# MEROMORPHIC FUNCTIONS WITH SEVERAL ESSENTIAL SINGULARITIES

A. A. Kondratyuk

Lviv National University,

Universitetska street 1, 79000, Lviv, Ukraine

July 8, 2008

Key words:
Annulus, meromorphic function, Nevanlinna characteristic, index, value
distribution theory.
Mathematics Subject Classification (2000)
30D35
Preprint
Lviv National University,
Faculty of Mathematics and Mechanics

Abstract

A two-parameter characteristic of functions meromorphic on annuli is
introduced and an extension of the Nevanlinna value distribution theory for
such functions is proposed.

## Introduction

Meromorphic functions with $m+1$ possible essential singularities
$\\{c_{j}\\},$ $c_{j}\in\mathbb{C},$ $j=1,2,...,m,$ $c_{m+1}=\infty,$ are
considered. In other words, we consider meromorphic functions in a
$m$-punctured plane. The example of such a function can be given by the
composition $f\circ\mathcal{R}$ of a function $f,$ transcendental meromorphic
in $\mathbb{C},$ and a rational function $\mathcal{R}$ with $m+1$ distinct
poles in $\overline{\mathbb{C}}.$

We introduce a $m+1$-parameter characteristic of meromorphic functions in a
$m$-punctured plane and investigate the distribution of their values. The
introduced characteristic possesses the properties similar to these ones of
the classical Nevanlinna characteristic.

To begin consider meromorphic functions with two possible essential
singularities. Up to a linear fractional transformation we have $0$ and
$\infty.$ That is, we consider meromorphic functions in the plane punctured at
the origin. We will approach to both singularities by the annuli.
$A_{sr}=\\{z:\,s<|z|<r\\}.$ In other words,
$\mathbb{C}\setminus\\{0\\}=\bigcup\limits_{0<s<r<+\infty}A_{sr}.$

It seems that consideration of annuli $A_{\frac{1}{\tau}r},\,1\leq\tau,\,1\leq
r,$ will be more convenient.

Note, that there are essential differences between disks and annuli in the
topoligal sence which are reflected in the theory of meromorphic functions.
Firstly, the fundamental (Poincré) group of a disk is trivial, while for an
annulus, we have a group isomorphic to the additive group $\mathbb{Z}.$
Secondly, the group of automorphisms of the unit disk is rich. It consists of
the Möbius transformations. Therefore, the Poisson integral formula is an
invariant form of the Gauss mean theorem and the Poisson-Jensen formula is an
invariant form of Jensen’s formula. More over, all disks are conformally
equivalent. The situation with annuli is another. The group of automorphisms
of an annulus is poor. If $s\neq\frac{1}{r},$ it consists of rotations only.
Annuli $A_{1r}$ and $A_{1\rho}$ is conformally equivalent iff $r=\rho.$
Therefore, the theory of meromorphic functions on annuli is more complicated
that this one in disks. But many results for meromorphic functions in
$\mathbb{C}$ have their counterparts for these ones in a punctured plane
$\mathbb{C}^{*}.$ For example, the Picard theorem may be reformulated in the
form. ”For every non-constant meromorphic function in $\mathbb{C}^{*}$ there
is a linear fractional transformation $\omega(\mathbb{C}^{*})\subset
f(\mathbb{C}^{*}).$” If both, $\omega$ and $f,$ are identical maps, we have
the coincidence.

In order to introduce a two-parameter characteristic we use the notion of
index of a meromorphic function along a circle.

##  $\bf 1^{0}.$ Index of $f$ along a circle

######  Lemma 1.

Let $f$ be a function meromorphic on $\\{z:\,|z|=t\\},$ non identical zero.
Then

$\displaystyle{}\displaystyle\nu(t,f)=\frac{1}{\pi}\int\limits_{|z|=t}{\rm
Im}\left(\frac{f^{\prime}(z)}{f(z)}\,dz\right)$ (1)

is an integer.

