# Complex-doubly periodic function in two variables?

I am looking for a function $$f:C2→C2f:\mathbb C^2 \rightarrow \mathbb C^2$$ that satisfies the two equations

$$∂z2f1(z1,z2)+∂z1f2(z1,z2)=0 and \partial_{z_2}f_1(z_1,z_2) + \partial_{z_1} f_2(z_1,z_2)=0 \text{ and }$$
$$∂ˉz1f1(z1,z2)−∂ˉz2f2(z1,z2)=0\partial_{\bar z_1}f_1(z_1,z_2) - \partial_{\bar z_2} f_2(z_1,z_2)=0$$

and in addition, is doubly-periodic in both its complex variables $$z1,z2z_1,z_2$$. 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

$$f1(z1+1,z2)=f1(z1,z2+1)=f1(z1,z2)f_1(z_1+1,z_2)=f_1(z_1,z_2+1)=f_1(z_1,z_2)$$ and

$$f1(z1+i,z2)=e2πik1f1(z1,z2)f_1(z_1+i,z_2) = e^{2\pi i k_1}f_1(z_1,z_2)$$
and

$$f1(z1,z2+i)=e2πik2f1(z1,z2)f_1(z_1,z_2+i) = e^{2\pi i k_2}f_1(z_1,z_2)$$

for some fixed $$k1,k2∈R.k_1,k_2 \in \mathbb R.$$ Please let me know if you do have any questions. I had some typos in there, but hopefully everything is coherent now.

## Answer

The answer is that the only solutions have the form
$$f=(f1,f2)=(c,h(¯z1,z2)) f = (f_1,f_2) = \bigl(c, h(\,\overline{z}_1, z_2)\bigr)$$
where $$h:C2→Ch:\mathbb{C}^2\to\mathbb{C}$$ is holomorphic and $$cc$$ is a constant,
which must equal zero unless $$k1k_1$$ and $$k2k_2$$ are integers.

The argument is as follows: The first equation implies that there exists a function $$g:C2→Cg:\mathbb{C}^2\to\mathbb{C}$$ such that
$$f1=∂g∂z1andf2=−∂g∂z2. f_1 = \frac{\partial g}{\partial z_1} \quad\text{and}\quad f_2 = -\frac{\partial g}{\partial z_2}.$$
Substituting this into the second equation implies that $$gg$$ must satisfy
$$∂2g∂z1∂¯z1+∂2g∂z2∂¯z2=0. \frac{\partial^2 g}{\partial z_1\partial\overline{z}_1} + \frac{\partial^2 g}{\partial z_2\partial\overline{z}_2} = 0.$$
In other words $$gg$$ is a harmonic function on $$C2\mathbb{C}^2$$. Since $$gg$$ is harmonic, so is its derivative with respect to $$z1z_1$$, i.e., $$f1f_1$$.

The periodicity conditions imposed on $$f1f_1$$ imply that $$f1f_1$$ is bounded, and a bounded harmonic function on $$C2\mathbb{C}^2$$ is constant. Thus, $$f1=cf_1 = c$$ for some constant $$c∈Cc\in\mathbb{C}$$. Obviously, $$cc$$ must be zero unless $$k1k_1$$ and $$k2k_2$$ are integers.

Since $$f1f_1$$ is constant, the given equations on $$f2f_2$$ reduce to
$$∂f2∂z1=∂f2∂¯z2=0. \frac{\partial f_2}{\partial z_1} = \frac{\partial f_2}{\partial\overline{z}_2} = 0.$$
Hence $$f2=h(¯z1,z2)f_2 = h(\overline{z}_1,z_2)$$ for some homorphic function $$h:C2→Ch:\mathbb{C}^2\to\mathbb{C}$$.

Remark: It wasn’t clear from the OP’s question whether the OP wanted $$ff$$ to be ‘doubly-periodic’ or just $$f1f_1$$, 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.

Attribution
Source : Link , Question Author : Sascha , Answer Author : Robert Bryant