Factoring a complex function

Question

How can I write the function $f(z)=1-0.7z-0.3z^{-1}$, $z$ complex, as $f^+(z) f^-(z)$, where

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 $f^-(z)=1$.

Answer 1

Since $f^+ (1) f^- (1) = f(1) = 0$, either $f^+(1) = 0$, or $f^-(1) = 0$. Either case contradicts your requirements, so such a factoring is not possible.

Answer 2

$$f(z)=1-0.7z-0.3z^{-1}=\dfrac{-7z^2+10z-3}{10z}=\dfrac{-7(z-1)(z-\frac37)}{10z}$$ $$f(z)=\dfrac{-7}{10}\cdot(z-\frac37)\left(1-\dfrac{1}{z}\right)$$

then $$f^+=\dfrac{-7}{10}~~~,~~~f^-=(z-\frac37)\left(1-\dfrac{1}{z}\right)$$