1. $f^+(z)$ is free of zeros and singularities **outside and on** the unit circle

2. $f^-(z)$ is free of zeros and singularities **inside and on** the unit circle.

As can be verified, $f$ has a single pole at $z=0$ and two zeros at $z=1$ and $z=3/7$, so I cannot simply take $f^+(z)=f(z)$ and $f^-(z)=1$.