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Treat $F$ as a vector-valued solution; matrix solutions are handled columnwise. It is important to distinguish the coefficient convention from the indicial exponent. In a fixed Fuchsian frame, a finite singularity at $a$ meanswhere $B_a$ is holomorphic near $a$. The residue is $R_a$, equivalently the residue of the matrix-valued one-form $A(z)dz$. Such a Fuchsian linear differential system has a regular singular point: on sectors, its solutions have at most power and logarithmic growth. Intrinsically a regular singular system can be made Fuchsian by a meromorphic change of frame; a higher coefficient pole in some other frame does not itself prove irregularity. Residues here refer to the displayed fixed coefficient one-form.

At infinity use $w=1/z$ and $\tilde{F}(w)=F(1/w)$. The equation becomesA Fuchsian singularity at infinity means $\tilde{A}$ has at most a simple pole at $w=0$, and $R_{\infty }={\mathrm{Res}}_{w=0}\tilde{A}(w)$. The residue theorem applied entrywise on the Riemann sphere givesFor a direct calculation when all coefficient poles are simple, write $A(z)=P(z)+{\sum }_j{R}_j/(z-a_j)$ by rational partial fractions. Regularity at infinity in this frame requires $A(z)=O(1/z)$, so $P=0$. Transforming coordinates gives $R_{\infty }=-{\sum }_j{R}_j$. More generally the entrywise residue theorem remains valid even when a different meromorphic frame has higher-order poles.

The printed nonconstant-solution claim is false. A counterexample satisfying its spectral hypothesis isIts residue has eigenvalues $0,-1/3$, with zero of largest real part. The equations are $u^{\prime }=0$ and $v^{\prime }-v/(3z)=0$. If $v$ is analytic at zero, coefficient comparison gives $(n-1/3)v_n=0$ for every nonnegative integer $n$, so $v=0$. Hence every analytic solution is $(c,0{)}^T$, which is constant. This disproves nonconstancy, rather than silently replacing it with a weaker conclusion.

A useful correct conclusion is that a nonzero analytic solution exists under the stated hypotheses. Write $A=R/z+{\sum }_{j\ge 0}{B}_j{z}^j$, and let $m$ be the largest nonnegative integer with $-m\in \mathrm{spec}R$; such an $m$ exists because $0\in \mathrm{spec}R$. Choose a nonzero $h_0\in \ker (mI+R)$ and seekThe Frobenius method givesAll left-hand matrices are invertible by maximality of $m$. For large $n$ their inverse norm is $O(1/n)$ by a geometric-series inverse; enlarging the constant bounds every $n\ge 1$ by $C/n$. If $\Vert B_j\Vert \le M{r}^{-j}$ on a smaller disc, the scalar majorant with generating functionsatisfies $v^{\prime }=CM(1-t/r{)}^{-1}v$ and dominates the recurrence coefficientwise. It converges for $|t|<r$, proving convergence of $H$. Thus the analytic solution at a maximal integer Fuchsian exponent is nonzero, and is nonconstant if $m>0$. When $m=0$, nonconstancy requires additional information, as the counterexample shows.

There is also a sign problem in the printed exponent terminology. Substitution of $F=z^{\mu }H$ with $H(0)\ne 0$ into the actual equation givesThus the indicial exponents are eigenvalues of $-R$, not $R$. With the printed sign, an exponent of maximal real part admits a pure Frobenius solution, because the corresponding coefficient matrices are invertible at all subsequent positive integer steps. Lesser exponents can meet resonance.

The supplied example does not demonstrate such failure. Solving it directly givesEvery solution is analytic. The residue eigenvalue $-1$ corresponds to the true exponent $1$, and setting $c_1=0$ gives $F=z(-c_{2}z/2,c_2{)}^T$ with nonzero leading coefficient. Even with the literal printed factor $z^{-1}$, every displayed solution has the form $z^{-1}H$ with $H=zF$ analytic, though $H(0)=0$.

Nevertheless the last existence-of-a-counterexample request, read literally, can be fulfilled by the diagonal counterexample above. For its nonmaximal residue eigenvalue $\lambda =-1/3$, substitution of $F=z^{-1/3}H$ yieldsfor every integer $n\ge 0$. Neither diagonal entry, $n-1/3$ or $n-2/3$, vanishes, so $H=0$. Therefore there is no nonzero solution of that literal form, even without imposing $H(0)\ne 0$.

For the intended logarithmic resonance phenomenon with correctly signed exponents, take insteadIts equation gives $v=c_2/z$ and $u=c_1+c_2\mathrm{Log}z$ on a chosen branch. The exponent $-1$ would require $H(0)=(0,c_2{)}^T\ne 0$, but $z\mathrm{Log}z$ prevents $H=zF$ from being analytic when $c_2\ne 0$. The exponent zero has the constant solution $(1,0{)}^T$. This supplies an explicit resonance obstruction under the correct convention and completes the qualified resolution of the printed defects.

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 $a$,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 $\Vert A(z)\Vert \le C/|z-a|$ 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 isDifferentiating the matrix exponential verifies ${\Phi }^{\prime }=R\Phi /z$. A counterclockwise circuit replaces $\log z$ by $\log z+2\pi i$, soThis 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 isBeing 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 $a_1,\dots ,a_s$, subtract their principal parts: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 yieldsThus 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 $1,\omega ,{\omega }^2$, andA point $\zeta$ is removable exactly when $B+\zeta C=0$ as a matrix. At infinity,which is holomorphic at $t=0$. Consequently the actual set is $S=\{\zeta :{\zeta }^3=1,B+\zeta C\ne 0\}$, 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 $N\ne 0$ 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.