
  • @Mr Y: There's no real theory: it's just the laws of exponents which is the same for the composition of a function with itself and for a number. Now, I think it is obvious that $;f^{2n}(z)=\bigl(f^2\bigr)^n(z)=(\operatorname{id})^n(z)=\operatorname{id}(z)=z$. – Bernard Dec 20 '15 at 15:47
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$f(z)=\cfrac{z+1}{z-1} \Rightarrow f(f(z))=f^2(z)=z$

Similarly $f^3(z)=\cfrac{z+1}{z-1}$

So proceeding in this manner we get , $f^{2n}(z)=z$ and $f^{2n+1}(z)=\cfrac{z+1}{z-1}$

So $f^{1991}(z)=\cfrac{z+1}{z-1}$

Therefore $f^{1991}(2+i)=\cfrac{3+i}{1+i}=2-i$

Angelo Mark
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