Let \(U\) be an open subset of \(\mathbb{C}\). Let \(z_0\) be a point in \(U\), and suppose that \(f\) is a meromorphic function on \(U\) with a pole at \(z_0\). Prove that there is no holomorphic function \(g : U \setminus \{z_0\} \to \mathbb{C}\) such that \(e^{g(z)} = f(z)\) for all \(z \in U \setminus \{z_0\}\).
Solution
1. The logarithmic derivative of \(f\) has a nonzero integralaround a sufficiently small circle about its pole.
Let \(m\geq1\) be the order of the pole. The local Laurent expansion gives \[ f(z)=(z-z_0)^{-m}h(z), \] where \(h\) is holomorphic near \(z_0\) and \(h(z_0)\ne0\) [@SS03]. Choose \(r>0\) so that \(\overline{D(z_0,r)}\subset U\) and \(h\) is holomorphic and nonzero on a neighborhood of this closed disk. On its punctured interior, \[ \frac{f'(z)}{f(z)}=-\frac{m}{z-z_0}+\frac{h'(z)}{h(z)}. \] For the positively oriented circle \(\gamma:|z-z_0|=r\), Cauchy’s theorem applied to the holomorphic function \(h'/h\) and the circle parametrization give \[ \int_\gamma\frac{f'}f\,dz=-2\pi i m\ne0. \]
2. A holomorphic logarithm would force the same integral to vanish.
Suppose the required \(g\) existed. On the punctured disk, differentiation of \(e^g=f\) gives \(f'/f=g'\). But \(g'\) has the single-valued primitive \(g\) there, so \[ \int_\gamma g'(z)\,dz=0. \] Indeed, parametrizing the circle makes the integrand the derivative of \(g(\gamma(t))\), and the starting and ending values agree. This uses the primitive, not simple connectedness of the punctured domain. The zero integral contradicts step 1, proving that no such \(g\) exists.