# When is $f^{-1}=1/f\,$?

I seem to remember a nice article that I read many years ago (perhaps in the American Mathematical Monthly) which investigated the question, "Under what conditions on $f:\mathbb{C}\to\mathbb{C}$ is $f^{-1}=1/f\,$?"

Despite my online searches, I cannot locate it. Does anyone have a reference? Or answer to the question---which to the best of my memory was nontrivial (and took a complex variables approach)?

Edit: Just to set the record straight, the assumption the authors used in the article was actually this:

ASSUMPTION. The function $f$ is one-to-one from the positive half-line $(0, \infty)$ onto itself and satisfies $f^{-1}(x)= 1/f(x)$ for all $x$ in $(0,\infty)$.

-
What is $f$, exactly? $f^{-1}=1/f$ when $f\circ f=1/x$, but I'm not sure if that's what you're asking about. –  tomasz Dec 16 '12 at 21:51
Sorry, but if I recall correctly, in that article they simply took $f:\mathbb{C}\to\mathbb{C}$. I will edit the question. Also, they were looking for the most general conditions on $f$ under which the result held, not just particular examples. –  JohnD Dec 16 '12 at 21:53

If you're interested in complex solutions, the function $f(z) = z^i$ (using the principal branch) satisfies $f(f(z)) = 1/z$ for $e^{-\pi} < |z| < e^{\pi}$.