For all complex $|z| \neq 1$ : $\frac{z}{1+z+z^{2}+z^{3}+z^{4}} = \sum_{n=0}^{\infty} T_n \frac{z^n}{1+z^n+z^{2n}}$?

Question

Inspired by this one

For all complex $|z| \neq 1$ : $\sum_{n=0}^{\infty} w_n \frac{z^n}{1+z^n+z^{2n}+z^{3n}+z^{4n}} = \sum_{n=0}^{\infty} u_n \frac{z^n}{1+z^n+z^{2n}}$?

It made sense to me, to take the terms apart and split the problem into 2 subproblems.

Allow me to explain.

if $\sum_n a_n x_n = \sum_n b_n y_n$ would be true, it would make sense that

$$a_1 x_1 = \sum_n c_n y_n ,a_2 x_2 = \sum_n d_n y_n$$,...

AND

$$b_1 y_1 = \sum_n e_n x_n ,b_2 y_2 = \sum_n f_n x_n$$,...

are all solvable

would imply that $\sum_n a_n x_n = \sum_n b_n y_n$.

So within the spirit of that idea, the 2 logical questions become :

1.

For all complex $|z| \neq 1$ : $\frac{z}{1+z+z^{2}+z^{3}+z^{4}} = \sum_{n=0}^{\infty} T_n \frac{z^n}{1+z^n+z^{2n}}$

For some sequence $T_n$ and

2.

For all complex $|z| \neq 1$ : $\sum_{n=0}^{\infty} V_n \frac{z^n}{1+z^n+z^{2n}+z^{3n}+z^{4n}} = \frac{z}{1+z+z^{2}}$?

For some sequence $V_n$

If both 1. and 2. are true it seems much more believable that the main question in the link is true.

Domain, Range, convergeance, uniqueness, existance are subtle things here, but at least it would be intuitively possible.

Keep in mind that the idea of the transitive property has been used implicitly :

if $A = B$ and $B = C$ then $A = C$,

where $A$ is the taylor series and $B$ and $C$ are the sums over rational functions.