§14.2 Properties of $𝑇 (𝑟, 𝑓)$ and the First Fundamental Theorem

14.2 Properties of $𝑇 (𝑟, 𝑓)$ and the First Fundamental Theorem

Proposition 14.2.1.

If $𝑓_{𝑘}$ ( $𝑘 = 1, \dots, 𝑛$ ) are meromorphic in $𝐷 (0, 𝑅)$ , then for $0 < 𝑟 < 𝑅$ ,

$𝑁 (𝑟, \prod_{𝑘 = 1}^{𝑛} 𝑓_{𝑘}) \leq \sum_{𝑘 = 1}^{𝑛} 𝑁 (𝑟, 𝑓_{𝑘})$ .
$𝑁 (𝑟, \sum_{𝑘 = 1}^{𝑛} 𝑓_{𝑘}) \leq \sum_{𝑘 = 1}^{𝑛} 𝑁 (𝑟, 𝑓_{𝑘})$ (subadditivity).
$𝑚 (𝑟, \prod_{𝑘 = 1}^{𝑛} 𝑓_{𝑘}) \leq \sum_{𝑘 = 1}^{𝑛} 𝑚 (𝑟, 𝑓_{𝑘})$ .
$𝑚 (𝑟, \sum_{𝑘 = 1}^{𝑛} 𝑓_{𝑘}) \leq log 𝑛 + \sum_{𝑘 = 1}^{𝑛} 𝑚 (𝑟, 𝑓_{𝑘})$ .
$𝑇 (𝑟, \prod_{𝑘 = 1}^{𝑛} 𝑓_{𝑘}) \leq \sum_{𝑘 = 1}^{𝑛} 𝑇 (𝑟, 𝑓_{𝑘})$ .
$𝑇 (𝑟, \sum_{𝑘 = 1}^{𝑛} 𝑓_{𝑘}) \leq log 𝑛 + \sum_{𝑘 = 1}^{𝑛} 𝑇 (𝑟, 𝑓_{𝑘})$ .

Proof.

Let $𝑧$ be an arbitrary point in the disk. If $𝑚_{𝑘}$ is the pole order of $𝑓_{𝑘}$ at $𝑧$ (0 if it is not a pole), then the pole order at $𝑧$ of $\sum_{𝑘} 𝑓_{𝑘}$ does not exceed ${max}_{𝑘} 𝑚_{𝑘}$ . It is hence clear that $𝑛 (𝑟, \sum_{𝑘} 𝑓_{𝑘}) \leq \sum_{𝑘} 𝑛 (𝑟, 𝑓_{𝑘})$ .

Moreover, if ${𝑏_{𝑘}^{(𝑗)}}_{𝑗}$ are the poles of $𝑓_{𝑘}$ in the disk counted according to orders, and ${𝑏^{(𝑗)}}_{𝑗}$ are the poles of $\sum_{𝑘} 𝑓_{𝑘}$ , then

\sum_{𝑘 = 1}^{𝑛} \sum_{𝑗} log | \frac{𝑟}{𝑏_{𝑘}^{(𝑗)}} | \geq \sum_{𝑗} log | \frac{𝑟}{𝑏^{(𝑗)}} | .

Recognizing that

\begin{matrix} 𝑁 (𝑟, \sum_{𝑘} 𝑓_{𝑘}) = \sum_{𝑗} log | \frac{𝑟}{𝑏^{(𝑗)}} | + 𝑛 (0, \sum_{𝑘} 𝑓_{𝑘}) log 𝑟 \\ \sum_{𝑘 = 1}^{𝑛} 𝑁 (𝑟, 𝑓_{𝑘}) = \sum_{𝑘 = 1}^{𝑛} \sum_{𝑗} log | \frac{𝑟}{𝑏_{𝑘}^{(𝑗)}} | + \sum_{𝑘} 𝑛 (0, 𝑓_{𝑘}) log 𝑟, \end{matrix}

Part 2 $Part 2 Part 2$ then follows. Since the pole order at $𝑧$ of $\prod_{𝑘} 𝑓_{𝑘}$ does not exceed $\sum_{𝑘} 𝑚_{𝑘}$ , it also follows that $𝑛 (𝑟, \prod_{𝑘} 𝑓_{𝑘}) \leq \sum_{𝑘} 𝑛 (𝑟, 𝑓_{𝑘})$ . The same logic can be applied henceforth to prove Part 1 $Part 1 Part 1$ .

