Consider the function $$ g(t) = \frac{1+it}{1-it} = \frac{1-t^2}{1+t^2} + i \frac{2t}{1+t^2}. $$ (The second equality holds except when $t=i$.)

It seems to be widely known that this function is the only linear fractional transformation having these properties:

  • $g$ maps $\mathbb R\cup\{\infty\}$ to $\{z\in\mathbb C : |z|=1\}$;
  • $g(0)=1$, $g(1)=i$, $g(-1)=-i$, $g(\infty)=-1$;
  • If $g(t)=x+iy$ and $x,y\in\mathbb R$ then \begin{align} g(-t) & = x-iy, \\ g(1/t) & = -x+iy. \end{align}

That's more than enough conditions to narrow it down to just one function.

(I can't resist mentioning that there's also a nice identity for $g(t_1\cdots t_n)$, but it would require me to talk about real and imaginary parts, and I don't think it's widely known.)

I'm wondering what simple sets of conditions are enough to completely characterize this function if one drops the condition that it be a linear fractional transformation? (A somewhat open-ended question.) (An answer dealing only with functions from $\mathbb R\cup\{\infty\}$ to the circle, rather than from $\mathbb C\cup\{\infty\}$ into itself, might be acceptable.)

  • $\begingroup$ Three values of a linear fractional transformation are enough to determine all of its values, so your statement beginning "it seems to be widely known that ..." is rather inconsequential ($g(0)= 1$, $g(1) = i$ and $g(\infty) = -1$ are enough to determine $g$). Hence it is a bit difficult to see what kind of answer you are expecting to your somewhat open-ended question. $\endgroup$ – Rob Arthan Aug 25 '13 at 23:49
  • $\begingroup$ How about, for example, functional equations having this function as their only solution? $\endgroup$ – Michael Hardy Aug 26 '13 at 1:19
  • 1
    $\begingroup$ OK, I may be completely off here, but I just recognized the Weierstrass substitutions in your given problem. So it can be written as cosz + isinz with tan(z/2) = t But then we really have the complex e-power. Does this help in anyway?? $\endgroup$ – imranfat Aug 26 '13 at 2:03

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