3.10 Meromorphic functions and the argument principle
A function \(f\) with an isolated singularity at \(z_0\) is said to be meromorphic at \(z_0\) if the singularity at \(z_0\) is a pole. If \(U\subseteq \mathbb {C}\) is open and \(f\in A(U)\), we say that \(f\) is meromorphic on \(U\) if every singularity of \(f\) on \(U\) is a pole.
Here’s another piece of notation, which should be useful.
Suppose \(f\) is meromorphic at \(z_0\in \mathbb {C}\). If \(f\) is analytic at \(z_0\), then we’ve already defined \(\operatorname{\mathrm{ord}}_{z_0} f\) to be the order of vanishing of \(f\) at \(z_0\) and \(+\infty \) if \(f\) is identically \(0\) in a neighborhood of \(z_0\). If \(f\) is not analytic at \(z_0\), then let’s set \(\operatorname{\mathrm{ord}}_{z_0} f = -m\) where \(m\) is the order of the pole of \(f\) at \(z_0\).
A function \(f\) is meromorphic at \(z_0\) if and only if
where \(g\) and \(h\) are analytic at \(z_0\) and \(\operatorname{\mathrm{ord}}_{z_0} h {\lt}\infty \). In this case,
Moreover, if \(f\) is not identically \(0\) in a neighborhood of \(z_0\), can write
where \(f_1\) is analytic at \(z_0\) and \(f_1(z_0)\neq 0\).
We already know the first assertion.
For the second, suppose \(f = g/h\) with \(g\) and \(h\) analytic at \(z_0\) and \(\operatorname{\mathrm{ord}}_{z_0} h {\lt} \infty \). If \(\operatorname{\mathrm{ord}}_{z_0} g = \infty \), then \(g = 0\). So \(f = 0\) as well. Otherwise, we can write
where \(g_1\) and \(h_1\) are analytic at \(z_0\) and \(g_1(z_0) h_1(z_0) \neq 0\). The Proposition follows directly by setting \(f_1 = g_1/h_1\).
Suppose \(f\) is meromorphic and nonzero at \(z_0\). Then \(\operatorname{\mathrm{ord}}_{z_0} f = \operatorname*{\mathrm{Res}}_{z_0} f'/f\).
Write \(f = (z-z_0)^{\operatorname{\mathrm{ord}}_{z_0} f} f_1\) with \(f_1\) analytic and with \(f_1(z_0)\neq 0\). Then
Since \(f_1\) is nonvanishing at and analytic at \(z_0\), \(f_1'/f_1\) is analytic at \(z_0\). So
[Argument principle] Suppose \(D\subseteq \mathbb {C}\) is a domain and \(C\) is a simple closed contour in \(D\). Let \(f\in A(D)\) be a meromorphic function, which is analytic at every point in \(C\). Then
There are only finitely many zeroes and poles of \(f\) in the region enclosed by \(C\).
If \(z_1,\ldots , z_n\) denotes the set of zeroes and poles inside of \(C\), then
\[ \frac{1}{2\pi i}\int _C \frac{f'}{f}\, dz = \sum _{i=1}^n \operatorname{\mathrm{ord}}_{z_i} f. \]
Let’s skip the detailed proof of the finiteness. Here’s the idea though: Since the zeros of holomorphic functions cannot have a limit point inside of \(D\) (unless the corresponding functions are identically \(0\) on \(D\)), the set of zeros and poles has no limit point in \(D\). But the union of \(C\) together with the region enclosed by \(C\) is compact. So any infinite set would have a limit point in this region.
Now to prove the principle just use the residue theorem together with Lemma 3.55.
If \(f\) and \(C\) are as in Theorem 3.56, let’s write \(N(C,f) = \sum _{i=1}^n \operatorname{\mathrm{ord}}_{z_i} f\) for the number of zeros and poles of \(f\) inside of \(C\) counted with multiplicities. So, in this language, the Theorem 3.56 says that
Here’s another related theorem.
Suppose \(C\) is a simple, closed contour contained in a domain \(D\subseteq \mathbb {C}\) and \(f,g\in A(D)\) with \(|f(z)| \geq |g(z)|\) for all \(z\in C\). Then
For each \(t\in [0,1]\), define a function \(h_t\in A(D)\) by setting \(h_t(z) = f(z) + tg(z)\). Note that, by our assumption, \(h_t\) has no poles or zeros on \(C\) itself. It follows that the function \([0,1]\to \mathbb {Z}\) given by \(t\mapsto N(C,h_t)\) is continuous. But, since \(\mathbb {C}\) is discrete, this implies (by the Intermediate Value Theorem) that \(N(C,h_t)\) is constant on \([0,1]\). So \(N(C, f+g) = N(C, h_1) = N(C, h_0) = N(C, f)\).