Part 3 $Part 3 Part 3$ and Part 4 $Part 4 Part 4$ are clear from Proposition 14.1.2 $Proposition 14.1.2 Proposition 14.1.2$ . Part 5 $Part 5 Part 5$ and Part 6 $Part 6 Part 6$ follow the previous inequalities. ⁠ $□$

Theorem 14.2.2.

Let $𝑓 : 𝐷 (0, 𝑅) \to \hat{ℂ}$ be meromorphic such that $𝑓 (𝑧) = 𝑐 𝑧^{𝑘} + 𝒪︀ (𝑧^{𝑘 + 1})$ . For $0 < 𝑟 < 𝑅$ ,

\begin{aligned} 𝑇 (𝑟, 𝑓) & = \frac{1}{2 π} \int_{0}^{2 π} 𝑁 (𝑟, e^{i 𝜃}, 𝑓) d 𝜃 + \frac{1}{2 π} \int_{0}^{2 π} log | 𝑐_{𝜃} | d 𝜃 \\ = \frac{1}{2 π} \int_{0}^{2 π} 𝑁 (𝑟, e^{i 𝜃}, 𝑓) d 𝜃 + {\begin{cases} log | 𝑐 | & if 𝑘 < 0 \\ {log}^{+} | 𝑐 | & if 𝑘 = 0 \\ 0 & if 𝑘 > 0 \end{cases}, \end{aligned}

where

𝑐_{𝜃}

is the first nonzero coefficient of the innermost Laurent expansion of

𝑓 - e^{i 𝜃}

at the origin.

Proof.

For $𝜃 \in [0, 2 π]$ , from (14.1.2) $(14.1.2) (14.1.2)$ on $𝑓 - e^{i 𝜃}$ (let $𝑐_{𝜃}$ be the first nonzero coefficient of the innermost Laurent series) we obtain

log | 𝑐_{𝜃} | = 𝑁 (𝑟, 𝑓) - 𝑁 (𝑟, e^{i 𝜃}, 𝑓) + \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜙} - e^{i 𝜃}) | d 𝜙 .

Therefore, by integrating in $𝜃$ ,

\begin{aligned} \frac{1}{2 π} \int_{0}^{2 π} log | 𝑐_{𝜃} | d 𝜃 & = 𝑁 (𝑟, 𝑓) - \frac{1}{2 π} \int_{0}^{2 π} 𝑁 (𝑟, e^{i 𝜃}, 𝑓) d 𝜃 \\ \begin{matrix}  \end{matrix} + \frac{1}{2 π} \int_{0}^{2 π} [\int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜙} - e^{i 𝜃}) | d 𝜙] d 𝜃 . (14.2 . 1) \end{aligned}

For any $𝑤 \in ℂ^{*}$ ,

{log}^{+} | 𝑤 | = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑤 - e^{i 𝜃} | d 𝜃, (14.2 . 2)

which follows directly from Jensen’s formula on the function $𝑧 \mapsto 𝑤 - 𝑧$ (by considering when $𝑤 \in \overline{𝔻}$ and $| 𝑤 | > 1$ ). Letting $𝜙 \in [0, 2 π]$ , we have by setting $𝑤 = 𝑓 (𝑟 e^{i 𝜙})$ ,

{log}^{+} | 𝑓 (𝑟 e^{i 𝜙}) | = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜙}) - e^{i 𝜃} | d 𝜃 .

