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For a nonconstant meromorphic function on the plane, the displayed estimate holds outside an exceptional set of radii of finite linear measure, for large $r$. The logarithmic derivative has only simple principal parts at zeros and poles; Poisson-Jensen formula estimates and control of increasing auxiliary growth functions produce the logarithmic error. It supplies the error term in the Nevanlinna second main theorem.

Let $f$ be a nonzero meromorphic function in a neighborhood of the closed disk $|z|\le R$, with no boundary zero or pole. List its interior zeros $a_j$ with multiplicities $m_j$ and its interior poles $b_k$ with orders $n_k$. Define the disk Green function and Poisson kernel byAt every interior point that is neither a zero nor a pole, the Poisson-Jensen formula isIn particular, at $z=0$ when $f(0)$ is finite and nonzero, this becomes Jensen's formula, with each zero contributing $-m_j\log (R/|a_j|)$ and each pole contributing $n_k\log (R/|b_k|)$.

Here is a proof that also explains the signs. The disk Blaschke factorshave a simple zero at $a$, no poles in the disk and boundary modulus one. Thereforeextends to a zero-free holomorphic function throughout the disk. Its logarithmic modulus is a harmonic function and agrees with $\log |f|$ on the boundary. The Poisson integral represents $\log |F(z)|$ by that boundary data. Since $\log |B_a|=-G_R(\cdot ,a)$, substitution gives the displayed Poisson-Jensen formula. Boundary zeros or poles can be handled by taking limiting radii: their logarithmic singularities are integrable. When zero is itself a zero or pole, factor out its signed power of $z$ before evaluating Jensen's formula there.

If $\varphi$ is identically zero there are no exceptions. Otherwise choose $x_0$ with $\varphi (x_0)>0$ and put $I_{+}=\{x\ge x_0:{\varphi }^{\prime }(x)>\varphi (x{)}^2\}$. Monotonicity implies $\varphi (x)>0$ thereafter. On this set, $1<{\varphi }^{\prime }/{\varphi }^2$, whenceAdding the finite initial interval $[0,x_0]$ provesThis derivative growth outside a finite-measure set controls exceptional radii in Nevanlinna estimates. For example let $T_s$ be the smooth spherical characteristic and $A(r)=r{T}_s^{\prime }(r)$ its increasing area-counting function. Apply the lemma to $T_s+1$ and $A+1$: outside the union of two finite-measure sets,Their logarithms are consequently $O(\log r+\log (T_s+1))$. Such derivative/area bounds in estimates from the Poisson-Jensen formula give the Nevanlinna logarithmic derivative lemma, $m(r,f^{\prime }/f)=O({\log }^{+}{T}_f+\log r)$ outside a finite-length set. The second main theorem uses that estimate as its error term. The elementary growth lemma controls that error and its exceptional set; it is not itself a substitute for the analytic estimates.

We use the truncated second main theorem stated in the preceding solution, together with the fact that a transcendental meromorphic function has $T_f(r)/\log r\to \infty$. Here is a justification of the latter fact. If there are infinitely many poles, fixing any arbitrarily large finite pole count gives $N(r;\infty )\ge A\log r-O_A(1)$. If there are only finitely many poles, multiply by a polynomial $p$ removing them to obtain a transcendental entire function $h=pf$. Its Taylor series has nonzero coefficients of arbitrarily high degrees $n$. Cauchy's coefficient estimate gives $\log M_h(r/2)\ge n\log r-O_n(1)$; the subharmonic function Poisson estimate gives $\log M_h(r/2)\le 3m(r,h)$. Since $T_h\le T_f+O(\log r)$, letting $n$ be arbitrarily large proves the claimed ratio limit. Thus the second-theorem error is $o(T_f)$ along large nonexceptional radii.

Let $N_{\mathrm{simple}}(r;a)$ count just the simple $a$-points. A multiple $a$-point of degree $m\ge 2$ contributes one to the truncated count and at least two to the full count; a simple point contributes one to each. Hence, with the same positive logarithmic weights for large radii,If each of the five values had only finitely many simple preimages, each simple counting function would be $O(\log r)$. Summing and using the first main theorem would give ${\sum }_{j=1}^5\overline{N}(r;a_j)\le (5/2)T_f+O(\log r)$. But the second theorem requires $3T_f\le \sum \overline{N}+o(T_f)$, a contradiction. This proves the stronger result five values force infinitely many simple preimages:In particular more than one simple zero is necessary, but more than one target with simple zeros is not necessary.

For the last distinction, take a nonsingular Weierstrass elliptic function scaled so that ${\wp }^{\prime 2}=4\wp (\wp -1)(\wp +1)$; it has branch values $-1,0,1,\infty$. These are the four totally ramified Weierstrass values: finite branch preimages have local degree two, and all poles are double. SetThe first four targets are the images of those branch values under the Möbius transformation. Their preimages remain double; at a lattice pole, for example, $f(z)\sim (z-z_{\ast }{)}^2$. The last target corresponds to $\wp =3$, where ${\wp }^{\prime 2}=96\ne 0$, so every such preimage is simple. The degree-two torus map attains that regular value, and periodicity produces infinitely many preimages in the plane. This nonconstant elliptic function is transcendental because it has infinitely many poles. Exactly one of the five selected finite targets therefore has simple preimages.