###### Proof.

Denote $\gamma(\theta)=te^{i\theta},$ $0\leq\theta\leq 2\pi.$ If $f$ is
holomorphic on the circle $\\{z:\,|z|=t\\}$ without zeroes on this circle then
$\Gamma(\theta)=(f\circ\gamma)(\theta),$ $0\leq\theta\leq 2\pi,$ is a closed
path, and

$\displaystyle\displaystyle\frac{1}{2\pi
i}\int\limits_{|z|=t}\frac{f^{\prime}(z)}{f(z)}\,dz=\frac{1}{2\pi
i}\int\limits_{0}^{2\pi}\frac{f^{\prime}(\gamma(\theta)))}{f(\gamma(\theta)))}\,\gamma^{\prime}(\theta)\,d\theta=$
$\displaystyle\displaystyle=\frac{1}{2\pi
i}\int\limits_{0}^{2\pi}\frac{\Gamma^{\prime}(\theta)}{\Gamma(\theta)}\,d\theta=\frac{1}{2\pi
i}\int\limits_{\Gamma}\frac{d\zeta}{\zeta}={\rm Ind}_{\Gamma}(0).$

Taking the real parts of both sides we have

$\displaystyle\displaystyle\nu(t,f)=2\,{\rm Ind}_{\Gamma}(0).$

As index of $\zeta=0$ with respect to $\Gamma$ is an integer we obtain the
needful conclusion in this case.

If $f$ has a unique simple zero $a=te^{i\alpha},$ then $f(z)=g(z)(z-a),$
$\displaystyle\frac{f^{\prime}}{f}=\frac{g^{\prime}}{g}+\frac{1}{z-a},$ and

$\displaystyle\displaystyle{\rm Im}\left(\frac{1}{z-a}\,dz\right)={\rm
Im}\left(\frac{ite^{i\theta}d\theta}{t(e^{i\theta}-e^{i\alpha})}\right)=$
$\displaystyle\displaystyle={\rm
Re}\frac{1}{1-e^{i(\alpha-\theta)}}\,d\theta=\frac{1}{2}\,d\theta,$

because the transformation $\displaystyle w=\frac{1}{1-z}$ maps the unit
circle on the straight line ${\rm Re}\,w=1/2.$ Hence, in this case

$\displaystyle\displaystyle\nu(t,f)=\nu(t,g)+1.$

If $f$ has a simple pole on the considered circle, then we obtain the similar
equality with ”-” instead of ”+”. The final conclusion in the general case now
follows by induction.

The value $\nu(t,f)$ is said to be index of $f$ along the circle
$\\{z:\,|z|=t\\}.$

If a branch of $\log f$ may by determined on the circle, then $d\log
f=\frac{f^{\prime}}{f}\,dz.$ Thus, ${\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right)=d\arg f(z)$ and $\nu(t,f)$ is the
increment of $\frac{1}{\pi}\,\arg f$ along the circle.

∎

##  $\bf 2^{0}.$ Version 1 of Jensen’s theorem

######  Lemma 2.

Let $f$ be a function meromorphic on the closure of the annulus
$A_{sr}=\\{z:\,s<|z|<r\\},$ and non identical zero. Then

$\displaystyle{}\displaystyle\int\limits_{s}^{r}\frac{\nu(t,f)}{t}\,dt=\frac{1}{\pi}\int\limits_{0}^{2\pi}\log|f(re^{i\theta})|\,d\theta-\frac{1}{\pi}\int\limits_{0}^{2\pi}\log|f(se^{i\theta})|\,d\theta.$
(2)

###### Proof.

Assume that there are neither zeroes nor poles of $f$ on the interval
$\\{z=te^{i\theta},\,s\leq t\leq r\\}.$ Then a branch of $\log f$ can by
determined on this interval, and we have

$\displaystyle{}\displaystyle\log f(re^{i\theta})-\log
f(se^{i\theta})=\int\limits_{s}^{r}\frac{f^{\prime}(z)}{f(z)}\,dz.$

This implies

$\displaystyle{}\displaystyle\log|f(re^{i\theta})|-\log|f(se^{i\theta})|=\int\limits_{s}^{r}{\rm
Re}\left(\frac{f^{\prime}(te^{i\theta})}{f(te^{i\theta})}\,e^{i\theta}\right)\,dt.$

The last equality is valid for all $\theta$ from $[0,2\pi]$ except for a
finite number of $\theta.$ The integration over $\theta$ yields

