# Harmonic functions are holomorphic and antiholomorphic (can't find my mistake)?!

Could you please help me find the mistake in the following reasoning? I may have become dumb since I've spent a lot of time thinking about it but can't see what's wrong.

Let $$U\subset \mathbb{C}$$ be open and let $$f=(u,v):U\to \mathbb{C}$$ be a function. It is easy to check that $$f$$ is harmonic in $$U$$ iff $$\frac{\partial}{\partial \overline{z}}\frac{\partial}{\partial z} f=0 \qquad\text{in U.}$$ Now, from this it seems to me that one can deduce that $$f$$ is holomorphic (and even antiholomorphic, and thus constant!), which is clearly wrong since there are no reasons for two arbitrary harmonic functions $$u$$ and $$v$$ to satisfy the Cauchy-Riemann equations. The reasoning is the following: from $$\frac{\partial}{\partial \overline{z}}\frac{\partial}{\partial z} f=0$$ we see that $$\frac{\partial}{\partial z} f$$ is holomorphic in $$U$$, then locally it has a holomorphic primitive, which must coincide with $$f$$, thus $$f$$ is holomorphic.

When writing down the question, I realized that the problem is that it is not true that the holomorphic primitive of $$\frac{\partial}{\partial z}f$$ must coincide with $$f$$. In fact, a function $$g$$ is such that $$\frac{\partial}{\partial z}g=\frac{\partial}{\partial z}f$$ if and only if $$g=f+h$$ for some antiholomorphic function $$h$$.
• correct and you deduced the local representation of harmonic functions $f=g+\bar h$ where $g,h$ are holomorphic Feb 7, 2022 at 21:55