While reading an article, I came across the following statement.
Moreover, locally every holomorphic function $f(x + iy)$ may be written as $(\partial_{x} - i \partial_{y})h(x, y)$, for some scalar harmonic function $h(x, y)$.
I see that if such a function exists, it must be harmonic due to the Cauchy-Riemann equations. But is it true that such a function exists locally? If we say that $f(x, y) = u(x, y) + i v(x, y)$, then this is equivalent to saying that the system $$\begin{cases} \partial_{x} h(x, y) &= u(x, y) \\ \partial_{y} h(x, y) &= - v(x, y) \end{cases}$$ has a local solution.
First we note two theorems.
Theorem 1. Let $D \subset \mathbb{C}$ be a simply connected region, $f$ be a holomorphic function, $\gamma_1$ and $\gamma_2$ be two contours in $D$ connecting the points $z_1$ and $z_2$ in $D$. Then $$\int_{\gamma _1} f(z)\, dz= \int_{\gamma _2} f(z)\, dz .$$ Proof. By Cauchy theorem, we have $\int_C f(z) \, dz=0$, where $C=\gamma _1-\gamma _2$. Therefore $\int_{\gamma _1} f(z)\, dz= \int_{\gamma _2} f(z)\, dz $.
Theorem 2. Let $D$ and $f$ be as above. Then there is a holomorphic function $F$ such that $F^\prime (z) = f(z)$ for any $z \in D$.
Proof. Let $\gamma $ be a contour connecting $z_0$ and $z$ in $D$, then $\int_\gamma f(z)\, dz$ is determined by $z_0$, $z$ and not $\gamma $ by the above theorem. This allows us to introduce $F(z)=\int_{z_0}^{z} f(\zeta ) d\zeta$ . It is easy to prove that $F^\prime (z)=f(z)$.
Now let $f(z)$ be holomorphic in a simply connected neighborhood of $z_0$. Then $f$ has an indefinite integral $F(z)$, $F^\prime(z)=f(z)$. If we say that $F(x+iy)=h(x, y)+ik(x,y)$, then $F^\prime(z)=\partial_x h(x,y)+i\partial_x k(x, y)=\partial_x h(x,y)-i\partial_y h(x, y)$ due to the Cauchy-Riemann equations. Thus, locally, $f$ may be written as $\partial_x h-i\partial_y h$ for a harmonic $h=$Re $F$ with $F^\prime =f$.