$\forall z \in \mathbb C, |z|>1$, calculate $\sum_{(p,q)\in\Delta} (z^{p+q}-1)^{-1}$ Let $\Delta = \{(p,q)\in(\mathbb N^*)^2|p\land q =1\}$ and $z\in\mathbb C$ with $|z|>1$,
Calculate : $$S(z)=\sum_{(p,q)\in\Delta}\frac{1}{z^{p+q}-1}$$
My try :
Let $n\in \mathbb N^*\backslash\{1\}$,
we know that : $\forall k \in [1,n-1], k \land (n-k) = k \land n$
So : $$\Delta = \bigcup_{n\ge2}\{(p,q)\in[1,n-1]^2|p\land q = 1 \text{ and } p+q = n\}
=\bigcup_{n\ge2}\{(k,n-k)|k\land n = 1 \text{ and } k\in[1,n-1]\}$$
Let $A_n= \{(k,n-k)|k\land n = 1 \text{ and } k\in[1,n-1]\}$ (Also $A_n$ are obviously disjoint),
We get : $$S(z) = \sum_{n\ge 2}\sum_{(p,q)\in A_n}\frac{1}{z^{n}-1}$$
However : $\text{Card}(A_n) = \varphi(n)$ (the euler totient function).
We have :
$$S(z) = \sum_{n\ge 2} \frac{\varphi(n)}{z^{n}-1} = \sum_{n\ge 2} \frac{\varphi(n)}{z^{n}}\frac{1}{1-z^{-n}} = \sum_{n\ge 2} \sum_{k\ge 0}\frac{\varphi(n)}{z^{n(k+1)}} $$
I don't know what I could do next.
Thanks for your help !

Thanks to @reun, we have our answer !
Using : $$S(z) = \sum_{n\ge 2} \sum_{k\ge 0}\frac{\varphi(n)}{z^{n(k+1)}} = \sum_{n\ge 2} \sum_{k\ge 1}\frac{\varphi(n)}{z^{nk}} $$
We have : $$\sum_{n\ge 2} \sum_{k\ge 1}\frac{\varphi(n)}{z^{nk}} = \sum_{m\ge 2} \sum_{d|m \text{ and } d\ge 2}\frac{\varphi(d)}{z^{m}}=\sum_{m\ge 2}(m-1)z^{-m}=\frac{1}{(z-1)^2} $$