$\displaystyle\displaystyle\frac{1}{\pi}\int\limits_{0}^{2\pi}\log|f(re^{i\theta})|\,d\theta-\frac{1}{\pi}\int\limits_{0}^{2\pi}\log|f(se^{i\theta})|\,d\theta=$
$\displaystyle\displaystyle=\frac{1}{\pi}\int\limits_{0}^{2\pi}d\theta\int\limits_{s}^{r}\frac{1}{t}{\rm
Im}\left(\frac{f^{\prime}}{f}\,ite^{i\theta}\right)dt.$

Using the Fubini theorem and changing the order of integration we obtain
(LABEL:eq2). ∎

##  $\bf 3^{0}.$ Version 2 of Jensen’s theorem

Suppose $s\leq 1\leq r$ and $f$ meromorphic on the closure of the annulus
$A_{sr}$ and non identical zero. Put in (LABEL:eq2) $r=1,$ $\tau=\frac{1}{s}$
and $t=\frac{1}{u}.$ Then

$\displaystyle\displaystyle-\int\limits_{1/s}^{1}\frac{\nu(\frac{1}{u},f)}{u}\,du=\int\limits_{1}^{\tau}\frac{\nu(\frac{1}{u},f)}{u}\,du=$
$\displaystyle\displaystyle=\frac{1}{\pi}\int\limits_{0}^{2\pi}\log|f(e^{i\theta})|\,d\theta-\frac{1}{\pi}\int\limits_{0}^{2\pi}\log\left|f\left(\frac{1}{\tau}\,e^{i\theta}\right)\right|\,d\theta.$
(3)

Putting now in (LABEL:eq2) $s=1$ and subtracting (LABEL:eq10) we have

$\displaystyle\displaystyle\int\limits_{1}^{r}\frac{\nu(t,f)}{t}\,dt-\int\limits_{1}^{\tau}\frac{\nu(\frac{1}{t},f)}{t}\,dt=\frac{1}{\pi}\int\limits_{0}^{2\pi}\log|f(re^{i\theta})|\,d\theta+$
$\displaystyle\displaystyle+\frac{1}{\pi}\int\limits_{0}^{2\pi}\log\left|f\left(\frac{1}{\tau}\,e^{i\theta}\right)\right|\,d\theta-\frac{2}{\pi}\int\limits_{0}^{2\pi}\log\left|f\left(e^{i\theta}\right)\right|\,d\theta.$
(4)

In order to obtain Version 2 of Jensen’s theorem we are going to connect the
notion of index $\nu(t,f)$ with the counting functions of zeroes and poles of
$f.$

Let $\mathbb{T}$ be the unit circle and $n(s,r;f)$ be the number of poles of
$f$ in $A_{sr}.$ It follows from the argument principle that

$\displaystyle\displaystyle\nu(t,f)-\nu(1,f)=$
$\displaystyle\displaystyle=2\,n(1,t;\frac{1}{f})+n(\mathbb{T},\frac{1}{f})-2\,n(1,t;f)-n(\mathbb{T},f),$
(5)

and

$\displaystyle\displaystyle\nu(1,f)-\nu(\frac{1}{t},f)=$
$\displaystyle\displaystyle=2\,n(\frac{1}{t},1;\frac{1}{f})+n(\mathbb{T},\frac{1}{f})-2\,n(\frac{1}{t},1;f)-n(\mathbb{T},f)$
(6)

under the assumption that neither zeroes nor poles of $f$ lie on the circles
$\\{z:\,|z|=t\\}$ and $\\{z:\,z=1/t\\}.$

Define

$\displaystyle\displaystyle
N(\tau,r;f)=\int\limits_{1}^{\tau}\frac{n(\frac{1}{t},1;f)}{t}\,dt+\int\limits_{1}^{r}\frac{n(1,t;f)}{t}\,dt+n(\mathbb{T},f)\log\sqrt{\tau
r}.$

Relations (LABEL:eq11) – (LABEL:eq13) yield the following lemma.

