I am working on solving for all complex-valued solutions $f: \mathbb{C} \rightarrow \mathbb{C}$ satisfying the following functional equation: $$f(x)\cdot f\left(\frac{1}{1-x}\right)=x, \; \forall x \neq 0,1.$$

I tried a linear solution $f(x)=A+Bx$ as a simple guess, which yields $(A+Bx)(A+B(\frac{1}{1-x}))=x \implies (A+Bx)(A+B-Ax)=x-x^2$. Since this holds $\forall x\in \mathbb{C}$ (other than 0 and 1), the coefficients of the polynomial on the left hand side must be the same as those of the corresponding terms in the polynomial on the right hand side. Equating $x^2$ terms $\implies -AB=-1$, equating $x$ terms $\implies (A+B)B-A^2=1$, and equating constants $\implies A(A+B)=0$. Calling these three equations (1), (2), and (3) respectively, (1) and (3) together imply that $A^2=-1 \implies A=\pm i \implies B=\mp i$ (from substituting $A$ into (1) or (3)). Thus there are two linear solutions, jointly given by $f(x)=\pm i \mp i x= \pm i (x-1)$. Are these the only solutions? If not, how can I obtain others? As an aside, I notice these values of $A$ and $B$ satisfy (2), although (2) was never used in obtaining them. Why is this the case?

  • $\begingroup$ Another class of well known functions are the Möbius transformations $z\mapsto \frac{az+b}{cz+d}$. Since we have Möbius trasnformations everywhere in this equation that might be worth checking out. $\endgroup$ – flawr Sep 28 '16 at 22:38
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    $\begingroup$ Possible duplicate of Find $f$ if $ f(x)+f\left(\frac{1}{1-x}\right)=x $ $\endgroup$ – user236182 Sep 28 '16 at 22:45
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    $\begingroup$ That is not a duplicate, as there it is $f(x) + f(1/(1-x)) = x$ and here it is $f(x) \cdot f(1/(1-x)) = x$. $\endgroup$ – TMM Sep 29 '16 at 0:42
  • $\begingroup$ Take a log(f), and that would make it an exact duplicate. $\endgroup$ – Ivan Neretin Sep 29 '16 at 7:58

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