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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.

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    $\begingroup$ You, probably, want more because as written, you can just take any function $f=(f_1,f_2)$ in which $f_1$ depends only on $z_1$ and $f_2$ only on $z_2$ with any periodicity in each variable you like. $\endgroup$
    – fedja
    Jan 28 at 21:59
  • $\begingroup$ @fedja you are right, there were some additional constraints missing to give this question a meaning. $\endgroup$
    – Sascha
    Jan 28 at 22:31
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    $\begingroup$ By (an analog of) Cauchy-Riemann, $f_1+if_2$ only depends on $z_1+iz_2$ and $f_1-if_2$ only on $z_1-iz_2$, is this intended? $\endgroup$ Jan 31 at 15:39
  • $\begingroup$ @მამუკაჯიბლაძე what does it mean for $f_1+if_2$ to only depend on $z_1+iz_2$? What kind of independent coordinate system are you considering here? $\endgroup$ Feb 1 at 19:53
  • $\begingroup$ @PritamBemis If I am not confused (and I believe something similar is done in the answer by Robert Bryant below), if you change variables to $z_1+iz_2=z$, $z_1-iz_2=w$, and denote $f_1+if_2=g$, $f_1-if_2=h$, then $\partial g/\partial w=\partial h/\partial z=0$, no? $\endgroup$ Feb 1 at 19:55
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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.

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