######  Lemma 3.

Let $f$ be a meromorphic function on the closure of $A_{\frac{1}{\tau}r}$ and
non identical zero. Then

$\displaystyle\displaystyle
N(\tau,r;f)-N(\tau,r;f)=\frac{1}{2\pi}\int\limits_{0}^{2\pi}\log\left|f\left(\frac{e^{i\theta}}{\tau}\right)\right|\,d\theta+$
$\displaystyle\displaystyle+\frac{1}{2\pi}\int\limits_{0}^{2\pi}\log|f\left(r{e^{i\theta}}\right)|\,d\theta-\frac{1}{\pi}\int\limits_{0}^{2\pi}\log|f\left({e^{i\theta}}\right)|\,d\theta+$
(7) $\displaystyle\displaystyle+\nu(1,f)\log\sqrt{\frac{\tau}{r}}.$

##  $\bf 4^{0}.$ Characteristic of $f$

Now we are able to introduce a two-parameter characteristic of meromorphic
functions in an annulus which possesses the properties like to its classical
counterpart. We follow the Cartan idea. Applying (LABEL:eq15) to the function
$f(z)-e^{i\varphi},$ and integrating over $\varphi$ on $[0,2\pi]$ we obtain

$\displaystyle\displaystyle\frac{1}{2\pi}\int\limits_{0}^{2\pi}N\left(\tau,r;\frac{1}{f-e^{i\varphi}}\right)\,d\varphi-N(\tau,r;f)=$
$\displaystyle\displaystyle=\frac{1}{2\pi}\int\limits_{0}^{2\pi}\log^{+}\left|f\left(\frac{e^{i\theta}}{\tau}\right)\right|\,d\theta+\frac{1}{2\pi}\int\limits_{0}^{2\pi}\log^{+}|f\left(r{e^{i\theta}}\right)|\,d\theta-$
(8)
$\displaystyle\displaystyle-\frac{1}{\pi}\int\limits_{0}^{2\pi}\log^{+}|f\left({e^{i\theta}}\right)|\,d\theta+\frac{1}{4\pi}\int\limits_{0}^{2\pi}\nu(1,f-e^{i\varphi})\,d\varphi\log{\frac{\tau}{r}},$

of course, if the last integral exists.

We will show that and evaluate the integral.

Let $f$ be a meromorphic function on $\mathbb{T}.$

Denote $E_{f}^{+}=\left\\{\mathbb{T}\ni z:\,|f(z)|>1\right\\},$
$E_{f}^{0}=\left\\{\mathbb{T}\ni z:\,|f(z)|=1\right\\}$ and
$E_{f}^{-}=\left\\{\mathbb{T}\ni z:\,|f(z)|<1\right\\}.$

######  Lemma 4.

Let $f$ be a meromorphic function on the unit circle $\mathbb{T},$
$f(z)\not\equiv 0.$ Then

$\displaystyle\displaystyle\frac{1}{4\pi}\int\limits_{0}^{2\pi}\nu(1,f-e^{i\varphi})\,d\varphi=\frac{1}{2\pi}\int\limits_{E_{f}^{+}}{\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right)+\frac{1}{4\pi}\int\limits_{E_{f}^{0}}{\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right).$

The proof of Lemma 4 needs some auxiliary results.

######  Lemma 5.

For each function $f$ meromorphic on $\\{z:\,|z|=t\\},$ $f\not\equiv 0,$ and
each $\zeta\in\mathbb{C}$ the relation

$\displaystyle\displaystyle\nu(t,f-\zeta)=\nu(t,f)-\frac{1}{\pi}\int\limits_{|z|=t}{\rm
Im}\left(\frac{\zeta f^{\prime}}{f(\zeta-f)}\,dz\right)$ (9)

holds.

###### Proof.

Relation (LABEL:eq18) follows immediately from (1) and the idenity

$\displaystyle\displaystyle\frac{f^{\prime}}{f-\zeta}=\frac{f^{\prime}}{f}\left(1-\frac{\zeta}{\zeta-f}\right).$

∎

######  Lemma 6.

Let $f$ be a meromorphic function on the unit circle $\mathbb{T},$
$f(z)\not\equiv 0.$ Then

$\displaystyle\displaystyle\frac{1}{4\pi}\iint\limits_{\mathbb{T}\times\mathbb{T}}{\rm
Re}\left(\frac{f^{\prime}}{f(\zeta-f)}\,dz\,d\zeta\right)=$
$\displaystyle\displaystyle=-\frac{1}{2\pi}\int\limits_{E_{f}^{-}}{\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right)-\frac{1}{4\pi}\int\limits_{E_{f}^{0}}{\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right).$ (10)

