Meromorphic Jensen identity with a zero or pole at the centre

Fix a radius $r$ with no zero or pole of $f-a$ on its boundary, and write $m_0=n(0,a;f)$ and ${\nu }_0=n(0,\mathrm{\infty };f)$. Applying [F1] to $f-a$ gives its boundary Poisson mean, a negative sum over $a$-points, and a positive sum over poles; the central Green factor is $G_r(z,0)=\log (r\mathrm{/}\mathrm{\mid }z\mathrm{\mid })$.

Let $P$ be the pole set and $\mathrm{\Omega }=\mathbb{C}\setminus P$. By [F8] and [F10], $\mathrm{\Omega }$ is open and $f$ is holomorphic there; it is nonempty because a discrete pole set cannot equal $\mathbb{C}$. For any two points of $\mathrm{\Omega }$, take a bounded closed disc containing a polygonal path between them in its interior. This disc meets $P$ in finitely many points because $P$ is closed and discrete. Choose small disjoint discs around those finitely many poles, avoiding the path endpoints and containing no other poles; replacing portions of the polygonal path through these discs by arcs in the punctured discs gives a path in $\mathrm{\Omega }$. Thus $\mathrm{\Omega }$ is connected.

If $f$ has a pole at $0$ of order ${\nu }_0$, [F3] gives a Laurent expansion on a punctured disc and [F7] makes its first nonzero exponent $-{\nu }_0$; subtracting finite $a$ does not change that leading exponent.

If $f(0)$ is finite and not equal to $a$, then $f-a$ is nonzero at $0$, so its leading exponent is $k_a=0=m_0-{\nu }_0$.

If $f(0)=a$ and the zero of $f-a$ at $0$ had infinite order, [F6] would make $f-a$ vanish near $0$. By [F9] on the connected domain from step 1.2, it would then vanish on all of $\mathrm{\Omega }$; if $P$ is empty this makes $f$ constant, while if $P$ is nonempty it contradicts [F7] at each pole. Thus the order is finite, and [F6] gives $f-a=z^{m_0}h$ with $h(0)\mathrm{\ne }0$, so its leading exponent is $k_a=m_0$.

The three cases in steps 1.3, 1.4, and 2.1 show that $f(z)-a=c_a{z}^{k_a}(1+o(1))$ and $k_a=m_0-{\nu }_0$; hence $\log \mathrm{\mid }f(z)-a\mathrm{\mid }=\log \mathrm{\mid }{c}_a\mathrm{\mid }+k_a\log \mathrm{\mid }z\mathrm{\mid }+o(1)$.

Let $z\to 0$ in step 1.1 through nondivisor points. The boundary Poisson mean tends to $M_r\log \mathrm{\mid }f-a\mathrm{\mid }$; every noncentral Green term tends to $\log (r\mathrm{/}\mathrm{\mid }b\mathrm{\mid })$ or $\log (r\mathrm{/}\mathrm{\mid }p\mathrm{\mid })$; and the central contribution is $-k_a(\log r-\log \mathrm{\mid }z\mathrm{\mid })$. Comparing with step 3.1 and cancelling $k_a\log \mathrm{\mid }z\mathrm{\mid }$ yields $M_r\log \mathrm{\mid }f-a\mathrm{\mid }=\log \mathrm{\mid }{c}_a\mathrm{\mid }+k_a\log r+{\sum }_{0<\mathrm{\mid }b\mathrm{\mid }<r}{m}_b\log (r\mathrm{/}\mathrm{\mid }b\mathrm{\mid })-{\sum }_{0<\mathrm{\mid }p\mathrm{\mid }<r}{\nu }_p\log (r\mathrm{/}\mathrm{\mid }p\mathrm{\mid })$.

By [F2] and the finite divisor lists in [F4], integrating each counting step gives $N(r,a;f)=m_0\log r+{\sum }_{0<\mathrm{\mid }b\mathrm{\mid }<r}{m}_b\log (r\mathrm{/}\mathrm{\mid }b\mathrm{\mid })$ and $N(r,\mathrm{\infty };f)={\nu }_0\log r+{\sum }_{0<\mathrm{\mid }p\mathrm{\mid }<r}{\nu }_p\log (r\mathrm{/}\mathrm{\mid }p\mathrm{\mid })$. Since $k_a=m_0-{\nu }_0$, step 3.1 is exactly $\log \mathrm{\mid }{c}_a\mathrm{\mid }+N(r,a;f)-N(r,\mathrm{\infty };f)$.

For any radius meeting a divisor, the pointwise chordal identity gives $M_r\log \mathrm{\mid }f-a\mathrm{\mid }=m(r,\mathrm{\infty };f)+\frac{1}{2}\log (1+\mathrm{\mid }a{\mathrm{\mid }}^2)-m(r,a;f)$. By [F4]–[F5], this mean is finite and continuous in $r$; the two counting functions are continuous as well. Taking regular radii to the divisor radius in step 4.2 proves the same identity there by continuous radial limit.