Prove $\sum_{k=1}^{m}\frac{1}{1-e^{2i\pi jk/m}z}=\frac{m}{1-z^{m/\gcd(m,j)}}$ 
Let $m\in\Bbb Z_{\ge1}$ be given. Let $j$ be an integer between $0$ and $m-1$. Prove
  $$R_{m,j}(z):=\sum_{k=1}^{m}\frac{1}{1-\zeta_m^{kj}z}=\frac{m}{1-z^{g(m,j)}},\tag1$$
  where $\zeta_m=e^{2i\pi/m}$, and $g(m,j)=m/\gcd(m,j)$.

Context:
For notational convenience, set $$u_m^n=\frac1{1-\zeta_m^n z}.$$ I noticed, while playing around with various similar algebraic identities involving roots of unity, that $(1)$ seems to hold. Most of my observations were made numerically, but after a bit I was able to come up with $(1)$. I have gotten three quarters of the proof of $(1)$ so far. 
Case $1$: $j=0$. 
Since $\zeta_m^0=1$, we have $u_m^0=1/(1-z)$, giving $R_{m,0}(z)=m/(1-z)$, which agrees with $(1)$.
Case $2$: $j=1$. 
This is the classic factorization formula
$$S_m:=\sum_{k=1}^{m}\frac{1}{1-\zeta_m^kz}=\frac{m}{1-z^m},$$
which agrees with $(1)$.
Case $3$: $2\le j\le m-1$ and $j|m$. 
Since $j|m$, we can write $m=jd$, so that $\zeta_{m}^{jk}=\zeta_d^k$ and thus $u_m^{jk}=u_d^k$. 
Also notice, that for all integers $a$, $\zeta_m^a=\zeta_m^{M(a,m)}$, where $M(p,q)=p-q\lfloor\tfrac{p}{q}\rfloor$.
We then have 
$$\begin{align}
R_{m,j}(z)&=\sum_{k=1}^{jd}u_d^k\\
&=\sum_{k=1}^{jd}u_d^{M(k,d)}\\
&=j\sum_{k=1}^{d}u_d^{k}\\
&=jS_d\\
&=\frac{jd}{1-z^d}.
\end{align}$$
Recalling that $m=jd$, we have that $\gcd(m,j)=j$ thus $g(m,j)=m/j=d$. Hence we have agreement with $(1)$.
Case $4$: $2\le j\le m-1$ and $j\not | m$.
I do not know how to proceed in this case.
My issue is that there are two sub-cases, and the first relies on the second which I do not know how to resolve. Specifically, the first sub case is $1<\gcd(m,j)<j$. I have a neat trick for this.
Let $d=\gcd(m,j)$, then write $m=ad$ and $j=bd$ for some integers $a,b$ both greater than $1$. We then have 
$$\begin{align}
R_{ad,bd}(z)&=\sum_{k=1}^{ad}u_{ad}^{bdk}\\
&=\sum_{k=1}^{ad}u_{a}^{bk}\\
&=\sum_{r=1}^{d}\sum_{k=1+(r-1)a}^{ra}u_{a}^{bk}\\
&=\sum_{r=1}^{d}\sum_{k=1+(r-1)a}^{ra}u_{a}^{b(k-(r-1)a)}\qquad \text{because }(1-r)ab\in\Bbb Z\\
&=\sum_{r=1}^{d}\sum_{k=1}^{a}u_{a}^{bk}\\
&=d\sum_{k=1}^{a}u_{a}^{bk}\\
R_{ad,bd}(z)&=d\cdot R_{a,b}(z).
\end{align}$$
As a consequence, we can basically reduce every $R_{m,j}$ to $(\text{integer})\cdot R_{m',j'}$, where $\gcd(m',j')=1$.
The problem is, I do not know how to prove $(1)$ for $\gcd(m,j)=1$. Could I have some help? Thanks.

Update (4/14/20):
To prove $(1)$ for coprime $m,j$, it would be sufficient to show that 
$$\gcd(m,j)=1\Rightarrow \{M(kj,m):k=1,...,m\}=[0,m-1]\cap\Bbb Z.$$
 A: You seem to be happy with the following (which is not so hard to prove): if $\zeta = \exp(2 \pi i/m)$, then
$$\sum_{k=1}^{m} \frac{1}{1 - \zeta^k x} = \frac{m}{1 - x^m}.$$
The first claim is that this holds for any primitive $m$th root of unity, so $\zeta = \exp(2 \pi i j/m)$ for any $(j,m) = 1$. The reason is that if $(j,m) = 1$, then the set $jk \pmod m$ as $k=1,\ldots,m$ is precisely the same as the set $k \pmod m$, because $j \in (\mathbf{Z}/m \mathbf{Z})$ is invertible. Since $\zeta^n$ only depends on $n \pmod m$, we just get the identical sum in a different order.
Now suppose that $m = M d$, and $j = J d$ where $(M,J) = 1$ and $(m,j) = d$. Then $\zeta^d = \xi$, where $\xi = \exp(2 \pi i/M)$, and $\xi^{n}$ only depends on $n \pmod M$. Then
$$\sum_{k=1}^{m} \frac{1}{1 - \zeta^{k j} z} = 
\sum_{k=1}^{m} \frac{1}{1 - \xi^{k J} z}= d \sum_{k=1}^{M} \frac{1}{1 - \xi^{k J} z} = d \cdot \frac{M}{1 - x^{M}} = \frac{m}{1 - x^{M}},$$
where the second last equality is an application of what was proved above since $(M,J) = 1$.

Of course, you didn't prove the first equality; the proof of that equality works in general. Let $F(x)$ be a power series, and let $\zeta$ be a primitive $M$th root of unity. Then:
$$\sum_{k=1}^{RM} \zeta^{nk} = \begin{cases} RM, & n \equiv 0 \pmod M, \\
0, & \text{otherwise}. \end{cases}$$
If $n \equiv 0 \pmod M$ then $\zeta^n = 1$ and this is obvious. Otherwise, we have $\zeta^n \ne 1$, and we get a geometric progression whose sum is zero (it has the factor $\zeta^{RM} - 1 = 0$ as the numerator).
It follows that if $F(X) = \sum_{n=0}^{\infty} a_n X^n$ is any power series, then
$$\sum_{k=1}^{RM} F(X \zeta^k) = RM \sum_{n=0}^{\infty} a_{nM} X^{nM}.$$
So for example if $F(X) = 1/(1-X)$ the sum is $RM/(1 - X^M)$, which also recovers your result.