###### Proof.

If $z\in\mathbb{T}\setminus E_{f}^{0},$ i.e. $|f(z)|\neq 1,$ we have

$\displaystyle\displaystyle\frac{1}{2\pi
i}\int\limits_{\mathbb{T}}\frac{d\zeta}{\zeta-f(z)}=\begin{cases}0,&{\text{i}f}\
\ |f(z)|>1,\\\ 1,&{\text{i}f}\ \ |f(z)|<1.\end{cases}$

This is true for each $z$ from $\mathbb{T}\setminus E_{f}^{0}$ except for a
finite number of $z.$ Multiplying the last equality by
$i\frac{f^{\prime}dz}{f}$ and taking the real parts we obtain

$\displaystyle\displaystyle\frac{1}{2\pi
i}\int\limits_{\mathbb{T}_{\zeta}}{\rm
Re}\left(\frac{f^{\prime}}{f(\zeta-f)}\,d\zeta\,dz\right)=\begin{cases}0,&{\text{i}f}\
\ z\in E_{f}^{+},\\\ -{\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right),&{\text{i}f}\ \ z\in
E_{f}^{-}.\end{cases}$

for each $z$ from $\mathbb{T}\setminus E_{f}^{0}$ except for a finite number
of $z.$ Therefore,

$\displaystyle\displaystyle\frac{1}{4\pi^{2}}\int\limits_{\mathbb{T}_{\zeta}}\int\limits_{\mathbb{T}_{z}\setminus
E_{f}^{0}}{\rm Re}\left(\frac{f^{\prime}}{f(\zeta-f)}\,dz\,d\zeta\right)=$
(11)
$\displaystyle\displaystyle=\frac{1}{2\pi}\int\limits_{\mathbb{T}_{z}\setminus
E_{f}^{0}}\frac{1}{2\pi}\int\limits_{\mathbb{T}_{\zeta}}{\rm
Re}\left(\frac{f^{\prime}}{f(\zeta-f)}\,dz\,d\zeta\right)=-\frac{1}{2\pi}\int\limits_{E_{f}^{-}}{\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right).$

If $|f(z)|=1,$ i.e. $z\in E_{f}^{0},$ then $f(z)=e^{i\alpha(\theta)}.$ We will
use the identity

$\displaystyle\displaystyle{\rm
Re}\left(\frac{f^{\prime}}{f(\zeta-f)}\,dz\,d\zeta\right)={\rm
Re}\left(\frac{f^{\prime}}{f}\,dz\right){\rm
Re}\left(\frac{d\zeta}{\zeta-f}\right)-$ $\displaystyle\displaystyle-{\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right){\rm
Im}\left(\frac{d\zeta}{\zeta-f}\right)$ (12)

If $E_{f}^{0}$ contains an open arc then $d\log|f(z)|={\rm
Re}\left(\frac{f^{\prime}}{f}\,dz\right)=0$ on this arc, because
$\log|f(z)|=0.$ By continuity

$\displaystyle\displaystyle{\rm Re}\left(\frac{f^{\prime}}{f}\,dz\right)=0$

on the closure of the arc. Thus, the intersection of $E_{f}^{0}$ with the set
$\left\\{z:\,{\rm Re}\left(\frac{f^{\prime}}{f}\,dz\right)\neq 0\right\\}$ is
empty or consists of isolated points. Besides this,

$\displaystyle\displaystyle{\rm Re}\frac{d\zeta}{\zeta-f}={\rm
Im}\frac{ie^{i\varphi}}{e^{i\varphi}-e^{i\alpha(\theta)}}={\rm
Re}\frac{1}{1-e^{i(\alpha(\theta)-\varphi)}}=\frac{1}{2}.$

Hence, relation (LABEL:eq24) implies

$\displaystyle\displaystyle\frac{1}{4\pi^{2}}\int\limits_{\mathbb{T}\times
E_{f}^{0}}{\rm
Re}\left(\frac{f^{\prime}}{f(\zeta-f)}\,dz\,d\zeta\right)=-\frac{1}{4\pi}\int\limits_{E_{f}^{0}}{\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right).$ (13)

Relations (LABEL:eq23) and (LABEL:eq27) yield (LABEL:eq20). ∎

###### Proof of Lemma 4.

Applying Lemma 5 we have

$\displaystyle\displaystyle\frac{1}{4\pi}\int\limits_{0}^{2\pi}\nu(1,f-e^{i\varphi})\,d\varphi=\frac{1}{2}\nu(1,f)-$
$\displaystyle\displaystyle-\frac{1}{4\pi^{2}}\int\limits_{0}^{2\pi}\int\limits_{\mathbb{T}}{\rm
Im}\left(\frac{f^{\prime}e^{i\varphi}dz}{f(e^{i\varphi}-f)}\right)\,d\varphi.$
(14)

