§4.3 Entireness and Meromorphy

4.3 Entireness and Meromorphy

We have previously defined the concept of an entire function in the chapter on complex differentiation. Let $𝑓$ be entire with the unique Taylor expansion $\sum_{𝑛 = 0}^{\infty} 𝑐_{𝑛} 𝑧^{𝑛}$ . Since $𝑧 = \infty$ is an isolated singularity, by the uniqueness of the Laurent expansion, the expansion at $𝑧 = 0$ has the same form as the expansion at $𝑧 = \infty$ . We will now analyze the implications on the entire function $𝑓$ given an isolated singularity.

If the infinity point is a removable singularity, then ${lim}_{𝑧 \to \infty} 𝑓 (𝑧)$ exists and is finite.
Proposition 4.3.1.
If $𝑓 (𝑧)$ is entire and has a removable singularity at $𝑧 = \infty$ , then $𝑓$ is constant.
Proof.
Let $𝑧 = \frac{1}{𝜁}$ , and let $𝑔 (𝜁) = 𝑓 (\frac{1}{𝜁})$ , which has a removable singularity at $𝜁 = 0$ . By Theorem 3.2.9 $Theorem 3.2.9 Theorem 3.2.9$ , $𝑔$ can be analytically continued to all of $ℂ$ , especially at $𝜁 = 0$ . Let $𝑤 = 𝑔 (0)$ . Then, $\forall 𝜀 > 0$ , $\exists 𝛿 > 0$ such that $\forall 𝜁 \in 𝐷 (0, 𝛿)$ , $| 𝑔 (𝜁) - 𝑤 | < 𝜀$ . It follows that $\forall | 𝑧 | > \frac{1}{𝛿}$ , $| 𝑓 (𝑧) | < | 𝑤 | + 𝜀$ , and is bounded. For the complement, $\forall 𝑧 \in \overline{𝐷 (0, \frac{1}{𝛿})}$ , $𝑓 (𝑧)$ is continuous on a compact set, and by Theorem 1.2.22 $Theorem 1.2.22 Theorem 1.2.22$ , is also bounded.
Then by Liouville’s Theorem (Theorem 3.2.5 $Theorem 3.2.5 Theorem 3.2.5$ ), $𝑓$ is constant. ⁠ $□$
If $𝑓 (𝑧)$ has a pole at $𝑧 = \infty$ of order $𝑚 \in ℕ$ , then $𝑓$ is a polynomial of degree $𝑚$ .
Proof.
By the classification of a pole at $\infty$ , $𝑓$ can be written as
$𝑓 (𝑧) = 𝑐_{𝑚} 𝑧^{𝑚} + 𝑐_{𝑚 - 1} 𝑧^{𝑚 - 1} + \dots + 𝑐_{0} + \frac{𝑐_{- 1}}{𝑧} + \dots .$
Since $𝑓 (𝑧)$ is entire, it is holomorphic at $𝑧 = 0$ and has a convergent Taylor expansion. By the uniqueness of Laurent expansions (Theorem 4.1.3 $Theorem 4.1.3 Theorem 4.1.3$ ), the two expansions are equivalent and therefore all terms with negative exponents vanish, and
$𝑓 (𝑧) = 𝑐_{𝑚} 𝑧^{𝑚} + 𝑐_{𝑚 - 1} 𝑧^{𝑚 - 1} + \dots + 𝑐_{0},$
and since $𝑐_{𝑚} \neq 0$ , the statement is confirmed. ⁠ $□$
If $𝑓 (𝑧)$ has an essential singularity at $𝑧 = \infty$ , $𝑓 (𝑧)$ is known as a transcendental entire function.

Example 4.3.2.

The entire functions

sin (𝑧)

cos (𝑧)

sinh (𝑧)

cosh (𝑧)

, and

exp (𝑧)

are transcendental.

Definition 4.3.3 (Meromorphy).

Let

𝑈 \subseteq ℂ

be open, and let

{𝑎_{𝑛}}_{𝑛 \in ℕ} \subset 𝑈

be a set of isolated points. Suppose

𝑓 : 𝑈 ∖ {𝑎_{𝑛}}_{𝑛 \in ℕ} \to ℂ

is holomorphic and has a pole at each of

𝑧 \in {𝑎_{𝑛}}

. Then

𝑓

is meromorphic in

𝑈

Similar to holomorphy, meromorphy on a compact set can be defined as meromorphy on a neighborhood of the set. In general, we imply for the set to be open unless stated otherwise. If the set is not implicitly specified, we assume meromorphy on $ℂ$ .

