I am looking for a function f:C2→C2 that satisfies the two equations

∂z2f1(z1,z2)+∂z1f2(z1,z2)=0 and

∂ˉz1f1(z1,z2)−∂ˉz2f2(z1,z2)=0and in addition, is doubly-periodic in both its complex variables z1,z2. 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) and

f1(z1+i,z2)=e2πik1f1(z1,z2)

andf1(z1,z2+i)=e2πik2f1(z1,z2)

for some fixed k1,k2∈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))

where h:C2→C is holomorphic and c is a constant,

which must equal zero unless k1 and k2 are integers.

The argument is as follows: The first equation implies that there exists a function g:C2→C such that

f1=∂g∂z1andf2=−∂g∂z2.

Substituting this into the second equation implies that g must satisfy

∂2g∂z1∂¯z1+∂2g∂z2∂¯z2=0.

In other words g is a harmonic function on C2. Since g is harmonic, so is its derivative with respect to z1, i.e., f1.

The periodicity conditions imposed on f1 imply that f1 is bounded, and a bounded harmonic function on C2 is constant. Thus, f1=c for some constant c∈C. Obviously, c must be zero unless k1 and k2 are integers.

Since f1 is constant, the given equations on f2 reduce to

∂f2∂z1=∂f2∂¯z2=0.

Hence f2=h(¯z1,z2) for some homorphic function h:C2→C.

**Remark:** It wasn’t clear from the OP’s question whether the OP wanted f to be ‘doubly-periodic’ or just f1, 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*