If $f:\Bbb C\to\Bbb C$ satisfies $f(f(x))=\left(x\overline x-x-\overline x\right)^2$ and $f(1)=0$, find $f(\mathrm i)$.
If $f(a+b\mathrm i)=a^\prime+b^\prime\mathrm i$, define $g:\Bbb R^2\to\Bbb R^2$ such that $g((a,b))=\left(a^\prime,b^\prime\right)$. Since \[\left[(a+b\mathrm i)\overline{(a+b\mathrm i)}-a-b\mathrm i-\overline{(a+b\mathrm i)}\right]^2=\left(a^2-2a+b^2\right)^2,\] we have $g(g((a,b)))=\left(\left(a^2-2a+b^2\right)^2,0\right)$. I don't know what to do then?
By the way, I found that $g((a,b))=\left((a-1)^2+b^2,0\right)$ satisfies.