All holomorphic functions are meromorphic functions (with poles on $⌀$ ). Consequently, entire functions are meromorphic on $ℂ$ . All rational functions (including polynomials) are also meromorphic on $ℂ$ . In the study of meromorphic functions with an isolated singularity at $\infty$ , rational functions are of important interest.

Let $𝑓 (𝑧)$ be rational, written as $𝑓 (𝑧) = \frac{𝑝 (𝑧)}{𝑞 (𝑧)}$ , where $𝑝$ and $𝑞$ are polynomials. Let

\begin{matrix} 𝑝 (𝑧) = 𝑎_{𝑛} 𝑧^{𝑛} + 𝑎_{𝑛 - 1} 𝑧^{𝑛 - 1} + \dots + 𝑎_{0} \\ 𝑞 (𝑧) = 𝑏_{𝑚} 𝑧^{𝑚} + 𝑏_{𝑚 - 1} 𝑧^{𝑚 - 1} + \dots + 𝑏_{0}, \end{matrix}

where $𝑎_{𝑛}, 𝑏_{𝑚} \neq 0$ . Trivially, the poles of $𝑓$ are the zeros of $𝑞$ . Since

𝑓 (𝑧) = \frac{𝑧^{𝑛}}{𝑧^{𝑚}} \cdot \frac{𝑎_{𝑛} + \frac{𝑎_{𝑛 - 1}}{𝑧} + \dots + \frac{𝑎_{0}}{𝑧^{𝑛}}}{𝑏_{𝑚} + \frac{𝑏_{𝑚 - 1}}{𝑧} + \dots + \frac{𝑏_{0}}{𝑧^{𝑚}}},

we have

lim_{𝑧 \to \infty} 𝑓 (𝑧) = {\begin{cases} \frac{𝑎_{𝑛}}{𝑏_{𝑚}} & if 𝑛 = 𝑚 \\ 0 & if 𝑛 < 𝑚 \\ \infty & if 𝑛 > 𝑚 . \end{cases}

Conversely, we have:

Theorem 4.3.4.

𝑓 (𝑧)

is meromorphic on

ℂ

and has a pole or removable singularity at

𝑧 = \infty

, then

𝑓

is a rational function.

Proof.

Since $𝑓$ is meromorphic on $ℂ$ , its singularities are isolated poles. The assumption that $𝑓$ has either a pole or a removable singularity at $\infty$ implies that this singularity is also isolated. Thus, there exists some $𝑅 > 0$ such that $𝑓$ is holomorphic on the punctured neighborhood ${𝑧 \in ℂ : 𝑅 < | 𝑧 | < \infty}$ of $\infty$ .

Consider the Laurent expansion of $𝑓$ at $\infty$ , obtained by substituting $𝑤 = \frac{1}{𝑧}$ and expanding around $𝑤 = 0$ :

𝑓 (𝑧) = \sum_{𝑛 = - \infty}^{\infty} 𝑎_{𝑛} 𝑧^{𝑛},

where the series converges for sufficiently large $| 𝑧 |$ . If $\infty$ is a removable singularity, the coefficients $𝑎_{𝑛} = 0$ for all $𝑛 > 0$ . If $\infty$ is a pole of order $𝑚$ , then $𝑎_{𝑛} = 0$ for all $𝑛 > 𝑚$ , and $𝑎_{𝑚} \neq 0$ . In either case, the principal part at $\infty$ is

𝜓_{\infty} (𝑧) = \sum_{𝑛 = 1}^{𝑚} 𝑎_{𝑛} 𝑧^{𝑛},

which is a polynomial (identically zero if degree is $0$ ).

Next, observe that $𝑓$ has only finitely many poles in the closed disk $\overline{𝐷 (0, 𝑅)} = {𝑧 : | 𝑧 | \leq 𝑅}$ . Suppose otherwise. Then the set of poles in $\overline{𝐷 (0, 𝑅)}$ would be infinite. By Bolzano–Weierstrass (Theorem 1.1.10 $Theorem 1.1.10 Theorem 1.1.10$ ), this set would have an accumulation point in $\overline{𝐷 (0, 𝑅)}$ . At such an accumulation point, $𝑓$ would have a non-isolated singularity, a contradiction of the meromorphy of $𝑓$ on $ℂ$ .

