# Reverse continued fraction

The problem doesn't seem that difficult but I am unable to come up with a solution :

Let $$\theta=\overline{[a_0,a_1 ....... a_n]}$$ be purely periodic continued fraction. Then find the continued fraction $$\overline{[a_n,a_{n-1}....... a_1]}$$ in terms of theta and it's conjugate.

If $$y=a_0+\frac 1{a_1+}\cdots\frac 1{a_n+x}$$ then $$-x=a_n+\frac 1{a_{n-1}+}\cdots\frac 1{a_0-y}$$ hence the reversed continued fraction of $$\theta$$ gives the opposite of the conjugate of $$\theta$$.