I want to construct a function $f$ with the following properties: $f$ has a singularity at $z=1$, and for any $\zeta = e^{2\pi i\frac{a}{b}}$ with $(a,b)=1$, then $$\lim\limits_{x\to1^-}\frac{f(\zeta x)}{f(x)}=\frac{1}{b}$$ E.g. $f(-x)/f(x)\to1/2, f(ix)/f(x)\to1/4$ etc...
I think it is likely such a function exists. For instance, let
$$f(z)=\sum\limits_{k=1}^{\infty}\frac{z^k}{1-z^k}$$
Then plotting the graph of $|(f(0.99\exp(2\pi ix))|$ for $0<x<1$ shows
Which appears linear! (linear by denominators, i.e. satisfying my conditions). Yet when I try to compute in individual cases, the ratios $\frac{f(\zeta x)}{f(x)}$, they do not seem to obey my conditions. (Or perhaps a software error. Maple has given me tantrums before). My question is, can anybody find a function satisfying my conditions? If not, any related information would be welcome.
I know deep down, that the growth rate of a function near roots of unity is wholey determined by the size of the denominator in $$\zeta = e^{2\pi i\frac{a}{b}}$$ But I would like to prove it! So any insights I would like to hear.