Hence, from absolute convergence we have

\begin{aligned} 𝑚 (𝑟, 𝑓) & = \frac{1}{2 π} \int_{0}^{2 π} [\frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜙}) - e^{i 𝜃} | d 𝜃] d 𝜙 \\ = \frac{1}{2 π} \int_{0}^{2 π} [\frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜙}) - e^{i 𝜃} | d 𝜙] d 𝜃 . \end{aligned}

From (14.2.1) $(14.2.1) (14.2.1)$ , it follows that

𝑇 (𝑟, 𝑓) = \frac{1}{2 π} \int_{0}^{2 π} 𝑁 (𝑟, e^{i 𝜃}, 𝑓) d 𝜃 + \frac{1}{2 π} \int_{0}^{2 π} log | 𝑐_{𝜃} | d 𝜃 .

If $𝑓 (0) = \infty$ , then subtracting $e^{i 𝜃}$ from $𝑓$ does not modify the first nonzero coefficient of the innermost Laurent series ( $𝑐 = 𝑐_{𝜃}$ ). Thus, in this case

𝑇 (𝑟, 𝑓) = \frac{1}{2 π} \int_{0}^{2 π} 𝑁 (𝑟, e^{i 𝜃}, 𝑓) d 𝜃 + log | 𝑐 |

Otherwise, $𝑐_{𝜃} = 𝑓 (0) - e^{i 𝜃}$ unless this quantity is 0 (cancels), which can happen for at most one $𝜃$ value, which is negligible when integrated. Thus by (14.2.2) $(14.2.2) (14.2.2)$ ,

\frac{1}{2 π} \int_{0}^{2 π} log | 𝑐_{𝜃} | d 𝜃 = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (0) - e^{i 𝜃} | d 𝜃 = {\begin{cases} {log}^{+} | 𝑐 | & if 𝑓 (0) \neq 0, \\ 0 & otherwise . \end{cases} □

Theorem 14.2.3.

For (non-constant) meromorphic

𝑓

𝐷 (0, 𝑅)

𝑇 (𝑟, 𝑓)

is a non-decreasing convex (not necessarily strictly convex) function of

log 𝑟

(for

0 < 𝑟 < 𝑅

Proof.

Applying $d / d log 𝑟$ to Theorem 14.2.2 $Theorem 14.2.2 Theorem 14.2.2$ , we have

\begin{aligned} \frac{d 𝑇 (𝑟, 𝑓)}{d log 𝑟} & = \frac{1}{2 π} \frac{d}{d log 𝑟} \int_{0}^{2 π} 𝑁 (𝑟, e^{i 𝜃}, 𝑓) d 𝜃 \\ = \frac{1}{2 π} \int_{0}^{2 π} [\frac{𝑛 (𝑟, e^{i 𝜃}, 𝑓) - 𝑛 (0, e^{i 𝜃}, 𝑓)}{𝑟} \frac{d 𝑟}{d log 𝑟} + 𝑛 (0, e^{i 𝜃}, 𝑓)] d 𝜃 \\ = \frac{1}{2 π} \int_{0}^{2 π} 𝑛 (𝑟, e^{i 𝜃}, 𝑓) d 𝜃 \geq 0 \end{aligned}

by Lebesgue’s Dominated Convergence Theorem. Because the counting function

𝑛

is non-decreasing in

𝑟

, it is also non-decreasing in

log 𝑟

. ⁠

□

Theorem 14.2.4 (First Fundamental Theorem of Nevanlinna Theory).

Let $𝑓$ be (non-constant) meromorphic on $𝐷 (0, 𝑅)$ where $0 < 𝑟 \leq \infty$ . Then for any $𝑎 \in ℂ$ ,

𝑇 (𝑟, 𝑎, 𝑓) = 𝑇 (𝑟, 𝑓) - log | 𝑐 | + 𝜀 (𝑟, 𝑎, 𝑓),

where

| 𝜀 (𝑟, 𝑎, 𝑓) | \leq {log}^{+} | 𝑎 | + log 2

for

0 < 𝑟 < 𝑅

and

𝑐

is the first nonzero coefficient of the innermost Laurent series expansion of

𝑓 (𝑧) - 𝑎

about the origin.

Proof.