Let $𝑧_{1}, \dots, 𝑧_{𝑛}$ denote these finitely many poles in $\overline{𝐷 (0, 𝑅)}$ . For each $𝑘 = 1, \dots, 𝑛$ , the Laurent expansion of $𝑓$ at $𝑧_{𝑘}$ has principal part

𝜓_{𝑘} (𝑧) = \sum_{𝑗 = 1}^{𝑚_{𝑘}} \frac{𝑐_{𝑘, - 𝑗}}{{(𝑧 - 𝑧_{𝑘})}^{𝑗}},

where $𝑚_{𝑘}$ is the order of the pole at $𝑧_{𝑘}$ . Define the auxiliary function

Φ (𝑧) = 𝑓 (𝑧) - 𝜓_{\infty} (𝑧) - \sum_{𝑘 = 1}^{𝑛} 𝜓_{𝑘} (𝑧),

which is meromorphic on $ℂ$ , with potential singularities only at $𝑧_{1}, \dots, 𝑧_{𝑛}$ and $\infty$ .

We now show that each of these singularities is removable. First, fix $𝑗 \in {1, \dots, 𝑛}$ arbitrarily. Since the poles are isolated, there exists $𝜀_{𝑗} > 0$ such that the punctured disk $𝐷^{*} (𝑧_{𝑗}, 𝜀_{𝑗}) = {𝑧 : 0 < | 𝑧 - 𝑧_{𝑗} | < 𝜀_{𝑗}}$ contains no other poles $𝑧_{𝑘}$ for $𝑘 \neq 𝑗$ .

Since $𝑓 (𝑧) - 𝜓_{𝑗} (𝑧)$ is the holomorphic part of the Laurent expansion at $𝑧_{𝑗}$ , it is holomorphic on $𝐷 (𝑧_{𝑗}, 𝜀_{𝑗})$ (including at $𝑧_{𝑗}$ ).
$\sum_{𝑘 \neq 𝑗} 𝜓_{𝑘} (𝑧)$ is holomorphic on $𝐷 (𝑧_{𝑗}, 𝜀_{𝑗})$ , as each $𝜓_{𝑘}$ has its singularity elsewhere.
$𝜓_{\infty} (𝑧)$ is a polynomial, hence entire.

Thus,

Φ (𝑧) = [𝑓 (𝑧) - 𝜓_{𝑗} (𝑧)] - 𝜓_{\infty} (𝑧) - \sum_{𝑘 \neq 𝑗} 𝜓_{𝑘} (𝑧)

is holomorphic on $𝐷 (𝑧_{𝑗}, 𝜀_{𝑗})$ , including at $𝑧_{𝑗}$ . Therefore, we can define $Φ (𝑧_{𝑗})$ to make $Φ$ holomorphic at $𝑧_{𝑗}$ .

Since $𝑓 (𝑧) - 𝜓_{\infty} (𝑧)$ is the holomorphic part of the expansion at $\infty$ , consisting of terms with nonpositive powers of $𝑧$ , ${lim}_{𝑧 \to \infty} 𝑓 (𝑧) - 𝜓_{\infty} (𝑧)$ exists and is finite. Additionally, each $𝜓_{𝑘} (𝑧)$ consists of negative powers of $𝑧 - 𝑧_{𝑘}$ , so ${lim}_{𝑧 \to \infty} 𝜓_{𝑘} (𝑧) = 0$ for each $𝑘$ , and thus ${lim}_{𝑧 \to \infty} \sum_{𝑘 = 1}^{𝑛} 𝜓_{𝑘} (𝑧) = 0$ . Therefore, ${lim}_{𝑧 \to \infty} Φ (𝑧)$ exists and is finite, so $\infty$ is a removable singularity of $Φ$ . Without the finite singularities at each $𝑧_{𝑘}$ , $Φ$ is entire. Since $Φ$ has a finite limit at $\infty$ , it is bounded on $ℂ$ . By Liouville’s theorem, $Φ (𝑧) \equiv 𝑐$ for some constant $𝑐$ .

Hence,

𝑓 (𝑧) = 𝑐 + 𝜓_{\infty} (𝑧) + \sum_{𝑘 = 1}^{𝑛} 𝜓_{𝑘} (𝑧) .

The right-hand side is a sum of a constant, a polynomial, and finitely many principal parts (each a rational function with a single pole), so $𝑓$ is rational. ⁠ $□$

If $𝑧 = \infty$ is not a pole or removable singularity of a meromorphic function $𝑓 (𝑧)$ , then it is either an essential singularity or an accumulation point of poles. In this case, $𝑓$ is not rational and is known as a transcendental meromorphic function.