Let $\Omega \subset \mathbb{C}$ be a region( open + connected), and suppose $p dx + q dy$ is a locally exact differential in $\Omega$ ($p,q$ real-valued and continuous, say), suppose $f(z) = \int_{\gamma_{z_0 \rightarrow z}} du + i( p dx + q dy)$, where $\gamma_{z_0 \rightarrow z}$ is any path from $z_0$ to $z$, and assume $f$ is multi-valued in the sense that for any two such paths, the difference of that integral is $2 \pi i k$ for some integer $k$ depending on the paths.
Moreover , suppose $u$ is a harmonic function in $\Omega$, such that locally, $du + i (pdx+qdy)$ is of the form $g dz$ for some holomorphic $g$, (i.e. $pdx + q dy$ is the conjugate harmonic differential of $u$), then why is $exp(f)$ holomorphic in $\Omega$?
I can see that $exp(f)$ is now unambiguously defined, but I fail to see why it is holomorphic.
Thanks in advance!
Best Answer
Fix some $x_0 \in \Omega$, then in some small disk around $x_0$ contained in $\Omega$, we have that $du + i (pdx + qdy) = gdz$ where $g$ is holomorphic, for any $z$ in this disk, $f(z) = f(x_0) + \int_{\gamma_{x_0 \rightarrow z}} g dz \mod 2 \pi i$, where $\gamma_{x_0 \rightarrow z}$ is the straight line path from $x_0$ to $z$, $g$ has a primitive $G$ in this disk, so $f(z) = f(x_0) + G(z) - G(x_0) \mod 2 \pi i$, and $\exp(f(z)) = \frac{\exp(f(x_0))}{\exp(G(x_0))} \exp(G(z))$ for $z$ in this disk, and the RHS is holomorphic in this disk, so $\exp(f(z))$ is holomorphic in $\Omega$.