Since for $𝑧 \in 𝜕 𝐷 (0, 𝑟)$ , by the properties of ${log}^{+}$ as in Proposition 14.1.2 $Proposition 14.1.2 Proposition 14.1.2$ ,

{log}^{+} | 𝑓 (𝑧) - 𝑎 | \leq {log}^{+} (| 𝑓 (𝑧) | + | 𝑎 |) \leq log 2 + {log}^{+} | 𝑓 (𝑧) | + {log}^{+} | 𝑎 | .

Integrating over $𝜕 𝐷 (0, 𝑟)$ implies

𝑚 (𝑟, 𝑓 - 𝑎) \leq log 2 + 𝑚 (𝑟, 𝑓) + {log}^{+} | 𝑎 | .

From

{log}^{+} | 𝑓 (𝑧) | \leq {log}^{+} (| 𝑓 (𝑧) - 𝑎 | + | 𝑎 |) \leq log 2 + {log}^{+} | 𝑓 (𝑧) - 𝑎 | + {log}^{+} | 𝑎 |,

we obtain

𝑚 (𝑟, 𝑓) \leq log 2 + 𝑚 (𝑟, 𝑓 - 𝑎) + {log}^{+} | 𝑎 | .

Letting $𝜀 (𝑟, 𝑎, 𝑓) = 𝑚 (𝑟, 𝑓 - 𝑎) - 𝑚 (𝑟, 𝑓)$ , the two proximity-related inequalities give

| 𝜀 (𝑟, 𝑎, 𝑓) | \leq log 2 + {log}^{+} | 𝑎 | .

Then by Proposition 14.1.5 $Proposition 14.1.5 Proposition 14.1.5$ , and the fact that the poles of $𝑓$ match those of $𝑓 - 𝑎$ (more importantly, $𝑁 (𝑟, 𝑓 - 𝑎) = 𝑁 (𝑟, 𝑓)$ ),

\begin{aligned} 𝑇 (𝑟, 0, 𝑓 - 𝑎) & = 𝑁 (𝑟, 0, 𝑓 - 𝑎) + 𝑚 (𝑟, 0, 𝑓 - 𝑎) \\ = 𝑁 (𝑟, 𝑓 - 𝑎) + 𝑚 (𝑟, 𝑓 - 𝑎) - log | 𝑐 | \\ = 𝑁 (𝑟, 𝑓) + 𝑚 (𝑟, 𝑓 - 𝑎) - log | 𝑐 | \\ = 𝑁 (𝑟, 𝑓) + 𝑚 (𝑟, 𝑓) + 𝜀 (𝑟, 𝑎, 𝑓) - log | 𝑐 | \\ = 𝑇 (𝑟, 𝑓) + 𝜀 (𝑟, 𝑎, 𝑓) - log | 𝑐 | . □ \end{aligned}

For $𝑅 = \infty$ , we have

𝑁 (𝑟, 𝑎, 𝑓) + 𝑚 (𝑟, 𝑎, 𝑓) = 𝑇 (𝑟, 𝑓) + 𝒪︀ (1),

which shows that $𝑇$ is essentially independent of the choice of $𝑎$ , except for a bounded $𝒪︀ (1)$ term. Thus, this motivates why $𝑇$ is characteristic of $𝑓$ .

As we have seen before, it is often pedantic and somewhat annoying to account for cases where $𝑓 (0)$ is either zero or infinity, which is a minor grievance in Nevanlinna theory. Expositions from [5] and [3] tend to leave out this tedious treatment. However, here we will make this distinction whenever necessary for completeness.

Corollary 14.2.4.1.

Let $𝑓$ be (non-constant) meromorphic on $𝐷 (0, 𝑅)$ . Then for $0 < 𝑟 \leq \infty$ ,

\frac{1}{2 π} \int_{0}^{2 π} 𝑚 (𝑟, e^{i 𝜃}, 𝑓) d 𝜃 \leq log 2 .

Proof.