As

$\displaystyle\displaystyle
d\varphi=\frac{-id\zeta}{\zeta},\qquad\zeta=e^{i\varphi},$

then

$\displaystyle\displaystyle-\frac{1}{4\pi^{2}}\int\limits_{0}^{2\pi}\int\limits_{\mathbb{T}}{\rm
Im}\left(\frac{\zeta f^{\prime}}{f(\zeta-f)}\,dz\,d\varphi\right)=$
$\displaystyle\displaystyle=\frac{1}{4\pi^{2}}\iint\limits_{\mathbb{T}\times\mathbb{T}}{\rm
Re}\left(\frac{f^{\prime}}{f(\zeta-f)}\,dz\,d\zeta\right).$

Applying Lemma 6, which evaluates the last integral, and (1) we obtain the
conclusion of Lemma 4 from (LABEL:eq28). ∎

Denote

$\displaystyle\displaystyle
m(\tau,r;f)=m\left(\frac{1}{\tau},f\right)+m(r,f)-2\,m(1,f),$

where $m(r,f)$ is usual Nevanlinna’s notation.

######  Definition 1.

Let $f$ be a meromorphic function on the closure of $A_{\frac{1}{\tau}r},$
$f\not\equiv 0.$ The function

$\displaystyle\displaystyle T(\tau,r;f)=N(\tau,r;f)+m(\tau,r;f)+$
$\displaystyle\displaystyle+c_{f}\log\frac{\tau}{r}\quad\tau\geq 1,\quad r\geq
1,.$ (15)

where

$\displaystyle\displaystyle c_{f}=\frac{1}{2\pi}\int\limits_{E_{f}^{+}}{\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right)+\frac{1}{4\pi}\int\limits_{E_{f}^{0}}{\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right)$ (16)

is called the characteristic of $f.$

A strange example. Let $m\in\mathbb{N}.$ Then

$\displaystyle\displaystyle
T(\tau,r;z^{m})=T(\tau,r;z^{-m})=m\,\frac{\log\tau+\log r}{2}.$

Note, that the introduced characteristic fits not only for meromorphic
functions in the punctured plane but in arbitrary annuli including punctured
disks.

######  Theorem 1.

Let $f,$ $f\not\equiv 0,$ be a meromorphic function on the annulus
$A_{s_{0}r_{0}},$ $s_{0}<1<r_{0}.$ Then the function $T(\tau,r;f)$ is non-
negative, continuous, non-decreasing and convex with respect to logarithm of
each variable for $1\leq\tau<1/s_{0},$ $1\leq r<r_{0},$

$\displaystyle\displaystyle T(1,1;f)=0\ \ {\mbox{a}nd}\ \
T(\tau,r;\frac{1}{f})=T(\tau,r;f).$

If $A_{s_{0}r_{0}}=\mathbb{C}\setminus\\{0\\},$ the function $f$ has the
meromorphic continuation at the origin and $T(r,f)$ is its Nevanlinna
characteristic, then

$\displaystyle\displaystyle T(r,f)-2\,T(1,f)\leq T(r,r;f)\leq T(r,f),\quad
r\geq 1.$ (17)

###### Proof.

The indicated properties follow immediately from (LABEL:eq16), Lemma 4 and
Lemma 3. Moreover, relation (LABEL:eq16) implies

$\displaystyle\displaystyle\tau\,\frac{\partial
T}{\partial\tau}=\frac{1}{2\pi}\int\limits_{0}^{2\pi}n\left(\frac{1}{\tau},1;\frac{1}{f-e^{i\varphi}}\right)\,d\varphi+\frac{1}{4\pi}\int\limits_{0}^{2\pi}n\left(\mathbb{T},\frac{1}{f-e^{i\varphi}}\right)\,d\varphi$

and

$\displaystyle\displaystyle r\,\frac{\partial T}{\partial
r}=\frac{1}{2\pi}\int\limits_{0}^{2\pi}n\left(1,r;\frac{1}{f-e^{i\varphi}}\right)\,d\varphi+\frac{1}{4\pi}\int\limits_{0}^{2\pi}n\left(\mathbb{T},\frac{1}{f-e^{i\varphi}}\right)\,d\varphi$

at the points of continuity of $n.$ Thus,

$\displaystyle\displaystyle\tau\,\frac{\partial
T}{\partial\tau}+r\,\frac{\partial T}{\partial
r}=\frac{1}{2\pi}\int\limits_{0}^{2\pi}n\left(\frac{1}{\tau},r;\frac{1}{f-e^{i\varphi}}\right)\,d\varphi.$

