Padé matching can be singular or create a Froissart doublet

For $f(z)=1$ and type $(1,1)$, the sole equation is $q_1{c}_1=-c_2$, namely $0q_1=0$. It is consistent with rank zero, so [F1] returns $\mathtt{\text{defective-nonunique}}$. If one arbitrarily selects $q_1=-1$, the convolution formula gives $p(z)=1-z$ and $q(z)=1-z$; after canceling their common factor the rational function is the constant $1$, so the apparent zero of the unreduced denominator is not a pole.

For $g(z)=z^2$ and type $(1,1)$, the sole equation is $q_1{c}_1=-c_2$, namely $0q_1=-1$. Thus [F1] returns $\mathtt{\text{full-order-inconsistent}}$: no normalized denominator attains the full matching order.

For $f_{\epsilon }(z)=1+\epsilon z+z^2$ and type $(1,1)$, the equation is $\epsilon q_1=-1$, so its unique solution is $q_1=-1\mathrm{/}\epsilon$. The numerator coefficients are $p_0=1$ and $p_1=\epsilon +q_1=-(1-{\epsilon }^2)\mathrm{/}\epsilon$. Thus $r_{11}(z)=\frac{1-(1-{\epsilon }^2)z\mathrm{/}\epsilon }{1-z\mathrm{/}\epsilon }\mathrm{.}$ Its denominator has a simple zero at $z=\epsilon$ while the numerator there equals ${\epsilon }^2\mathrm{\ne }0$, so this is a genuine pole. The numerator zero is $z=\epsilon \mathrm{/}(1-{\epsilon }^2)$, and its distance from the pole is ${\epsilon }^3\mathrm{/}(1-{\epsilon }^2)$. The residue at the pole is ${\epsilon }^2\mathrm{/}(-1\mathrm{/}\epsilon )=-{\epsilon }^3$. Hence the exact coefficient-matching rational function has a pole-zero pair with arbitrarily small separation and residue as $\epsilon \downarrow 0$, although the target $f_{\epsilon }$ is a polynomial.

For $f_{\epsilon }$ the system is solved, so [F2] gives only the stated local Taylor order at the origin. Step 1.3 shows that this does not prevent a nearby pole-zero pair, while step 1.1 shows that a singular system can leave denominator coefficients nonunique. Neither case supplies analytic continuation or a target pole-location guarantee.