I am trying to understand the identity $$(I-z^nT^n)^{-1} =\frac{1}{n}[(I-zT)^{-1}+(I-wzT)^{-1}+...+(I-w^{n-1}zT)^{-1}] \quad (*),$$ where $T \in \mathbb{C}^{n\times n},z\in \mathbb{C}$ and the spectral radius of $T$, $\rho(T)=\max\{|\lambda|: \exists v, Tv=\lambda v\}\leq 1$ and $|z|<1$ and $w$ is a primitive $n$th root of 1,i.e. $w =e^{i2\pi/n}$.
I have tried using the identity $$[I-A]^{-1} = I+A+A^2+A^3+... $$ which holds when $A^n$ becomes zero matrix. This works as $\rho(T)=\max\{|\lambda|: \exists v, Tv=\lambda v\}\leq 1$ and $|z|<1$ implies the convergence of $A^n$. So I expand both sides of $(*)$ and then compared the coefficient of the two power series.
But this approach seems to be a bit tedious, I wonder whether there is a more clean and simpler proof.
First we establish the identity: $$ \frac{n}{1-z^n}=\sum_{i}\frac{1}{1-\omega^iz},\qquad z\in \mathbb{C}. $$ (Hypothetically) expanding the RHS, we observe that the numerator is a polynomial of degree $\leq n-1$ and the denominator agrees with the LHS. Thus we are reduced to verifying a polynomial identity of degree $n-1$, so evaluating at $n$ points will suffice to establish equality.
For any $0\leq i\leq n-1$, evaluate both sides at $z=\omega^i$. The RHS has a pole with residue $-w^i$, coming from the term $$ \frac{1}{1-\omega^{-i}z}=\frac{w^i}{w^i-z}. $$ Meanwhile, the residue on the LHS is $$ \left.\frac{n}{(1-z^n)'}\right|_{z=w^{i}}=\left.\frac{n}{-nz^{n-1}}\right|_{z=w^{i}}=-w^i. $$ Thus the two expressions agree at $n$ points, whereupon the identity is established.
Applying the holomorphic functional calculus, we may formally replace $1$ with $I$ and $z$ with $zT$ (since $\rho(T)\leq 1$ and $|z|< 1$ everything converges), yielding your result.