= Solution
For a first-order system, it is important to specify the frame in which its coefficients are written. In the simple-pole convention intended here, a <Fuchsian linear differential system> has, at a finite singular point $a$,
$$
A(z)=\frac{R_a}{z-a}+H_a(z),\qquad H_a\text{ holomorphic near }a.
$$
The <residue> is the matrix $R_a=\lim_{z\to a}(z-a)A(z)$, computed entry by entry. Such a point is a <regular singular point>: on a sector, the coefficient bound $\|A(z)\|\le C/|z-a|$ and integration along rays give at most power growth of solutions. Intrinsically, a <regular singular point> means moderate power growth of all solutions on sectors, allowing logarithmic factors. This intrinsic definition is more general than requiring a simple coefficient pole in an arbitrary meromorphic frame.
For $A(z)=R/z$, a <fundamental matrix> on a chosen <complex logarithm> branch is
$$
\Phi(z)=\exp(R\log z).
$$
Differentiating the <matrix exponential> verifies $\Phi'=R\Phi/z$. A counterclockwise circuit replaces $\log z$ by $\log z+2\pi i$, so
$$
\Phi\longmapsto\Phi M,\qquad
\boxed{M=e^{2\pi iR},\quad \text{monodromy group }=\{M^n:n\in\mathbb Z\}.}
$$
This is the <residue and monodromy of a constant Fuchsian system>. A change of solution basis conjugates $M$. Its <eigenvalues> are $e^{2\pi i\lambda}$ for the residue <eigenvalues> $\lambda$; a nontrivial <Jordan block> supplies the corresponding logarithmic terms in $z^R$. The <monodromy> does not uniquely recover $R$, since integer shifts of residue <eigenvalues> can leave their exponentials unchanged.
The coordinate-independent coefficient is the matrix-valued one-form $A(z)\,dz$. With $t=1/z$, its coefficient is
$$
\widetilde A(t)=-t^{-2}A(1/t).
$$
Being Fuchsian at infinity means $A(z)=O(z^{-1})$, while being ordinary there means $A(z)=O(z^{-2})$. If the finite simple poles are $a_1,\ldots,a_s$, subtract their principal parts:
$$
H(z)=A(z)-\sum_{j=1}^s\frac{R_j}{z-a_j}.
$$
Each entry of $H$ is <entire> and tends to zero at infinity. <Liouville's theorem> therefore gives $H=0$. Expanding this <partial fraction decomposition> at infinity yields
$$
\boxed{A(z)=\sum_{j=1}^s\frac{R_j}{z-a_j},\qquad R_\infty=-\sum_{j=1}^sR_j.}
$$
Thus the sum of all <residues>, including the <residue> at infinity of the one-form, is zero. If infinity is ordinary, $R_\infty=0$ and the finite <residues> alone sum to zero. Treating $A$ merely as a scalar function of $z$, without transforming $dz$, would give the wrong infinity convention.
Put $\omega=e^{2\pi i/3}$. The possible finite singular points of $(B+zC)/(1-z^3)$ are $1,\omega,\omega^2$, and
$$
R_\zeta=-\frac{B+\zeta C}{3\zeta^2}\qquad(\zeta^3=1).
$$
A point $\zeta$ is removable exactly when $B+\zeta C=0$ as a matrix. At infinity,
$$
\widetilde A(t)=\frac{C+tB}{1-t^3},
$$
which is <holomorphic> at $t=0$. Consequently \b[the actual set is $S=\{\zeta:\zeta^3=1,\ B+\zeta C\ne0\}$], with $\{1,\omega,\omega^2\}$ the generic set. The finite <residues> sum to zero, as follows also from $\sum_{\zeta^3=1}\zeta^{-1}=\sum_{\zeta^3=1}\zeta^{-2}=0$.
Conversely, suppose a <Fuchsian linear differential system> has its finite singular points among these three roots and has no singularity at infinity. Then $Q(z)=(1-z^3)A(z)$ extends to an <entire> matrix function, and ordinariness at infinity gives $Q(z)=O(z)$. By the <Cauchy estimate>, each entry has zero derivatives of order two or higher, so $Q(z)=B+zC$. The same argument applies to any subset $S$, with additional cancellations at the omitted roots. This proves the requested normal form.
There is a genuine convention limitation: the normal form is not true under intrinsic regular singularity alone in an unrestricted frame. For example, if $N\ne0$ and $N^2=0$, then $A(z)=N/(z-1)^2$ has <fundamental matrix> $I-N/(z-1)$. Its solutions have only power growth, so $1$ is intrinsically a <regular singular point>, and infinity is ordinary. Nevertheless its coefficient has a double pole and cannot equal $(B+zC)/(1-z^3)$. The simple-pole convention makes the preceding residue and normal-form assertions correct.
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