Show that for any given rational functional $f(z)$, with poles in the unit disc and without poles in the unit circle, it is possible to find another rational function $g(z)$, with no poles in the unit disc, and such that $|f(z)|= |g(z)|$ if $|z| = 1$
I'm not really sure where to start here. A hint on how to begin the proof would be really appreciated.
Hint: If $f(z)$ has poles at $\{a_j\}_{j=1}^k$ (counted with their multiplicites), then $$g(z)=\left(\prod_{j=1}^k\frac{z-a_j}{1-\overline{a_j}z}\right)\cdot f(z)$$ has no poles and since the modulus of each term in the product is $1$ on the boundary, we have $|g(z)|=|f(z)|$, as desired.
The product above is known as a Blaschke factor, and can be read about at Wolfram MathWorld or Wikipedia.
Let $a_1,\dots,a_n$ be all the poles of $f$ in the unit disk(counting multiplicity). Consider the Blaschke product $B(z)=\prod_{i=1}^n\frac{z-a_i}{1-\bar{a_i}z}$ and $g$ can be chosen as $f\cdot B$.
