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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.