Find $f$ if $f(f(x))=\sqrt{1-x^2}$ Find $f$ if $f(f(x))=\sqrt{1-x^2} \land [-1; 1] \subseteq Dom(f)$
$$$$Please give both real and complex functions. Can it be continuous or not (if f is real)

I don't know how to approach the real case, but here's a way to solve it if you allow $f$ to be complex.
Start from the guess
$f(x) = \sqrt{1-x^2}$. It almost works, and suggests trying the larger family
$$f(x) = \sqrt{a-bx^2}.$$
For what values of $a,b$ does $f(f(x)) = \sqrt{1-x^2}$?
