Complex-doubly periodic function in two variables? 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.
 A: 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.