By the First Fundamental Theorem (Theorem 14.2.4 $Theorem 14.2.4 Theorem 14.2.4$ ), we have

𝑇 (𝑟, 𝑓) = 𝑁 (𝑟, e^{i 𝜃}, 𝑓) + 𝑚 (𝑟, e^{i 𝜃}, 𝑓) + log | 𝑐_{𝜃} | + 𝜀 (𝑟, e^{i 𝜃}, 𝑓),

where $| 𝜀 (𝑟, e^{i 𝜃}, 𝑓) | \leq log 2$ and $𝑐_{𝜃}$ is the first nonzero coefficient of the Laurent expansion for $𝑓 - e^{i 𝜃}$ . Hence,

𝑇 (𝑟, 𝑓) = \frac{1}{2 π} \int_{0}^{2 π} [𝑁 (𝑟, e^{i 𝜃}, 𝑓) + 𝑚 (𝑟, e^{i 𝜃}, 𝑓) + log | 𝑐_{𝜃} | + 𝜀 (𝑟, e^{i 𝜃}, 𝑓)] d 𝜃 .

By Theorem 14.2.2 $Theorem 14.2.2 Theorem 14.2.2$ , we thus have

\begin{aligned} 𝑇 (𝑟, 𝑓) & = 𝑇 (𝑟, 𝑓) - \frac{1}{2 π} \int_{0}^{2 π} log | 𝑐_{𝜃} | d 𝜃 + \frac{1}{2 π} \int_{0}^{2 π} 𝑚 (𝑟, e^{i 𝜃}, 𝑓) d 𝜃 \\ + \frac{1}{2 π} \int_{0}^{2 π} log | 𝑐_{𝜃} | d 𝜃 + \frac{1}{2 π} \int_{0}^{2 π} 𝜀 (𝑟, e^{i 𝜃}, 𝑓) d 𝜃 . \end{aligned}

This implies that

\int_{0}^{2 π} 𝑚 (𝑟, e^{i 𝜃}, 𝑓) d 𝜃 = - \int_{0}^{2 π} 𝜀 (𝑟, e^{i 𝜃}, 𝑓) d 𝜃 \leq 2 π log 2 . □

Remark.

Since

𝑚

is bounded in the integrated sense, if

𝑇

is generally large, then for most values,

𝑁

will be nearly equal to

𝑇

Proposition 14.2.5.

Let $𝑓, 𝑔$ be meromorphic on $𝐷 (0, 𝑅)$ and suppose $𝑚 \in ℕ$ . For any $0 < 𝑟 < 𝑅$ , the following properties hold:

$𝑇 (𝑟, 𝑓 𝑔) \leq 𝑇 (𝑟, 𝑓) + 𝑇 (𝑟, 𝑔)$ .
$𝑇 (𝑟, 𝑓 + 𝑔) \leq 𝑇 (𝑟, 𝑓) + 𝑇 (𝑟, 𝑔) + 𝒪︀ (1)$ .
$𝑇 (𝑟, \frac{1}{𝑓}) = 𝑇 (𝑟, 𝑓) + 𝒪︀ (1)$ .
$𝑇 (𝑟, 𝑓^{𝑛}) = 𝑛 𝑇 (𝑟, 𝑓)$ .

Proof.

Since

𝑁 (𝑟, 𝑓 𝑔), 𝑁 (𝑟, 𝑓 + 𝑔) \leq 𝑁 (𝑟, 𝑓) + 𝑁 (𝑟, 𝑔),

and by

{log}^{+} | 𝑓 𝑔 | \leq {log}^{+} | 𝑓 | + {log}^{+} | 𝑔 |, {log}^{+} | 𝑓 + 𝑔 | \leq log 2 + {log}^{+} | 𝑓 | + {log}^{+} | 𝑔 |,

it follows that

𝑚 (𝑟, 𝑓 𝑔) \leq 𝑚 (𝑟, 𝑓) + 𝑚 (𝑟, 𝑔), 𝑚 (𝑟, 𝑓 + 𝑔) \leq 𝑚 (𝑟, 𝑓) + 𝑚 (𝑟, 𝑔) + 𝒪︀ (1),

