Corollary 1 [Picard-Borel]
For a nonconstant meromorphic function there are at most two values of \(a\) for which the counting function \(N(r, a)\) [or \(n(r, a)\)] is of lower order (class, type) than the characteristic \(T(r)\).
Theorem [deficiency sum]
If the function \(w(z)\) is meromorphic for \(z \neq \infty\), then the deficiency \(\delta(a)\) defined by
\[\delta(a) = \varliminf\limits_{r = \infty} \frac{m(r,a)}{T(r)} = 1 - \varlimsup\limits_{r = \infty} \frac{N(r,a)}{T(r)} \]
vanishes for all but at most a countable set of values \(a\). The sum of all deficiencies is at most equal to \(2\):
\[\sum \delta(a_{i}) \leq 2\]
Corollary 2 [Picard]
If the curve \(f(x, y) = 0\) is of genus \(p > 1\), then there is no pair of meromorphic functions \(x(t), y(t)\) such that \(f(x(t), y(t)) = 0\).
Therorem [P'olya]
Suppose that \(f(z), g(z)\) are integral functions and that \(\phi(р) = g\left(f(z)\right)\) has finite order. Then either \(f(z)\) is a polynomial or \(g(z)\) has zero order.
Suppose that f(z) and g(z) are transcendental, since otherwise there is nothing to prove. In this case it follows from (2.13) that
\[T(r,\phi) \geq \frac{1}{3}T(r^{n+1},g)\]
for all sufficiently large \(r\), when \(N\) is a fixed positive integer. Since \(\phi(z)\) has finite order \(k\), say, we deduce that, for all sufficiently large \(r\), we have by (2.13)
\[T(r^{n+1},g) < 3r^{k+1}\]
and setting \(\rho = r^{N+1}\) we deduce for all sufficiently large \(\rho\)
\[T(\rho,g) < 3 \rho^{(k+1)/(N+1)}\]
Since \(k\) is fixed and \(N\) can be chosen as large as we please, \(g(z)\) must have zero order. This proves Theorem 2.9. \(\square\)
Reference
[1] Nevanlinna R. Le théorème de Picard-Borel et la théorie des fonctions méromorphes[M]. Chelsea Publishing Company, Incorporated, 1929.