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 guessed that $f$ is of the form $\sqrt{ax^2+b}$. Then, $f^2$ is $\sqrt{a^2x^2 + \frac{a^2-1}{a-1}b}$. From here on in, it is algebra:
$$
a^2 =-1 \implies a = i ~~~~\text{and}~~~~\frac{a^2-1}{a-1}b = 1 \implies b = \frac{1-i}{2}
$$
So we get $f(x) = \sqrt{ix^2 + \frac{1-i}{2}}$. I checked using Wolfram, and $f^2$ appears to be what we want. 
DISCLAIMER: This is not the only solution.
