Lecture-21 : Meromorphic Functions

LECTURE-21 : MEROMORPHIC FUNCTIONS

VED V. DATAR∗

A function on a domain Ω is called meromorphic, if there exists a sequence ∗ of points p1, p2, ··· with no limit point in Ω such that if we denote Ω = Ω \{p1, ···} ∗ • f :Ω → C is holomorphic. • f has poles at p1, p2 ··· . It is often useful to think of z = ∞ on the same footing as other points in the complex plane, and to deﬁne the extended complex plane Cˆ = C ∪ ∞. We can then think of meromorphic functions f :Ω \{p1, ··· , pj, · · · } → C, as functions f :Ω → Cˆ, by deﬁning

f(pj) = ∞ for all the poles pjs. Similarly, In studying meromorphic functions on C, it is also useful to consider the extension of the function themselves to Cˆ. We say that z = ∞ is a pole (resp. removable or essential singularity) if z = 0 is a pole (resp. removable or essential singularity) for the function fˆ(z) = f(1/z). We then say that a meromorphic function on C is meromorphic on the extended plane, if it does not have an essential singularity at z = ∞. It turns out that meromorphic functions on Cˆ can be classiﬁed. Recall that a rational function on C is a function of the form P (z) R(z) = Q(z) where both P (z) and Q(z) are polynomials. Lemma 0.1. A rational function has a pole or removable singularity at inﬁnity. It has a removable singularity if and only if deg Q ≥ deg P .

Theorem 0.1. The only meromorphic functions on Cˆ are rational functions.

Proof. Let F : Cˆ → Cˆ be a meromorphic function.

Claim-1. F has only ﬁnitely many poles {p1, ··· , pn} in the complex plane C.

Date: 24 August 2016. 1 To see this, note that F (1/z) has either a pole or zero at z = 0. In either case there is a small neighborhood |z| < ε which has no other pole. Which is the same as saying that F has no ﬁnite pole in |z| > 1/ε. But |z| ≤ 1/ε is compact, and since all poles are isolated, this shows that there are only ﬁnitely many poles. Now, corresponding to each of the poles pk ∈ C there exists a polynomial Pk (see Remark 0.2 in Lecture-20) such that 1 F (z) = Pk + Gk(z), z − pk where Gk is holomorphic on a whole neighborhood around pk (including at the point pk). Similarly, we can write 1 1 F = P + G (z), z ∞ z ∞ where as before, G(z) is holomorphic in a neighborhood of z = 0.

Claim-2. The function n X 1 H(z) = F (z) − P∞(z) − Pk z − pk k=1 is an entire and bounded function. Assuming the claim, by Liouville’s theorem, H(z is a constant, and hence F (z) must be rational, and the theorem is proved. To prove the claim, ﬁrst note that clearly, H(z) is holomorphic away from {p1, ··· , pn}. At some z = pk, Pj(1/z − pj) is holomorphic for all j 6= k. On the other hand, near pk, 1 F (z) − Pk = Gk(z) z − pk which is holomorphic. This shows that H(z) is entire. As a consequence, to show boundedness, we only need to show boundedness on |z| > R for some large R. To see, ﬁrst observe that since Pk are polynomials, 1 lim Pk = 0. z→∞ z − pk

Hence it is enough to show that F (z) − P∞(z) is bounded near inﬁnity. But this follows immediately from noting that 1 1 G (z) = F − P ∞ z ∞ z is holomorphic near z = 0 and hence is bounded on |z| < ε for some ε > 0. In particular F (z) − P∞(z) is bounded on |z| > 1/ε. This proves the claim, and hence completes the proof of the theorem. A simple consequence of the proof is the following theorem on partial fraction decomposition that we take for granted as an important tool in integration theory, but never see the proof of. 2 Corollary 0.1. For any rational function R(z) = P (z)/Q(z) has a partial fraction decomposition of the form n X 1 R(z) = P∞(z) + Pk , z − pk k=1 where pk is a root of Q(z) of order mk, Pk is a polynomial of degree mk, and deg P∞ = deg P − deg Q.

∗ Department of Mathematics, UC Berkeley E-mail address: [email protected]