It seems, that the last identity with some boundary condition can be a
definition of $T(\tau,r;f).$

Verify also the identity $T(\tau,r;1/f)=T(\tau,r;f).$ It follows from Version
2 of Jensen’s theorem (see (LABEL:eq15)) and the relation

$\displaystyle\displaystyle\frac{1}{2}\,\nu(1,f)=c_{f}-c_{\frac{1}{f}}.$ (18)

Relation (LABEL:eq35) is in [1]. ∎

##  $\bf 5^{0}.$ First Fundamental Theorem (FFT)

FFT. Let $f$ be a non-constant meromorphic function in
$\mathbb{C}\setminus\\{0\\}.$ Then for each $a\in\mathbb{C}$

$\displaystyle\displaystyle
N\left(\tau,r;\frac{1}{f-a}\right)+m\left(\tau,r;\frac{1}{f-a}\right)=$
$\displaystyle\displaystyle=T(\tau,r;f)+\varepsilon_{1}(\tau,r,a)+\varepsilon_{2}(a)\log\frac{\tau}{r},\quad\tau\geq
1,\quad r\geq 1$ (19)

where

$\displaystyle\displaystyle\left|\varepsilon_{1}(\tau,r,a)\right|\leq
4\,\log^{+}|a|+4\log 2,$

and

$\displaystyle\displaystyle\left|\varepsilon_{2}(a)\right|\leq C,\qquad C\
{\mbox{is a constant.}}$ (20)

###### Proof.

As in classical Nevanlinna theory we apply Version 2 of Jensen’s theorem (see
(LABEL:eq15)) for $f(z)-a.$ Then

$\displaystyle\displaystyle
N\left(\tau,r;\frac{1}{f-a}\right)+m\left(\tau,r;\frac{1}{f-a}\right)=N\left(\tau,r,f\right)+$
$\displaystyle\displaystyle+m\left(\tau,r;f-a\right)+\frac{1}{2}\,\nu(1,f-a)\log\frac{\tau}{r}=$
$\displaystyle\displaystyle=T\left(\tau,r;f\right)+m\left(\tau,r;f-a\right)-m\left(\tau,r;f\right)+$
$\displaystyle\displaystyle+\left(\frac{1}{2}\,\nu(1,f-a)-c_{f}\right)\log\frac{\tau}{r}.$

Set

$\displaystyle\displaystyle\varepsilon_{1}(\tau,r;a)=m(\tau,r;f-a)-m(\tau,r;f)$

and

$\displaystyle\displaystyle\varepsilon_{2}(a)=\frac{1}{2}\,\nu(1,f-a)-c_{f}.$

We have, as usually

$\displaystyle\displaystyle\left|\varepsilon_{1}(\tau,r,a)\right|\leq
4\log^{+}|a|+4\log 2.$

Applying (LABEL:eq18) for $\zeta=a$ we obtain

$\displaystyle\displaystyle\varepsilon_{2}(a)=\frac{1}{2\pi}\int\limits_{E_{f}^{-}}{\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right)+\frac{1}{4\pi}\int\limits_{E_{f}^{0}}{\rm
Im}\left(\frac{f^{\prime}}{f}\,dz\right)-$
$\displaystyle\displaystyle-\frac{1}{2\pi}\int\limits_{\mathbb{T}}{\rm
Im}\left(\frac{af^{\prime}}{f(a-f)}\,dz\right).$

Since the last integral is bounded as a function of $a$ then (LABEL:eq43) is
valid. ∎

## References

  * [1] Andriy Kondratyuk and and Ilpo Laine, Meromorphic functions in multiply connected domains, Fourier series methods in complex analysis (Mekrijärvi, 2005). Univ. Joensuu Dept. Math. Rep. Ser. No 10 (2006), pp. 9–111.