Solve $f(z) = z$ for $$f(z)=\frac{iz-1}{(z+1)^2}, \quad \forall z \in \mathbb{C}\backslash\{-1\}.$$
I found that $i$ is a solution. However, the exercise says there are 2 more. I tried using the algebraic expression but it's harder, and used arguments and got nothing.
HINT You have $$ iz-1 = z(z+1)^2 $$ which is equivalent to $$ p(z) = z^3+2z^2+(1-i)z + 1 = 0, $$ which is indeed a cubic equation and should have exactly 3 complex roots. Since you know $z=i$ is a root, $q(z) = p(z)/(z-i)$ is a quadratic polynomial, so solving $q(z)=0$ will give you the other 2 roots.