I am looking for a function $f:\mathbb C^2 \rightarrow \mathbb C^2$ that satisfies the two equations


$$\partial\_{z\_2}f\_1(z\_1,z\_2) + \partial\_{z\_1} f\_2(z\_1,z\_2)=0 \text{ and }$$
$$\partial\_{\bar z\_1}f\_1(z\_1,z\_2) - \partial\_{\bar z\_2} f\_2(z\_1,z\_2)=0$$


and in addition, is doubly-periodic in both its complex variables $z\_1,z\_2$. Does such a function exist and if not, why? I would not even know how to start building such a function.


In particular, I would like to have


$$f\_1(z\_1+1,z\_2)=f\_1(z\_1,z\_2+1)=f\_1(z\_1,z\_2)$$ and


$$f\_1(z\_1+i,z\_2) = e^{2\pi i k\_1}f\_1(z\_1,z\_2)$$
and


$$f\_1(z\_1,z\_2+i) = e^{2\pi i k\_2}f\_1(z\_1,z\_2)$$


for some fixed $k\_1,k\_2 \in \mathbb R.$ Please let me know if you do have any questions. I had some typos in there, but hopefully everything is coherent now.



Complex-doubly periodic function in two variables?
The answer is that the only solutions have the form
$$
f = (f\_1,f\_2) = \bigl(c, h(\,\overline{z}\_1, z\_2)\bigr)
$$
where $h:\mathbb{C}^2\to\mathbb{C}$ is holomorphic and $c$ is a constant,
which must equal zero unless $k\_1$ and $k\_2$ are integers.


The argument is as follows: The first equation implies that there exists a function $g:\mathbb{C}^2\to\mathbb{C}$ such that
$$
f\_1 = \frac{\partial g}{\partial z\_1}
\quad\text{and}\quad
f\_2 = -\frac{\partial g}{\partial z\_2}.
$$
Substituting this into the second equation implies that $g$ must satisfy
$$
\frac{\partial^2 g}{\partial z\_1\partial\overline{z}\_1}
+ \frac{\partial^2 g}{\partial z\_2\partial\overline{z}\_2} = 0.
$$
In other words $g$ is a harmonic function on $\mathbb{C}^2$. Since $g$ is harmonic, so is its derivative with respect to $z\_1$, i.e., $f\_1$.


The periodicity conditions imposed on $f\_1$ imply that $f\_1$ is bounded, and a bounded harmonic function on $\mathbb{C}^2$ is constant. Thus, $f\_1 = c$ for some constant $c\in\mathbb{C}$. Obviously, $c$ must be zero unless $k\_1$ and $k\_2$ are integers.


Since $f\_1$ is constant, the given equations on $f\_2$ reduce to
$$
\frac{\partial f\_2}{\partial z\_1} = \frac{\partial f\_2}{\partial\overline{z}\_2} = 0.
$$
Hence $f\_2 = h(\overline{z}\_1,z\_2)$ for some homorphic function $h:\mathbb{C}^2\to\mathbb{C}$.


**Remark:** It wasn't clear from the OP's question whether the OP wanted $f$ to be 'doubly-periodic' or just $f\_1$, nor was it clear exactly what the OP meant by 'doubly-periodic' because, normally, the 'doubly-periodic' condition wouldn't have the exponential factors in its definition.