and the conclusions of Part 1 $Part 1 Part 1$ and Part 2 $Part 2 Part 2$ . Part 3 $Part 3 Part 3$ is a corollary of the First Fundamental Theorem (Theorem 14.2.4 $Theorem 14.2.4 Theorem 14.2.4$ ). Lastly, because a pole of $𝑓^{𝑛}$ has order $𝑛$ times of that of $𝑓$ , it follows that $𝑁 (𝑟, 𝑓^{𝑛}) = 𝑛 𝑁 (𝑟, 𝑓)$ . If $| 𝑓 | \geq 1$ , it follows that ${log}^{+} | 𝑓^{𝑛} | = 𝑛 {log}^{+} | 𝑓 |$ exactly. If $| 𝑓 | < 1$ , ${log}^{+} | 𝑓^{𝑛} | = 𝑛 {log}^{+} | 𝑓 | = 0$ . Hence, $𝑚 (𝑟, 𝑓^{𝑛}) = 𝑛 𝑚 (𝑟, 𝑓)$ .

Adding the two quantities $𝑁$ and $𝑚$ together gives Part 4 $Part 4 Part 4$ . ⁠ $□$

We now provide some classical examples of the Nevanlinna characteristic:

Let
$𝑓 (𝑧) = 𝑐 \frac{𝑧^{𝑝} + 𝑎_{1} 𝑧^{𝑝 - 1} + \dots + 𝑎_{𝑝}}{𝑧^{𝑞} + 𝑏_{1} 𝑧^{𝑞 - 1} + \dots + 𝑏_{𝑞}}, 𝑐 \neq 0$
be an arbitrary rational function in reduced form. If $𝑝 > 𝑞$ , then for sufficiently large $𝑟$ , $𝑓 \to \infty$ in every direction. This implies ${log}^{+} | \frac{1}{𝑓 (𝑧) - 𝑎} | = 0$ for any complex $𝑎$ and sufficiently large $𝑟$ ( $| 𝑧 | > 𝑟$ ). This implies $𝑚 (𝑟, 𝑎, 𝑓) = 0$ . Additionally, the equation $𝑓 (𝑧) = 𝑎$ has $𝑝$ complex solutions by the Fundamental Theorem of Algebra (Theorem 3.3.1 $Theorem 3.3.1 Theorem 3.3.1$ ). Thus, for sufficiently large $𝑟$ ( $𝑟 > 𝑟^{'}$ ), any $| 𝑧 | > 𝑟$ has the property that $𝑛 (𝑟, 𝑎, 𝑓) = 𝑛 (| 𝑧 |, 𝑎, 𝑓)$ ,
$\begin{aligned} 𝑁 (𝑟, 𝑎, 𝑓) & = (\int_{0}^{𝑟^{'}} + \int_{𝑟^{'}}^{𝑟}) \frac{𝑛 (𝑥, 𝑎, 𝑓) - 𝑛 (0, 𝑎, 𝑓)}{𝑥} d 𝑥 + 𝑛 (0, 𝑎, 𝑓) log 𝑟 \\ = 𝒪︀ (1) + (𝑛 (𝑟^{'}, 𝑎, 𝑓) - 𝑛 (0, 𝑎, 𝑓)) (log 𝑟 - log (𝑟^{'})) \\ \begin{matrix} \end{matrix} + 𝑛 (0, 𝑎, 𝑓) log 𝑟 \\ = 𝑝 log 𝑟 + 𝒪︀ (1), 𝑎 \neq \infty . \end{aligned}$
Hence,
$𝑇 (𝑟, 𝑓) = 𝑝 log 𝑟 + 𝒪︀ (1) .$
Similarly, $𝑇 (𝑟, \frac{1}{𝑓}) = 𝑇 (𝑟, 𝑓) + 𝒪︀ (1)$ by a preceding result. It then follows that if $𝑝 < 𝑞$ ,
$𝑇 (𝑟, 𝑓) = 𝑞 log 𝑟 + 𝒪︀ (1) .$
