Rational degree appears as logarithmic characteristic

Polynomial division gives $R(z)=z+1+2\mathrm{/}(z-1)$. The numerator equals $2$ at $z=1$, so this is the unique pole and it is simple; the numerator and denominator are coprime and their maximum degree is $2$. Also $R(0)=-1$, so $n(0,\mathrm{\infty };R)=0$; then $n(t,\mathrm{\infty };R)=0$ for $t<1$ and $n(t,\mathrm{\infty };R)=1$ for $t\ge 1$. By [F3, F4], $N(1,\mathrm{\infty };R)=0$ (the boundary pole has logarithmic weight $\log 1=0$), while for every $r>1$, $N(r,\mathrm{\infty };R)={\int }_1^{r}dt\mathrm{/}t=\log r$.

For $\mathrm{\mid }z\mathrm{\mid }=r\ge 4$, the decomposition gives $\mathrm{\mid }R(z)\mathrm{\mid }\ge r-1-2\mathrm{/}(r-1)\ge r\mathrm{/}2$ and $\mathrm{\mid }R(z)\mathrm{\mid }\le r+1+2\mathrm{/}(r-1)\le 2r$. Hence [F2] bounds the pointwise proximity by $\log r-\log 2\le \frac{1}{2}\log (1+\mathrm{\mid }R(z){\mathrm{\mid }}^2)\le \log r+\frac{1}{2}\log (4+r^{-2})=\log r+O(1)$ uniformly on the circle. Averaging gives $m(r,\mathrm{\infty };R)=\log r+O(1)$.

By [F1] and steps 1.1–1.2, $T(r,R)=2\log r+O(1)$. Also, [F1, F2] give $T(r,z)=\frac{1}{2}\log (1+r^2)=\log r+O(1)$ because $z$ has no poles and $\mathrm{\mid }z\mathrm{\mid }=r$ on the averaging circle. Since $R$ has degree two, [F5] applied to the identity map gives the same $T(r,R)=2T(r,z)+O(1)$; this agrees with the exact rational degree law [F6].