State and prove a necessary and sufficient condition for a meromorphic \(1\)-form \[ \omega=\omega(z)\,dz \] on \(\mathbb C\) to be the logarithmic derivative of a meromorphic function; that is, for \[ \omega=d\log f=\frac{f'(z)}{f(z)}\,dz \] for some meromorphic \(f\).
Solution
A meromorphic \(1\)-form \(\omega\) on \(\mathbb C\) is of the form \[ \omega=\frac{f'}f\,dz \] for a nonzero meromorphic function \(f\) if and only if every pole of \(\omega\) is simple and has integer residue.
1. If \(\omega=(f'/f)\,dz\) for a nonzero meromorphic \(f\), then everypole of \(\omega\) is simple with integer residue.
Fix \(a\in\mathbb C\). Since \(f\) is meromorphic and not identically zero, in a neighborhood of \(a\) it can be written \[ f(z)=(z-a)^m u(z), \] where \(m\in\mathbb Z\) and \(u\) is holomorphic with \(u(a)\neq0\). Here \(m>0\) is the order of a zero, \(m<0\) is minus the order of a pole, and \(m=0\) means that \(f\) is holomorphic and nonzero at \(a\).
Taking the logarithmic derivative gives \[ \frac{f'(z)}{f(z)} =\frac{m}{z-a}+\frac{u'(z)}{u(z)}. \] The second term is holomorphic near \(a\). Hence, whenever \(f'/f\) has a pole at \(a\), that pole is simple and \[ \operatorname{Res}_a\left(\frac{f'}f\,dz\right)=m\in\mathbb Z. \]
2. Conversely, suppose every pole of \(\omega\) is simple and hasinteger residue. Let \(P\) be the discrete set of poles and choose\(z_0\in\mathbb C\setminus P\).
The set of poles of a meromorphic form is discrete, so \(\mathbb C\setminus P\) is path connected. On this punctured plane we may integrate \(\omega\) along piecewise smooth paths beginning at \(z_0\).
3. For \(z\in\mathbb C\setminus P\), the formula\[ f(z)=\exp\left(\int_{z_0}^z\omega\right) \]is independent of the chosen path.
If \(\gamma_1\) and \(\gamma_2\) are two paths from \(z_0\) to \(z\), their difference is a closed cycle \(\gamma\). By the residue theorem, \[ \int_\gamma\omega =2\pi i\sum_{a\in P}n(\gamma,a)\operatorname{Res}_a(\omega), \] where only finitely many terms are nonzero and each winding number \(n(\gamma,a)\) is an integer. By hypothesis each residue is also an integer, so \[ \int_\gamma\omega\in2\pi i\mathbb Z. \] Exponentiation therefore removes the ambiguity: \[ \exp\left(\int_{\gamma_1}\omega\right) =\exp\left(\int_{\gamma_2}\omega\right). \] Thus \(f\) is well-defined on \(\mathbb C\setminus P\).
4. On \(\mathbb C\setminus P\), the function \(f\) is holomorphic,nowhere zero, and satisfies \(f'/f\,dz=\omega\).
Locally on a simply connected disk avoiding \(P\), choose a primitive \(H\) of \(\omega\). The definition in 3 then has the form \[ f(z)=Ce^{H(z)} \] for some nonzero constant \(C\). Hence \(f\) is holomorphic and nowhere zero, and \[ \frac{f'}f=H'=\omega/dz. \]
5. The function \(f\) extends meromorphically across every point of \(P\).
Let \(a\in P\) and write \[ \operatorname{Res}_a(\omega)=m\in\mathbb Z. \] Because the pole is simple, \[ \omega =\left(\frac{m}{z-a}+h(z)\right)dz \] near \(a\), with \(h\) holomorphic. Choose a holomorphic primitive \(H\) of \(h\) on a small disk about \(a\). On the punctured disk, 4 implies \[ f(z)=C(z-a)^m e^{H(z)} \] for some \(C\neq0\).
If \(m\ge0\), this extends holomorphically across \(a\), with a zero of order \(m\) when \(m>0\). If \(m<0\), it extends meromorphically with a pole of order \(-m\). Thus \(f\) extends meromorphically across every pole of \(\omega\).
6. The extended meromorphic function satisfies\[ \omega=d\log f \]on all of \(\mathbb C\).
4. Near every\(a\in P\), the local expression from 5 gives\[ \frac{f'}f =\frac{m}{z-a}+h(z), \]which is exactly the local coefficient of \(\omega\). Hence the equality holdsmeromorphically across \(P\) as well.