Lastly, the case where $𝑝 = 𝑞$ (derived with $𝑎 \neq 𝑐$ ) is resolvable with
$𝑇 (𝑟, 𝑓) = 𝑝 log 𝑟 + 𝒪︀ (1) = 𝑞 log 𝑟 + 𝒪︀ (1) .$
In all cases,
$𝑇 (𝑟, 𝑓) = 𝑑 log 𝑟 + 𝒪︀ (1),$
where $𝑑 = max (𝑝, 𝑞)$ is defined to be the degree of the rational function $𝑓$ .
Unless $𝑓 (\infty) = 𝑎$ , the proximity function vanishes for sufficiently large $𝑟$ . In this case, let $𝛼$ be the multiplicity at which $𝑓$ attains $𝑎$ at $𝑧 = \infty$ . If $𝑎$ is finite,
$\begin{matrix} 𝑚 (𝑟, 𝑎, 𝑓) = \frac{1}{2 π} \int_{0}^{2 π} {log}^{+} | \frac{1}{{(𝑟 e^{i 𝜃})}^{- 𝛼} 𝒪︀ (1)} | d 𝜃 = 𝛼 log 𝑟 + 𝒪︀ (1) \\ 𝑁 (𝑟, 𝑎, 𝑓) = (𝑑 - 𝛼) log 𝑟 + 𝒪︀ (1) . \end{matrix}$
If $𝑓$ has a pole of order $𝛼$ at $\infty$ , we have likewise that
$𝑚 (𝑟, 𝑎, 𝑓) = 𝛼 log 𝑟 + 𝒪︀ (1), 𝑁 (𝑟, 𝑎, 𝑓) = (𝑑 - 𝛼) log 𝑟 + 𝒪︀ (1) .$
Consider
$exp (𝑧) = exp (𝑟 cos 𝜃 + i 𝑟 sin 𝜃), 𝑧 = 𝑟 e^{i 𝜃} .$
By its entireness, $𝑁 (𝑟, 𝑓) = 0$ for any $𝑟 > 0$ , and the proximity is given by
$𝑚 (𝑟, 𝑓) = \frac{1}{2 π} \int_{0}^{2 π} {log}^{+} | e^{𝑟 e^{i 𝜃}} | d 𝜃 = \frac{1}{2 π} (\int_{0}^{\frac{π}{2}} + \int_{\frac{3 π}{2}}^{2 π}) {(𝑟 cos 𝜃)}^{+} d 𝜃 = \frac{𝑟}{π} .$
Therefore,
$𝑇 (𝑟) = 𝑚 (𝑟, 𝑓) + 𝑁 (𝑟, 𝑓) = \frac{𝑟}{π} .$
Moreover, $𝑚 (𝑟, 0, 𝑓) = \frac{𝑟}{π}$ and $𝑁 (𝑟, 0, 𝑓) = 0$ , whereas for $𝑎 \in ℂ^{*}$ , the solutions to $e^{𝑧} = 𝑎$ are in the form of $𝑧_{𝑘} = 𝑧_{0} + 2 𝑘 π i$ , where $𝑧_{0} = 𝑥_{0} + i 𝑦_{0}$ . The condition $| 𝑧_{𝑘} | \leq 𝑟$ is equivalent to
${| 𝑦_{0} + 2 𝑘 π |}^{2} + {| 𝑥_{0} |}^{2} \leq 𝑟^{2} ⟹ | 𝑦_{0} + 2 𝑘 π | \leq 𝑟 + 𝒪︀ (1) .$
It follows that
$𝑛 (𝑟, 𝑎, 𝑓) = \frac{𝑟}{π} + 𝒪︀ (1) ⟹ 𝑁 (𝑟, 𝑎, 𝑓) = \frac{𝑟}{π} + 𝒪︀ (1), 𝑚 (𝑟, 𝑎, 𝑓) = 𝒪︀ (1) .$

§14.2 Properties of 𝑇(𝑟,𝑓) and the First Fundamental Theorem

14.2 Properties of 𝑇(𝑟,𝑓) and the First Fundamental Theorem

§14.2 Properties of $𝑇 (𝑟, 𝑓)$ and the First Fundamental Theorem

14.2 Properties of $𝑇 (𝑟, 𝑓)$ and the First Fundamental Theorem